Circuits

DC and AC circuit analysis: Ohm’s and Kirchhoff’s laws, nodal and mesh analysis, Thevenin and Norton, op amps, capacitors and inductors, RC, RL and RLC transients, phasors, AC power, power factor and three-phase, with six labs. Free, self-paced, and tied to LED lighting, solar panels, batteries, EV charging and the power grid.

NASA (ISS066-E-080762, Nov. 8, 2021)NASA/SDONASA (ISS072-E-159172, Oct. 30, 2024)NASA/Ben Smegelsky (KSC-20240124-PH-JBS01_0071)

Courses › ENGR 214 + ENGR 214L

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About this course

This is a free self-study course outline written by Saving Planets. It is not a university course, it carries no academic credit, and no school is involved.

Its scope follows a one-semester sophomore circuit analysis course with a lab. It is modeled on the topics and depth of MIT 6.002 Circuits and Electronics (free on MIT OpenCourseWare), Georgia Tech ECE 2040 Circuit Analysis and Purdue ECE 20001 Electrical Engineering Fundamentals I, and on the free OpenStax University Physics Volume 2 and PhET simulations, which we link where they help. None of those authors or schools endorses this page.

What we add: every unit ties to real clean-energy work with a cited public source (LED lighting, solar panels and inverters, battery storage, EV charging, data centers and the power grid), every number on the page is re-computed by a checking script before it goes live, and every lab has a free PhET simulation or a safe low-voltage home version.

Work through it in order. Each unit has goals, key ideas, equations, step-by-step examples, practice with full answers, common mistakes, flashcards and a quiz. Your progress is saved only in this browser.

How this course helps save the planet

Circuits are how clean energy reaches people. Solar panels make direct current that inverters turn into alternating current, batteries store it, transformers and power lines carry it, and efficient lights, motors and chargers use it.

Every percent of loss in a wire, transformer or converter is energy that had to be generated somewhere. The analysis in this course is what engineers use to cut those losses. Each unit ends with a short, sourced section. Values that are not from a source are labeled as sample values.

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Before you start

Prerequisites

  • Physics II: charge, electric field, potential, current, resistance, capacitance, inductance and Faraday’s law.
  • Calculus I and II: derivatives and integrals; Units 7 and 8 solve first- and second-order differential equations.
  • Complex numbers (reviewed in Unit 9) for phasors; Linear Algebra helps with nodal and mesh equations.
  • Differential Equations can be taken alongside this course.

How to study

  • Draw and label the circuit first: node voltages, branch currents with arrows, and polarity marks on every element.
  • Use the passive sign convention: current entering the + terminal means the element absorbs power.
  • Write KCL at nodes or KVL around meshes, count unknowns and equations, then solve (by hand for 2 × 2, by matrix for more).
  • Check every answer: power absorbed must equal power supplied, and limits (t = 0, t → ∞, ω → 0 or ∞) must make sense.
  • Keep units in SI: V, A, Ω, W, F, H, s, and watch prefixes (mA, kΩ, µF, mH).
  • Do every practice problem on paper before opening the answer, and retake the quizzes a few days later.

Lab equipment (for the wet labs)

  • A computer or tablet for the free PhET simulations (they run in a browser).
  • A digital multimeter (inexpensive models measure V, A and Ω).
  • A solderless breadboard, jumper wires, a 2 × AA battery holder (3 V) and AA batteries.
  • Resistors (100 Ω, 1 kΩ, 10 kΩ), a 1000 µF capacitor rated at least 6.3 V, and LEDs.
  • Optional: a plug-in power meter for comparing light bulbs and appliances (it plugs into the wall; do not open it).
  • Safety glasses, a stopwatch or phone, and spreadsheet software.

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Suggested 15-week schedule

About 6 to 9 hours a week: reading, practice problems and, in lab weeks, the lab.

WeekUnitTopicsReading
1Circuit Variables and Elements: Charge, Current, Voltage, Power, Energy and Ohm’s LawCharge, current, voltage, power, sign convention, sourcesMIT 6.002; OpenStax UP2 9.4
2Circuit Variables and Elements: Charge, Current, Voltage, Power, Energy and Ohm’s LawOhm’s law, resistivity, energy in kWh; Lab AUP2 9.4, 10.1
3Kirchhoff’s Laws, Series and Parallel Resistors, and Voltage and Current DividersKCL, KVL, series and parallel, dividers, meters; Lab BUP2 10.2–10.3
4Systematic Analysis: Node-Voltage and Mesh-Current MethodsNode-voltage method and supernodesMIT 6.002
5Systematic Analysis: Node-Voltage and Mesh-Current MethodsMesh-current method, supermeshes, dependent sourcesMIT 6.002; UP2 10.3
6Circuit Theorems: Linearity, Superposition, Source Transformation, Thevenin, Norton and Maximum Power TransferSuperposition, source transformation, Thevenin, Norton, maximum power; Lab CMIT 6.002
7Operational AmplifiersOperational amplifiers; midterm practiceMIT 6.002
8Capacitors and InductorsCapacitors, inductors, energy, integratorsUP2 8.1, 14.2
9First-Order Circuits: RC and RL TransientsRC and RL natural and step responsesUP2 10.5, 14.4
10First-Order Circuits: RC and RL TransientsSwitching problems and digital switching energy; Lab DMIT 6.002
11Second-Order Circuits: RLC Natural and Step ResponsesSeries and parallel RLC, damping, LC oscillationUP2 14.5–14.6
12Sinusoidal Steady State: Phasors and ImpedanceSinusoids, phasors, impedanceUP2 15.2–15.3
13Sinusoidal Steady State: Phasors and ImpedanceAC nodal, mesh and Thevenin analysis, filters; Lab EMIT 6.002; UP2 15.3, 15.5
14AC Power, Power Factor, Resonance, Transformers and Three-Phase SystemsAC power, power factor, transformers, three-phase; Lab FUP2 15.4–15.6
15ReviewReview and final practice setAll units

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Your progress

Tick units and labs as you finish them, and take the quizzes. Progress and quiz scores are saved only in this browser, never sent to us.

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Unit 1 · Weeks 1–2

Circuit Variables and Elements: Charge, Current, Voltage, Power, Energy and Ohm’s Law

Charge and current, voltage as energy per charge, power and energy, the passive sign convention, ideal independent and dependent sources, resistance and resistivity, Ohm’s law, and power in resistors.

Learning objectives

By the end of this unit you can:

  • Relate charge, current, voltage, power and energy with correct units.
  • Apply the passive sign convention to decide whether an element absorbs or delivers power.
  • Use Ohm’s law and R = ρL/A.
  • Compute energy use in kWh and compare devices.

Key concepts

  • Current is the rate of charge flow, i = dq/dt, in amperes (1 A = 1 C/s). Voltage is energy per unit charge between two points, in volts (1 V = 1 J/C).
  • Power p = vi. With the passive sign convention (current enters the + terminal), p > 0 means the element absorbs power and p < 0 means it delivers power. In any circuit, total power absorbed = total power delivered.
  • An ideal voltage source holds its voltage whatever the current; an ideal current source holds its current whatever the voltage. Dependent sources are controlled by a voltage or current elsewhere and model transistors and amplifiers.
  • Ohm’s law v = iR. A resistor always absorbs power p = i²R = v²/R. Resistance of a wire R = ρL/A grows with length and falls with cross-section; conductance G = 1/R in siemens.
  • Energy w = ∫p dt; at constant power w = Pt. Utilities bill in kilowatt-hours: 1 kWh = 3.6 MJ.

Key equations

i=dqdti=\frac{dq}{dt}
Current
p=vip=vi
Power
v=iR,  p=i2R=v2Rv=iR,\ \ p=i^2R=\frac{v^2}{R}
Ohm’s law and resistor power
R=ρLAR=\frac{\rho L}{A}
Resistance of a conductor

Worked examples, step by step

Example 1.1 A 4 Ω heater is connected to 12 V. Find the current and power.

Show the steps
  1. i = v/R = 12/4 = 3 A.
  2. p = vi = 12 × 3.

Answer: i = 3 A, p = 36 W.

Example 1.2 Find the resistance of 10 m of copper wire with a 1 mm² cross-section (ρ = 1.68 × 10⁻⁸ Ω·m).

Show the steps
  1. A = 1 mm² = 1 × 10⁻⁶ m².
  2. R = ρL/A = 1.68 × 10⁻⁸ × 10/10⁻⁶.

Answer: R = 0.168 Ω.

Example 1.3 A 60 W incandescent bulb is replaced by an LED that uses 75% less energy. Both run 3 hours a day for a year. Find the energy saved.

Show the steps
  1. LED power = 0.25 × 60 = 15 W.
  2. Incandescent: 60 × 3 × 365 = 65,700 Wh = 65.7 kWh; LED: 16.4 kWh.

Answer: About 49.3 kWh saved per bulb per year.

Practice problems

Work each on paper first, then check.

  1. A 2 A current flows for 30 s. How much charge passes?

    Show answer

    q = it = 60 C.

  2. A lamp draws 0.5 A from 120 V. Find its power and resistance.

    Show answer

    p = 60 W, R = 240 Ω.

  3. An element has v = 5 V and i = −2 A entering its + terminal. Does it absorb or deliver power?

    Show answer

    p = −10 W: it delivers 10 W.

  4. How many joules are in 2 kWh?

    Show answer

    7.2 × 10⁶ J.

Common misconceptions

  • Myth: Current is used up by a resistor.

    Actually: Charge is conserved; the same current leaves as enters. Energy is converted, not current.

  • Myth: Voltage exists at a point.

    Actually: Voltage is always between two points; a “node voltage” is measured from a chosen ground.

  • Myth: Power is always vi with a plus sign.

    Actually: Use the passive sign convention to get the sign right.

Flashcards

  • Current

    i = dq/dt, in A.

  • Voltage

    Energy per charge, in V = J/C.

  • Power

    p = vi.

  • Passive sign convention

    Current into + terminal: p > 0 absorbs.

  • Ohm’s law

    v = iR.

  • kWh

    3.6 MJ.

Self-check quiz

1. 1 A equals…

Show answer

1 C/s. Charge per second.

2. Doubling the voltage across a resistor multiplies its power by…

Show answer

4. p = v²/R.

3. Doubling a wire’s length (same area) changes its resistance by…

Show answer

2. R = ρL/A.

4. 1 kWh equals…

Show answer

3.6 MJ. 1000 W × 3600 s.

How this helps save the planet

Lighting is the simplest place to see power and energy at work. The U.S. Department of Energy’s guide to energy-efficient lighting says LEDs use at least 75 percent less energy than incandescent bulbs, emit virtually no heat, and last 25 times longer.

With p = vi and w = Pt, the worked example shows what that means for one bulb: replacing a 60 W incandescent with a 15 W LED, used 3 hours a day, saves about 49.3 kWh a year. Multiply by the dozens of bulbs in a home and the millions of homes in a country and the power plants that never have to run become visible.

On Saving Planets: Physics II · Solutions

Sources: U.S. DOE: Guide to energy-efficient lighting (PDF) · OpenStax UP2, 9.4 Ohm’s law

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Unit 2 · Week 3

Kirchhoff’s Laws, Series and Parallel Resistors, and Voltage and Current Dividers

Nodes, branches and loops, Kirchhoff’s current and voltage laws, series and parallel combinations, equivalent resistance, voltage and current dividers, delta–wye conversions, and measuring with voltmeters and ammeters.

Learning objectives

By the end of this unit you can:

  • Write KCL and KVL for any circuit.
  • Reduce series–parallel networks to an equivalent resistance.
  • Use voltage and current dividers.
  • Explain how meters are connected and how they load a circuit.

Key concepts

  • KCL: the sum of currents leaving any node is zero (charge is conserved). KVL: the sum of voltages around any closed loop is zero (energy is conserved).
  • Series elements carry the same current; resistances add: R_eq = R₁ + R₂ + …. Parallel elements share the same voltage; conductances add: 1/R_eq = 1/R₁ + 1/R₂ + …; for two resistors R₁R₂/(R₁ + R₂).
  • Voltage divider: v₂ = v·R₂/(R₁ + R₂). Current divider (two resistors): i₁ = i·R₂/(R₁ + R₂); the larger share goes through the smaller resistor.
  • Delta–wye (Δ–Y) conversion handles bridges that are neither series nor parallel: R_Y = (product of the two adjacent Δ resistors)/(sum of all three).
  • A voltmeter goes in parallel and should have very high resistance; an ammeter goes in series and should have very low resistance.

Key equations

∑iout=0\sum i_{\text{out}}=0
KCL
∑v=0 around a loop\sum v=0\ \text{around a loop}
KVL
Rs=R1+R2,  Rp=R1R2R1+R2R_s=R_1+R_2,\ \ R_p=\frac{R_1R_2}{R_1+R_2}
Series and parallel
v2=vR2R1+R2,  i1=iR2R1+R2v_2=v\frac{R_2}{R_1+R_2},\ \ i_1=i\frac{R_2}{R_1+R_2}
Voltage and current dividers

Worked examples, step by step

Example 2.1 Resistors of 100, 200 and 300 Ω are in series across 12 V. Find the current and the voltage across the 200 Ω resistor.

Show the steps
  1. R_eq = 600 Ω; i = 12/600 = 0.02 A.
  2. v = 0.02 × 200.

Answer: i = 20 mA, v₂₀₀ = 4 V.

Example 2.2 A 9 V source feeds a divider with R₁ = 10 kΩ on top and R₂ = 5 kΩ on the bottom. Find v_out across R₂.

Show the steps
  1. v_out = 9 × 5/(10 + 5).

Answer: v_out = 3 V.

Example 2.3 A 6 A current splits between 2 Ω and 4 Ω in parallel. Find each current.

Show the steps
  1. i₂Ω = 6 × 4/(2 + 4) = 4 A.
  2. i₄Ω = 6 × 2/6 = 2 A; check 4 + 2 = 6.

Answer: 4 A through 2 Ω and 2 A through 4 Ω.

Practice problems

Work each on paper first, then check.

  1. Find 6 Ω ∥ 3 Ω.

    Show answer

    18/9 = 2 Ω.

  2. Three 30 Ω resistors are in parallel. Find R_eq.

    Show answer

    10 Ω.

  3. Find R_eq of 10 Ω in series with (20 Ω ∥ 20 Ω).

    Show answer

    10 + 10 = 20 Ω.

  4. Three currents enter a node: 2 A, 3 A and −1 A. What current leaves?

    Show answer

    4 A.

Common misconceptions

  • Myth: Parallel resistance is larger than the smallest resistor.

    Actually: It is always smaller than the smallest one.

  • Myth: A voltage divider works with any load.

    Actually: A load across R₂ is in parallel with it and lowers v_out unless its resistance is much larger.

  • Myth: An ammeter goes across the element.

    Actually: Across a low-resistance ammeter is nearly a short circuit; it goes in series.

Flashcards

  • KCL

    Currents out of a node sum to 0.

  • KVL

    Voltages around a loop sum to 0.

  • Series

    R adds; same current.

  • Parallel

    G adds; same voltage.

  • Voltage divider

    v₂ = vR₂/(R₁ + R₂).

  • Current divider

    i₁ = iR₂/(R₁ + R₂).

Self-check quiz

1. KCL expresses conservation of…

Show answer

charge. Charge cannot pile up at a node.

2. Two equal resistors R in parallel give…

Show answer

R/2. R·R/(2R).

3. Series elements share the same…

Show answer

current. One path.

4. An ideal voltmeter has resistance…

Show answer

infinite. It must draw no current.

How this helps save the planet

Solar panels are series and parallel circuits. The U.S. Energy Information Administration explains that one PV cell can only produce 1 or 2 watts, so cells are electrically connected in weather-tight panels, panels are connected into arrays of two to hundreds of panels, and the number of panels sets how much electricity an array can generate. Cells make direct current, and inverters convert it to the alternating current used on the grid.

Series connection adds voltages and parallel connection adds currents. As a sample, cells making 0.5 V and 3 A (1.5 W, within EIA’s 1–2 W range) give 30 V at 3 A, or 90 W, when 60 are wired in series. KCL and KVL also explain why one shaded cell in a series string can drag down the whole string, which is why panels include bypass diodes.

On Saving Planets: Physics II · Parker Solar Probe

Sources: U.S. EIA: Photovoltaics and electricity · OpenStax UP2, 10.2 Resistors in series and parallel

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Unit 3 · Weeks 4–5

Systematic Analysis: Node-Voltage and Mesh-Current Methods

Choosing a reference node, writing KCL in terms of node voltages, supernodes, mesh currents and KVL, supermeshes, circuits with dependent sources, and solving the resulting linear systems by hand and with matrices.

Learning objectives

By the end of this unit you can:

  • Write and solve node-voltage equations, including supernodes.
  • Write and solve mesh-current equations, including supermeshes.
  • Handle dependent sources in both methods.
  • Choose the method that gives fewer equations.

Key concepts

  • Node-voltage method: pick a ground node, name the other node voltages, write KCL at each non-reference node using Ohm’s law (current from a to b through R is (v_a − v_b)/R), and solve.
  • A voltage source between two non-reference nodes makes a supernode: write one KCL for the combined region plus the source equation v_a − v_b = V_s.
  • Mesh-current method (planar circuits): assign a clockwise current to each window, write KVL around each mesh; a shared resistor carries the difference of the two mesh currents.
  • A current source shared by two meshes makes a supermesh: one KVL around the combined loop plus the source equation.
  • An N-node circuit needs N − 1 node equations; a circuit with M meshes needs M mesh equations. Choose the smaller set. The equations form Gv = i or Ri = v, a linear system.

Key equations

∑branchesvn−vkRnk=isource into n\sum_{\text{branches}}\frac{v_n-v_k}{R_{nk}}=i_{\text{source into }n}
KCL at node n
∑meshR (im−iadj)=∑vsources\sum_{\text{mesh}} R\,(i_m-i_{\text{adj}})=\sum v_{\text{sources}}
KVL around mesh m
Gv=iG\mathbf{v}=\mathbf{i}
Nodal matrix form
Ri=vR\mathbf{i}=\mathbf{v}
Mesh matrix form

Worked examples, step by step

Example 3.1 A 10 V source feeds node 1 through 2 Ω. Node 1 has 4 Ω to ground and 2 Ω to node 2; node 2 has 4 Ω to ground and a 1 A source flowing into it. Find v₁ and v₂.

Show the steps
  1. Node 1 (×4): 2(v₁ − 10) + v₁ + 2(v₁ − v₂) = 0, so 5v₁ − 2v₂ = 20.
  2. Node 2 (×4): 2(v₂ − v₁) + v₂ − 4 = 0, so −2v₁ + 3v₂ = 4.
  3. det = 15 − 4 = 11; v₁ = (60 + 8)/11, v₂ = (20 + 40)/11.

Answer: v₁ = 6.182 V, v₂ = 5.455 V.

Example 3.2 Mesh 1 has a 10 V source and 2 Ω; mesh 2 has 6 Ω; the meshes share 4 Ω. Find the mesh currents and the current in the shared resistor.

Show the steps
  1. Mesh 1: 6i₁ − 4i₂ = 10. Mesh 2: −4i₁ + 10i₂ = 0.
  2. det = 60 − 16 = 44; i₁ = 100/44, i₂ = 40/44.

Answer: i₁ = 2.273 A, i₂ = 0.909 A; shared resistor carries 1.364 A.

Example 3.3 A 5 V source sits between nodes a and b (v_a − v_b = 5). Node a has 1 Ω to ground, node b has 4 Ω to ground, and 2 A is injected into the supernode. Find v_a and v_b.

Show the steps
  1. Supernode KCL: v_a/1 + v_b/4 = 2.
  2. Substitute v_a = v_b + 5: v_b + 5 + v_b/4 = 2, so 1.25v_b = −3.

Answer: v_b = −2.4 V, v_a = 2.6 V.

Practice problems

Work each on paper first, then check.

  1. How many node equations does a circuit with 5 nodes need?

    Show answer

    4.

  2. A single node at v connects through 2 Ω to 6 V and through 2 Ω to ground. Find v.

    Show answer

    (v − 6)/2 + v/2 = 0: v = 3 V.

  3. Two meshes share a 3 Ω resistor; i₁ = 2 A and i₂ = 0.5 A, both clockwise. Find the current in the shared resistor.

    Show answer

    1.5 A in the direction of i₁.

  4. Write 5v₁ − 2v₂ = 20, −2v₁ + 3v₂ = 4 as Gv = i.

    Show answer

    G = [[5, −2], [−2, 3]], i = (20, 4).

Common misconceptions

  • Myth: KCL can be written at the reference node too.

    Actually: It gives a dependent equation; use N − 1 nodes.

  • Myth: Mesh currents are the branch currents.

    Actually: Only on outer branches; shared branches carry differences.

  • Myth: A voltage source current is zero because it has no resistance.

    Actually: Its current is set by the rest of the circuit; that is why supernodes are used.

Flashcards

  • Reference node

    Ground, v = 0.

  • Node-voltage method

    KCL at N − 1 nodes.

  • Supernode

    Voltage source between two nodes.

  • Mesh-current method

    KVL around each window.

  • Supermesh

    Current source shared by two meshes.

  • Gv = i

    Nodal equations in matrix form.

Self-check quiz

1. A circuit with 4 nodes needs how many node-voltage equations?

Show answer

3. N − 1.

2. A voltage source between two non-reference nodes calls for…

Show answer

a supernode. Combine the two nodes.

3. The current in a resistor shared by meshes 1 and 2 is…

Show answer

i₁ − i₂. Both flow clockwise in their own mesh.

4. Node equations form…

Show answer

a linear system Gv = i. Linear resistive circuits are linear systems.

How this helps save the planet

The power grid is one enormous nodal-analysis problem. EIA explains that the U.S. grid is made up of thousands of miles of high-voltage lines and millions of miles of low-voltage lines connecting thousands of power plants to hundreds of millions of customers. At the start of the 20th century more than 4,000 isolated utilities operated; today the Lower 48 run as three main interconnections, Eastern, Western and ERCOT, linked to provide multiple paths for power to flow.

EIA notes that supply must match demand at every moment and that balancing authorities keep them matched. Grid operators do that with power-flow models whose equations are the AC version of the node-voltage method in this unit, solved for thousands of buses, and the same analysis tells planners where new lines are needed to connect wind and solar far from cities.

On Saving Planets: Linear Algebra · Solutions

Sources: U.S. EIA: Delivery to consumers (the electric power grid) · OpenStax UP2, 10.3 Kirchhoff’s rules

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Unit 4 · Week 6

Circuit Theorems: Linearity, Superposition, Source Transformation, Thevenin, Norton and Maximum Power Transfer

Linearity and superposition, source transformations, Thevenin and Norton equivalent circuits, finding R_Th by deactivating sources or with a test source, and the maximum power transfer theorem.

Learning objectives

By the end of this unit you can:

  • Use superposition to analyze circuits with several sources.
  • Convert between voltage and current sources.
  • Find Thevenin and Norton equivalents.
  • Choose a load for maximum power transfer and compute that power.

Key concepts

  • Linearity: in a circuit of resistors and sources, every voltage and current is a linear combination of the source values. Superposition: find the response to each independent source alone (other voltage sources shorted, current sources opened) and add.
  • Source transformation: a voltage source V in series with R is equivalent, at its terminals, to a current source V/R in parallel with R.
  • Thevenin’s theorem: any linear two-terminal network looks like a source V_Th in series with R_Th. V_Th is the open-circuit voltage; R_Th is the resistance seen with independent sources turned off (or V_oc/i_sc).
  • Norton’s theorem: the same network is a current source I_N = i_sc in parallel with R_N = R_Th; V_Th = I_N R_Th.
  • Maximum power transfer: a load receives the most power when R_L = R_Th, and then P_max = V_Th²/(4R_Th). The efficiency at that point is only 50%, so power systems are designed for R_L ≫ R_source instead.

Key equations

v=∑kv(k)v=\sum_k v^{(k)}
Superposition
VTh=voc,  RTh=vociscV_{Th}=v_{oc},\ \ R_{Th}=\frac{v_{oc}}{i_{sc}}
Thevenin equivalent
IN=VThRThI_N=\frac{V_{Th}}{R_{Th}}
Norton current
Pmax⁡=VTh24RTh at RL=RThP_{\max}=\frac{V_{Th}^2}{4R_{Th}}\ \text{at}\ R_L=R_{Th}
Maximum power transfer

Worked examples, step by step

Example 4.1 A 12 V source in series with 4 Ω feeds terminals a–b, with 12 Ω across a–b. Find the Thevenin and Norton equivalents.

Show the steps
  1. V_Th = 12 × 12/(4 + 12) = 9 V (divider).
  2. R_Th = 4 ∥ 12 = 48/16 = 3 Ω (source shorted).
  3. I_N = 9/3.

Answer: V_Th = 9 V, R_Th = 3 Ω; I_N = 3 A.

Example 4.2 For that circuit, find the power to a 6 Ω load and the maximum possible load power.

Show the steps
  1. i = 9/(3 + 6) = 1 A; P = 1² × 6 = 6 W.
  2. Maximum at R_L = 3 Ω: P_max = 9²/(4 × 3).

Answer: 6 W; P_max = 6.75 W at 3 Ω.

Example 4.3 A node connects to a 10 V source through 5 Ω, to ground through 5 Ω, and receives a 2 A current source. Find the node voltage by superposition.

Show the steps
  1. 10 V alone (2 A opened): divider gives 5 V.
  2. 2 A alone (10 V shorted): 2 × (5 ∥ 5) = 5 V.
  3. Add the two.

Answer: v = 10 V. Check KCL: (10 − 10)/5 + 2 = 10/5.

Practice problems

Work each on paper first, then check.

  1. Convert a 6 V source in series with 2 Ω to a Norton form.

    Show answer

    3 A in parallel with 2 Ω.

  2. A network has v_oc = 15 V and i_sc = 5 A. Find R_Th.

    Show answer

    3 Ω.

  3. What load gets maximum power from V_Th = 20 V, R_Th = 5 Ω, and how much?

    Show answer

    R_L = 5 Ω, P = 400/20 = 20 W.

  4. Can superposition add powers?

    Show answer

    No: power is not linear; add voltages or currents, then compute power.

Common misconceptions

  • Myth: Turning off a voltage source means removing it.

    Actually: Replace it with a short circuit; a current source becomes an open circuit.

  • Myth: Maximum power transfer means maximum efficiency.

    Actually: At R_L = R_Th half the power is lost in R_Th.

  • Myth: Dependent sources are turned off for R_Th.

    Actually: Keep dependent sources; use a test source to find R_Th.

Flashcards

  • Superposition

    Add responses to each source alone.

  • Source transformation

    V in series R ↔ V/R in parallel R.

  • Thevenin

    V_Th in series with R_Th.

  • Norton

    I_N in parallel with R_N.

  • R_Th

    v_oc/i_sc.

  • Max power

    V_Th²/(4R_Th) at R_L = R_Th.

Self-check quiz

1. To find R_Th, independent voltage sources are replaced by…

Show answer

shorts. Zero volts is a short.

2. V_Th is the…

Show answer

open-circuit voltage. Measured with nothing connected.

3. Maximum power transfer occurs when…

Show answer

R_L = R_Th. Matched load.

4. Efficiency at maximum power transfer is…

Show answer

50%. Equal power in R_Th and R_L.

How this helps save the planet

Solar cells behave like sources whose Thevenin-style operating point changes with sunlight and temperature, so their electronics constantly hunt for the load that draws the most power. How much of the sunlight they turn into electricity has improved steadily: EIA reports that commercially available PV panels averaged less than 10% efficiency in the mid-1980s, about 15% by 2015, and are now approaching 25% for state-of-the-art modules, while experimental and space cells have reached nearly 50%.

EIA also reports that U.S. utility-scale PV plants generated about 292 billion kWh in 2025, up from 6 million kWh in 2004, and small-scale rooftop systems about 93 billion kWh. Matching each panel string to its best operating point, the idea behind maximum power transfer, is part of how those kilowatt-hours are squeezed out of every square meter.

On Saving Planets: Physics II · Space

Sources: U.S. EIA: Photovoltaics and electricity · MIT OpenCourseWare 6.002 Circuits and Electronics (Spring 2007)

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Unit 5 · Week 7

Operational Amplifiers

The ideal op amp and its two rules, inverting and non-inverting amplifiers, voltage followers, summing and difference amplifiers, saturation, cascaded stages, and op amps in sensors and measurement.

Learning objectives

By the end of this unit you can:

  • Apply the ideal op-amp rules with negative feedback.
  • Analyze inverting, non-inverting, follower, summing and difference circuits.
  • Recognize saturation and its limits.
  • Design a simple amplifier for a sensor signal.

Key concepts

  • An op amp amplifies the difference between its inputs by a very large gain. With negative feedback, the ideal rules are: no current flows into either input, and the two input voltages are equal (v₊ = v₋).
  • Inverting amplifier: v_o = −(R_f/R_in)v_in. Non-inverting amplifier: v_o = (1 + R_f/R_g)v_in. Voltage follower (buffer): v_o = v_in with almost no current drawn from the source.
  • Summing amplifier: v_o = −R_f(v₁/R₁ + v₂/R₂ + …). Difference amplifier with matched ratios: v_o = (R₂/R₁)(v₂ − v₁).
  • The output cannot exceed the supply rails; beyond them the op amp saturates and the linear rules stop applying.
  • Cascaded stages multiply their gains, and because an op-amp output acts like an ideal voltage source, a stage is not loaded by the next.

Key equations

v+=v−,  i+=i−=0v_+=v_-,\ \ i_+=i_-=0
Ideal op-amp rules (negative feedback)
vo=−RfRinvinv_o=-\frac{R_f}{R_{in}}v_{in}
Inverting amplifier
vo=(1+RfRg)vinv_o=\left(1+\frac{R_f}{R_g}\right)v_{in}
Non-inverting amplifier
vo=−Rf∑kvkRkv_o=-R_f\sum_k\frac{v_k}{R_k}
Summing amplifier

Worked examples, step by step

Example 5.1 An inverting amplifier has R_in = 10 kΩ and R_f = 100 kΩ. Find the gain and v_o for v_in = 0.2 V.

Show the steps
  1. Gain = −100/10 = −10.
  2. v_o = −10 × 0.2.

Answer: Gain −10, v_o = −2 V.

Example 5.2 Design a non-inverting amplifier with a gain of 10 using R_g = 1 kΩ.

Show the steps
  1. 1 + R_f/R_g = 10, so R_f/R_g = 9.

Answer: R_f = 9 kΩ.

Example 5.3 A summing amplifier has R_f = R₁ = R₂ = 10 kΩ, v₁ = 1 V and v₂ = 2 V. Find v_o. What if the supply rails are ±12 V and the gain is raised so the ideal answer is −15 V?

Show the steps
  1. v_o = −10k(1/10k + 2/10k) = −3 V.
  2. The output cannot go below the −12 V rail.

Answer: v_o = −3 V; at the higher gain it saturates near −12 V.

Practice problems

Work each on paper first, then check.

  1. Find the gain of an inverting amplifier with R_f = 47 kΩ and R_in = 4.7 kΩ.

    Show answer

    −10.

  2. A voltage follower has v_in = 1.5 V. Find v_o.

    Show answer

    1.5 V.

  3. Two stages with gains −10 and 5 are cascaded. Find the overall gain.

    Show answer

    −50.

  4. A difference amplifier with R₂/R₁ = 4 has v₂ = 1.2 V and v₁ = 1.0 V. Find v_o.

    Show answer

    4 × 0.2 = 0.8 V.

Common misconceptions

  • Myth: The op amp inputs draw the circuit current.

    Actually: Ideally no current enters the inputs; it flows through R_f.

  • Myth: v₊ = v₋ always.

    Actually: Only with negative feedback and when not saturated.

  • Myth: Gain can be as large as you like.

    Actually: The output is limited by the supply rails and real op-amp bandwidth.

Flashcards

  • Ideal op amp

    v₊ = v₋, no input current.

  • Inverting

    −R_f/R_in.

  • Non-inverting

    1 + R_f/R_g.

  • Follower

    Gain 1 buffer.

  • Summing

    −R_f Σ v_k/R_k.

  • Saturation

    Output limited by the rails.

Self-check quiz

1. With negative feedback, an ideal op amp makes…

Show answer

v₊ = v₋. Virtual short.

2. An inverting amplifier with R_f = 20 kΩ, R_in = 5 kΩ has gain…

Show answer

−4. −R_f/R_in.

3. A voltage follower is used to…

Show answer

buffer a source. Gain 1, no loading.

4. The minimum gain of a non-inverting amplifier is…

Show answer

1. 1 + R_f/R_g ≥ 1.

How this helps save the planet

A smarter grid starts with measurement, and measuring small signals is the op amp’s job. EIA explains that the smart grid combines digital technology with the traditional electrical system so utilities and customers can send and receive information, helping utilities reduce electricity losses and detect and fix problems faster, and that smart devices on power lines and at substations let utilities manage voltage more efficiently and quickly pinpoint where an outage is.

Every smart meter and line sensor scales a tiny current-sensor or voltage-divider signal up to a range a digital converter can read, using exactly the inverting, non-inverting and difference amplifiers in this unit. EIA adds that smart devices can tell customers when prices are high so they can shift use away from times of very high demand.

On Saving Planets: Intro to Programming · Solutions

Sources: U.S. EIA: Delivery to consumers (the electric power grid) · MIT OpenCourseWare 6.002 Circuits and Electronics (Spring 2007)

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Unit 6 · Week 8

Capacitors and Inductors

Capacitance and inductance, their current–voltage relations, stored energy, continuity of capacitor voltage and inductor current, series and parallel combinations, DC steady-state behavior, and op-amp integrators and differentiators.

Learning objectives

By the end of this unit you can:

  • Use i = C dv/dt and v = L di/dt.
  • Compute energy stored in capacitors and inductors.
  • Combine capacitors and inductors in series and parallel.
  • Find DC steady-state voltages and currents in circuits with C and L.

Key concepts

  • A capacitor stores energy in an electric field: q = Cv, i = C dv/dt, w = ½Cv². Its voltage cannot jump instantly; at DC steady state it acts as an open circuit.
  • An inductor stores energy in a magnetic field: v = L di/dt, w = ½Li². Its current cannot jump instantly; at DC steady state it acts as a short circuit.
  • Capacitors in parallel add; in series they combine like resistors in parallel. Inductors combine like resistors: series add, parallel combine as reciprocals.
  • An op amp with a capacitor as feedback integrates: v_o = −(1/RC)∫v_in dt. Swapping R and C gives a differentiator.
  • Ideal capacitors and inductors store and return energy without loss; real ones have small resistance and leakage.

Key equations

i=Cdvdt,  w=12Cv2i=C\frac{dv}{dt},\ \ w=\tfrac12Cv^2
Capacitor
v=Ldidt,  w=12Li2v=L\frac{di}{dt},\ \ w=\tfrac12Li^2
Inductor
Cp=C1+C2,  1Cs=1C1+1C2C_p=C_1+C_2,\ \ \frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}
Capacitor combinations
vo=−1RC∫vin dtv_o=-\frac{1}{RC}\int v_{in}\,dt
Op-amp integrator

Worked examples, step by step

Example 6.1 A 10 µF capacitor’s voltage rises at 2 V/ms. Find its current.

Show the steps
  1. dv/dt = 2000 V/s.
  2. i = 10 × 10⁻⁶ × 2000.

Answer: i = 20 mA.

Example 6.2 Find the energy in a 1000 µF capacitor at 3 V and in a 0.2 H inductor carrying 3 A.

Show the steps
  1. w_C = ½ × 10⁻³ × 3² = 4.5 × 10⁻³ J.
  2. w_L = ½ × 0.2 × 3².

Answer: 4.5 mJ and 0.9 J.

Example 6.3 A 12 V DC source feeds 2 Ω in series with an inductor, with a 4 Ω resistor in parallel with a capacitor across the output (steady state). Find the inductor current and capacitor voltage.

Show the steps
  1. At DC the inductor is a short and the capacitor is open.
  2. i = 12/(2 + 4) = 2 A; v_C = 2 × 4.

Answer: i_L = 2 A, v_C = 8 V.

Practice problems

Work each on paper first, then check.

  1. Find 6 µF and 3 µF in series.

    Show answer

    2 µF.

  2. A 50 mH inductor’s current changes at 100 A/s. Find its voltage.

    Show answer

    5 V.

  3. Find 4 mH and 6 mH in series.

    Show answer

    10 mH.

  4. How much charge does 100 µF hold at 9 V?

    Show answer

    q = 0.9 mC.

Common misconceptions

  • Myth: A capacitor passes DC.

    Actually: At DC steady state it blocks current (open circuit).

  • Myth: Inductor current can change instantly.

    Actually: That would need infinite voltage; current is continuous.

  • Myth: Capacitors combine like resistors.

    Actually: They combine the opposite way: parallel add.

Flashcards

  • Capacitor

    i = C dv/dt.

  • Inductor

    v = L di/dt.

  • Capacitor energy

    ½Cv².

  • Inductor energy

    ½Li².

  • DC steady state

    C open, L short.

  • Integrator

    v_o = −(1/RC)∫v_in dt.

Self-check quiz

1. At DC steady state an inductor acts as…

Show answer

a short. di/dt = 0 so v = 0.

2. Doubling a capacitor’s voltage multiplies its energy by…

Show answer

4. w ∝ v².

3. Which cannot change instantly?

Show answer

capacitor voltage. It would need infinite current.

4. Two 10 µF capacitors in parallel give…

Show answer

20 µF. Parallel capacitors add.

How this helps save the planet

Storing energy is what lets solar and wind power be used after sunset or when the wind drops. EIA reported in August 2026 that U.S. utility-scale battery storage capacity grew at an average 70% a year over the last three years, reaching 43.6 GW by the end of 2025, and that operators added another 8.3 GW in the first half of 2026, reaching nearly 52 GW, with about 54 GW more planned over the next two and a half years.

EIA notes that solar plants host the largest battery units and that solar-plus-storage lets operators store power when prices are low and discharge it when prices are high. Batteries store energy chemically, but the capacitors and inductors of this unit sit in every battery inverter and charger, smoothing currents and storing energy for microseconds at a time as power is converted between DC and AC.

On Saving Planets: Physics II · Solutions

Sources: U.S. EIA: Battery storage capacity averaged 70% growth over the last three years (Aug. 7, 2026) · OpenStax UP2, 8.1 Capacitors and capacitance

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Unit 7 · Weeks 9–10

First-Order Circuits: RC and RL Transients

Natural and step responses of RC and RL circuits, the time constant, the general solution x(t) = x(∞) + [x(0) − x(∞)]e^(−t/τ), switching problems, and energy in digital switching.

Learning objectives

By the end of this unit you can:

  • Find initial and final values across a switching event.
  • Compute time constants τ = RC and τ = L/R.
  • Write and use the general first-order solution.
  • Estimate switching energy and power in digital circuits.

Key concepts

  • A circuit with one capacitor or one inductor (plus resistors and DC sources) obeys a first-order differential equation. Its response moves exponentially from an initial value to a final value.
  • Time constant: τ = R_Th C for an RC circuit and τ = L/R_Th for an RL circuit, where R_Th is the resistance seen by the C or L.
  • General solution: x(t) = x(∞) + [x(0⁺) − x(∞)] e^(−t/τ). After one τ the change is 63.2% complete; after 5τ it is 99.3% complete.
  • Use continuity: v_C(0⁺) = v_C(0⁻) and i_L(0⁺) = i_L(0⁻). Find x(∞) from the DC steady state (C open, L short).
  • Charging a capacitance C to V through a resistor takes CV² from the source: ½CV² is stored and ½CV² is lost as heat. Digital chips charge tiny capacitances billions of times a second, so their dynamic power is about P = αCV²f.

Key equations

τ=RC,  τ=LR\tau=RC,\ \ \tau=\frac{L}{R}
Time constants
x(t)=x(∞)+[x(0+)−x(∞)]e−t/τx(t)=x(\infty)+[x(0^+)-x(\infty)]e^{-t/\tau}
General first-order response
vC(t)=V(1−e−t/RC)v_C(t)=V\left(1-e^{-t/RC}\right)
RC charging from zero
Pdyn≈αCV2fP_{dyn}\approx\alpha CV^2f
Digital switching power

Worked examples, step by step

Example 7.1 A 1000 µF capacitor charges through 10 kΩ from a 3 V source, starting at 0 V. Find τ, v_C at t = τ and the time to reach 99.3% of 3 V.

Show the steps
  1. τ = 10⁴ × 10⁻³ = 10 s.
  2. v_C(τ) = 3(1 − e⁻¹) = 3 × 0.632.
  3. 99.3% takes 5τ.

Answer: τ = 10 s, v_C(τ) = 1.90 V, about 50 s.

Example 7.2 A 0.5 H inductor in series with 10 Ω is switched onto 20 V at t = 0. Find τ, the final current and i(0.1 s).

Show the steps
  1. τ = 0.5/10 = 0.05 s; i(∞) = 20/10 = 2 A.
  2. i(0.1) = 2(1 − e^(−0.1/0.05)) = 2(1 − e⁻²).

Answer: τ = 50 ms, i(∞) = 2 A, i(0.1 s) = 1.73 A.

Example 7.3 A charged capacitor discharges through a resistor with τ = 10 s. When does its voltage reach half its starting value?

Show the steps
  1. V₀e^(−t/τ) = V₀/2 gives t = τ ln 2.

Answer: t = 6.93 s.

Practice problems

Work each on paper first, then check.

  1. Find τ for 2 kΩ and 50 µF.

    Show answer

    0.1 s.

  2. After 3τ, what fraction of the change is complete?

    Show answer

    1 − e⁻³ = 95.0%.

  3. Find τ for 20 mH and 4 Ω.

    Show answer

    5 ms.

  4. Sample chip: effective switched C = 1 nF, V = 1 V, f = 1 GHz, α = 1. Find P, and P at 0.8 V.

    Show answer

    1 W; 0.64 W, 36% less.

Common misconceptions

  • Myth: The capacitor voltage jumps when the switch closes.

    Actually: Capacitor voltage is continuous; its current may jump.

  • Myth: The response is complete after one time constant.

    Actually: One τ is 63.2%; use about 5τ for “done”.

  • Myth: τ uses the source resistance only.

    Actually: Use the Thevenin resistance seen by the capacitor or inductor.

Flashcards

  • RC time constant

    τ = RC.

  • RL time constant

    τ = L/R.

  • General solution

    x(∞) + [x(0) − x(∞)]e^(−t/τ).

  • One τ

    63.2% of the change.

  • Five τ

    99.3%.

  • Switching power

    P ≈ αCV²f.

Self-check quiz

1. τ for an RC circuit is…

Show answer

RC. Units of seconds.

2. After one time constant a charging capacitor reaches about…

Show answer

63%. 1 − e⁻¹.

3. Lowering a chip’s supply voltage by 20% cuts dynamic power by about…

Show answer

36%. 0.8² = 0.64.

4. Inductor current at t = 0⁺ equals…

Show answer

its value at 0⁻. Continuity.

How this helps save the planet

Every bit a computer flips charges and discharges a tiny capacitance through a resistance, which is why computing uses so much electricity. A Berkeley Lab report for the U.S. Department of Energy found that data centers consumed about 4.4% of total U.S. electricity in 2023, that their use climbed from 58 TWh in 2014 to 176 TWh in 2023, and that they are expected to use 325 to 580 TWh, or 6.7% to 12% of U.S. electricity, by 2028, with growth driven largely by SI servers.

The switching-power formula from this unit shows the main lever chip designers have: power grows with the square of the supply voltage, so a 20% lower voltage cuts dynamic power by 36% (sample example). The report’s lead researcher says knowing what drives the growth helps identify opportunities for efficiency.

On Saving Planets: Intro to Programming · Differential Equations

Sources: Berkeley Lab: Report evaluates increase in electricity demand from data centers (Jan. 2025) · OpenStax UP2, 10.5 RC circuits

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Unit 8 · Week 11

Second-Order Circuits: RLC Natural and Step Responses

Series and parallel RLC circuits, the characteristic equation, the neper frequency α and resonant frequency ω₀, overdamped, critically damped and underdamped responses, quality factor, and LC energy exchange.

Learning objectives

By the end of this unit you can:

  • Write the second-order equation for series and parallel RLC circuits.
  • Compute α, ω₀ and ω_d and classify the damping.
  • Find the response form and its constants from initial conditions.
  • Relate Q to ringing and energy loss.

Key concepts

  • A circuit with an inductor and a capacitor obeys a second-order equation. For a series RLC: α = R/(2L), ω₀ = 1/√(LC). For a parallel RLC: α = 1/(2RC), same ω₀.
  • Roots s = −α ± √(α² − ω₀²). Overdamped (α > ω₀): two real roots, no oscillation. Critically damped (α = ω₀): fastest return without overshoot. Underdamped (α < ω₀): decaying oscillation at ω_d = √(ω₀² − α²).
  • An ideal LC circuit (R = 0) swaps energy between the capacitor’s electric field and the inductor’s magnetic field forever at ω₀.
  • Quality factor for a series RLC: Q = ω₀L/R = (1/R)√(L/C). High Q means sharp resonance and long ringing.
  • Critical resistance for a series RLC: R = 2√(L/C).

Key equations

α=R2L,  ω0=1LC\alpha=\frac{R}{2L},\ \ \omega_0=\frac{1}{\sqrt{LC}}
Series RLC parameters
s1,2=−α±α2−ω02s_{1,2}=-\alpha\pm\sqrt{\alpha^2-\omega_0^2}
Characteristic roots
ωd=ω02−α2\omega_d=\sqrt{\omega_0^2-\alpha^2}
Damped frequency
Q=ω0LRQ=\frac{\omega_0L}{R}
Quality factor (series)

Worked examples, step by step

Example 8.1 A series RLC has R = 10 Ω, L = 10 mH and C = 10 µF. Find ω₀, f₀, α and the damping type.

Show the steps
  1. ω₀ = 1/√(10⁻² × 10⁻⁵) = 1/√10⁻⁷ = 3162 rad/s; f₀ = ω₀/2π.
  2. α = 10/(2 × 0.01) = 500 s⁻¹ < ω₀.

Answer: ω₀ = 3162 rad/s, f₀ = 503.3 Hz, α = 500 s⁻¹: underdamped.

Example 8.2 For the same circuit find ω_d and Q.

Show the steps
  1. ω_d = √(3162² − 500²) = √(10⁷ − 2.5 × 10⁵).
  2. Q = 3162 × 0.01/10.

Answer: ω_d = 3122.5 rad/s, Q = 3.16.

Example 8.3 What resistance would make that circuit critically damped?

Show the steps
  1. R = 2√(L/C) = 2√(0.01/10⁻⁵) = 2√1000.

Answer: R = 63.2 Ω.

Practice problems

Work each on paper first, then check.

  1. A series RLC has α = 200 s⁻¹ and ω₀ = 100 rad/s. Classify it.

    Show answer

    Overdamped.

  2. Find ω₀ for L = 1 mH, C = 1 µF.

    Show answer

    31,623 rad/s (5.03 kHz).

  3. A parallel RLC has R = 1 kΩ and C = 1 µF. Find α.

    Show answer

    1/(2 × 10³ × 10⁻⁶) = 500 s⁻¹.

  4. Doubling R in a series RLC changes Q by…

    Show answer

    A factor of ½.

Common misconceptions

  • Myth: α has the same formula for series and parallel RLC.

    Actually: Series α = R/2L; parallel α = 1/(2RC).

  • Myth: Critically damped means oscillating.

    Actually: It is the boundary: no oscillation, fastest settling.

  • Myth: ω_d equals ω₀.

    Actually: ω_d is lower whenever there is damping.

Flashcards

  • Neper frequency

    α = R/2L (series).

  • Resonant frequency

    ω₀ = 1/√(LC).

  • Overdamped

    α > ω₀.

  • Underdamped

    α < ω₀, rings at ω_d.

  • Critical R

    2√(L/C) (series).

  • Q

    ω₀L/R.

Self-check quiz

1. Underdamped means…

Show answer

α < ω₀. Complex roots.

2. ω₀ for an LC circuit is…

Show answer

1/√(LC). Resonant frequency.

3. Raising R in a series RLC makes the response…

Show answer

more damped. α grows.

4. In an ideal LC circuit energy…

Show answer

swaps between C and L. No resistance.

How this helps save the planet

Resonant circuits are how electric vehicles can charge without a cord. The U.S. Department of Energy’s Alternative Fuels Data Center explains that inductive charging uses an electromagnetic field to transfer electricity to an EV, that some wireless stations already operate at power levels comparable to Level 2, that it is more common for transit and fleets at higher power, and that DOE is researching high-powered wireless charging.

Wireless chargers tune a coil (L) and a capacitor (C) on each side to the same ω₀ = 1/√(LC) so that energy swaps efficiently across the air gap, and a high Q keeps losses low. That is the circuit idea behind the transit and fleet wireless chargers AFDC describes.

On Saving Planets: Physics II · Differential Equations

Sources: U.S. DOE Alternative Fuels Data Center: Electric vehicle charging stations · OpenStax UP2, 14.6 RLC series circuits

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Unit 9 · Weeks 12–13

Sinusoidal Steady State: Phasors and Impedance

Sinusoids, amplitude, frequency, phase and RMS values, complex numbers, phasors, impedance and admittance of R, L and C, AC circuit analysis with Ohm’s law, KCL, KVL, nodal and Thevenin methods in phasor form, and frequency response basics.

Learning objectives

By the end of this unit you can:

  • Convert between time-domain sinusoids and phasors.
  • Compute impedances of R, L and C at a given frequency.
  • Analyze AC circuits with phasor versions of the DC methods.
  • Describe how impedance changes with frequency in simple filters.

Key concepts

  • A sinusoid v(t) = V_m cos(ωt + φ) has amplitude V_m, angular frequency ω = 2πf and phase φ. Its RMS value is V_m/√2; U.S. household 120 V is an RMS value at 60 Hz, a peak of about 170 V.
  • A phasor is the complex number V = V_m∠φ that carries amplitude and phase. In sinusoidal steady state, d/dt becomes multiplication by jω.
  • Impedance Z = V/I: Z_R = R, Z_L = jωL, Z_C = 1/(jωC) = −j/(ωC). Inductors look like shorts at low frequency and opens at high frequency; capacitors do the opposite.
  • All the DC tools (series/parallel, dividers, nodal and mesh analysis, Thevenin and Norton) work with impedances and complex arithmetic.
  • An RC low-pass filter passes low frequencies and attenuates high ones; its cutoff (half-power) frequency is ω_c = 1/(RC).

Key equations

Vrms=Vm2V_{rms}=\frac{V_m}{\sqrt2}
RMS value of a sinusoid
ZR=R, ZL=jωL, ZC=1jωCZ_R=R,\ Z_L=j\omega L,\ Z_C=\frac{1}{j\omega C}
Impedances
V=IZ\mathbf{V}=\mathbf{I}Z
Phasor Ohm’s law
ωc=1RC\omega_c=\frac{1}{RC}
RC filter cutoff

Worked examples, step by step

Example 9.1 Express v(t) = 170 cos(377t) V as a phasor and find its RMS value and frequency.

Show the steps
  1. V = 170∠0° V (peak).
  2. V_rms = 170/√2; f = 377/(2π).

Answer: V_rms = 120.2 V at f = 60.0 Hz.

Example 9.2 At 60 Hz find the impedance of a 0.1 H inductor and a 100 µF capacitor.

Show the steps
  1. ω = 2π × 60 = 377.0 rad/s.
  2. Z_L = j × 377.0 × 0.1; Z_C = −j/(377.0 × 10⁻⁴).

Answer: Z_L = j37.70 Ω, Z_C = −j26.53 Ω.

Example 9.3 A 120 V (RMS), 60 Hz source drives 30 Ω in series with 0.1 H. Find the current and its phase.

Show the steps
  1. Z = 30 + j37.70 Ω; |Z| = √(30² + 37.70²) = 48.18 Ω, ∠51.49°.
  2. I = 120/48.18 at −51.49°.

Answer: I = 2.491 A RMS, lagging the voltage by 51.49°.

Practice problems

Work each on paper first, then check.

  1. Find the RMS value of a sinusoid with 10 V peak.

    Show answer

    7.071 V.

  2. Find the impedance of 1 µF at ω = 1000 rad/s.

    Show answer

    −j1000 Ω.

  3. Find the cutoff frequency of an RC filter with R = 1 kΩ and C = 1 µF.

    Show answer

    ω_c = 1000 rad/s (159.2 Hz).

  4. Convert 3 + j4 to polar form.

    Show answer

    5∠53.13°.

Common misconceptions

  • Myth: 120 V AC means a 120 V peak.

    Actually: It is the RMS value; the peak is about 170 V.

  • Myth: Impedances add like magnitudes.

    Actually: Add complex numbers: 30 + j40 has magnitude 50, not 70.

  • Myth: Phasors work for any waveform.

    Actually: They apply to sinusoidal steady state at one frequency.

Flashcards

  • Phasor

    V_m∠φ.

  • RMS

    V_m/√2.

  • Z_L

    jωL.

  • Z_C

    1/(jωC).

  • Phasor Ohm’s law

    V = IZ.

  • RC cutoff

    ω_c = 1/RC.

Self-check quiz

1. The impedance of an inductor is…

Show answer

jωL. v = L di/dt → jωL.

2. At very high frequency a capacitor acts like…

Show answer

a short. |Z_C| = 1/(ωC) → 0.

3. The RMS value of a sine wave is its peak times…

Show answer

0.707. 1/√2.

4. In an RL circuit the current…

Show answer

lags the voltage. Inductive phase angle.

How this helps save the planet

Home EV charging is an AC circuit problem. The Alternative Fuels Data Center explains that AC Level 1 charging uses a 120 V plug, assumes about 1.9 kW, and adds about 5 miles of range per hour, so 8 hours can replenish about 40 miles for a mid-size EV. AC Level 2 uses 240 V (homes) or 208 V (commercial); most residential Level 2 chargers run at up to 30 A, delivering 7.2 kW and about 25 miles of range per hour, and need a dedicated 40 A circuit under National Electrical Code Article 625.

Those are RMS values: 1.9 kW at 120 V means about 15.8 A, and 7.2 kW at 240 V is 30 A. AFDC adds that as of 2023 nearly 80% of public charging ports were Level 2, and that Level 2 units range from 2.9 to 19.2 kW. Most EV owners do most of their charging at home.

On Saving Planets: Physics II · Solutions

Sources: U.S. DOE Alternative Fuels Data Center: Electric vehicle charging stations · OpenStax UP2, 15.2 Simple AC circuits

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Unit 10 · Week 14

AC Power, Power Factor, Resonance, Transformers and Three-Phase Systems

Instantaneous and average power, real, reactive and apparent power, the power triangle, power factor and its correction, series resonance and bandwidth, ideal transformers, balanced three-phase systems, and why the grid transmits at high voltage.

Learning objectives

By the end of this unit you can:

  • Compute real, reactive and apparent power and the power factor.
  • Size a capacitor for power-factor correction.
  • Analyze ideal transformers and balanced three-phase loads.
  • Explain transmission losses with I²R.

Key concepts

  • Average (real) power P = V_rms I_rms cos θ in watts, where θ is the angle between voltage and current. Reactive power Q = V_rms I_rms sin θ in VAR. Apparent power S = V_rms I_rms in VA, and S² = P² + Q².
  • Power factor pf = cos θ = P/S. Motors are inductive (lagging pf). A parallel capacitor supplies reactive power locally, raising pf and cutting the current the utility must deliver: C = Q/(ωV²) for full correction.
  • At series resonance ω₀ = 1/√(LC) the reactances cancel, the impedance is just R and current is largest; the bandwidth is ω₀/Q.
  • An ideal transformer with turns ratio N₁:N₂ has V₂/V₁ = N₂/N₁ and I₂/I₁ = N₁/N₂, so power in equals power out.
  • Balanced three-phase: three sinusoids 120° apart. Total power P = √3 V_L I_L cos θ, constant in time. Line losses are I²R, so raising voltage by k for the same power cuts current by k and losses by k².

Key equations

P=VrmsIrmscos⁡θ,  Q=VrmsIrmssin⁡θP=V_{rms}I_{rms}\cos\theta,\ \ Q=V_{rms}I_{rms}\sin\theta
Real and reactive power
S2=P2+Q2,  pf=PSS^2=P^2+Q^2,\ \ \text{pf}=\frac{P}{S}
Power triangle
V2V1=N2N1=I1I2\frac{V_2}{V_1}=\frac{N_2}{N_1}=\frac{I_1}{I_2}
Ideal transformer
P3ϕ=3 VLILcos⁡θP_{3\phi}=\sqrt3\,V_LI_L\cos\theta
Balanced three-phase power

Worked examples, step by step

Example 10.1 A motor draws 20 A RMS from 240 V RMS at a power factor of 0.8 lagging. Find S, P and Q.

Show the steps
  1. S = 240 × 20 = 4800 VA.
  2. P = 4800 × 0.8; sin θ = 0.6, so Q = 4800 × 0.6.

Answer: S = 4.8 kVA, P = 3.84 kW, Q = 2.88 kVAR.

Example 10.2 What capacitor in parallel would correct that motor to unity power factor at 60 Hz?

Show the steps
  1. The capacitor must supply Q = 2880 VAR.
  2. C = Q/(ωV²) = 2880/(377.0 × 240²).

Answer: C = 132.6 µF; the line current falls from 20 A to 16 A.

Example 10.3 A line with 5 Ω resistance delivers 10 MW. Compare the I²R loss at 100 kV and at 20 kV (unity power factor, single-phase sample).

Show the steps
  1. At 100 kV: I = 10⁷/10⁵ = 100 A, loss = 100² × 5 = 50 kW (0.5%).
  2. At 20 kV: I = 500 A, loss = 500² × 5 = 1.25 MW (12.5%).

Answer: 5 times the voltage cuts losses 25 times.

Practice problems

Work each on paper first, then check.

  1. A load has P = 6 kW and Q = 8 kVAR. Find S and pf.

    Show answer

    S = 10 kVA, pf = 0.6.

  2. A transformer steps 7200 V down to 240 V. Find N₁:N₂.

    Show answer

    30:1.

  3. Find the power of a balanced 480 V three-phase load drawing 100 A at pf 0.9.

    Show answer

    √3 × 480 × 100 × 0.9 = 74.8 kW.

  4. A series RLC has ω₀ = 1000 rad/s and Q = 20. Find the bandwidth.

    Show answer

    50 rad/s.

Common misconceptions

  • Myth: Reactive power does no work, so it does not matter.

    Actually: It still raises the current, and current causes I²R losses in wires and transformers.

  • Myth: A transformer works on DC.

    Actually: It needs a changing flux; DC gives none.

  • Myth: Three-phase power is three times single-phase with the line voltage.

    Actually: P = √3 V_L I_L cos θ.

Flashcards

  • Real power

    P = VI cos θ (W).

  • Reactive power

    Q = VI sin θ (VAR).

  • Apparent power

    S = VI (VA).

  • Power factor

    P/S.

  • Transformer

    V₂/V₁ = N₂/N₁.

  • Three-phase

    P = √3 V_L I_L cos θ.

Self-check quiz

1. Power factor equals…

Show answer

P/S. cos θ.

2. Raising transmission voltage 10 times (same power) cuts I²R losses by…

Show answer

100×. Current falls 10×; loss ∝ I².

3. An inductive load is corrected with a…

Show answer

parallel capacitor. Capacitors supply reactive power.

4. A step-down transformer with 10:1 turns makes the secondary current…

Show answer

10× larger. Power is conserved.

How this helps save the planet

The grid moves power at high voltage to keep I²R losses down. EIA explains that high-voltage transmission lines carry electricity over long distances because higher voltage makes long-distance transmission more efficient and less expensive, lower-voltage distribution lines deliver it to customers, and transformers at substations step voltages up and down between the stages. EIA estimates that transmission and distribution losses averaged about 5% of the electricity transmitted and distributed in the United States in 2018 through 2022.

The worked example shows the physics: five times the voltage cuts losses twenty-five times. Power-factor correction, efficient transformers and new long-distance lines to reach wind and solar resources, which EIA lists among the grid’s challenges, all chip away at that 5%. AFDC notes that common U.S. three-phase voltages are 208/120 V and 480/277 V, common at commercial and industrial sites, and that DC fast chargers typically take a three-phase AC input.

On Saving Planets: Solutions · Physics II

Sources: U.S. EIA: Delivery to consumers (the electric power grid) · U.S. EIA: How much electricity is lost in transmission and distribution? · OpenStax UP2, 15.6 Transformers

Free learning resources

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The lab (Circuits lab)

The lab is the hands-on part of the course. Here it is free and for practice only, with no credit.

Every lab can be done in a free PhET simulation or with a 3 V battery pack. Never work on household wiring or wall outlets: mains voltage can kill. Do not short a battery, and put electrolytic capacitors in the right way round.

Safety reference: OpenStax University Physics Volume 2, 10.6 Household wiring and electrical safety

Lab A: Ohm’s law and the multimeter

Week 2

Goal: Measure voltage, current and resistance correctly, show that current is proportional to voltage for a resistor, and find R from the slope.

v=iR,p=viv=iR,\quad p=vi
Ohm’s law and power

Procedure

  1. Open PhET Ohm’s Law and Circuit Construction Kit: DC; build a battery and resistor and add a voltmeter (in parallel) and ammeter (in series).
  2. Home: on a breadboard, connect a 1 kΩ resistor to one AA cell (1.5 V) and then to two (3 V). Measure v across the resistor and i in series with the meter on its mA range.
  3. Measure the resistor alone with the Ω range (power disconnected) and compare with v/i.
  4. Plot i against v; the slope is 1/R. Compute the power in each case.

Data table

v (V)i (mA) for 1 kΩp (mW)
1.51.52.25
3.03.09.0

Ideal values for a 1 kΩ resistor; fresh cells read a little above 1.5 V each and resistors are within their tolerance (often ±5%).

Sample calculation

  1. R = v/i = 3.0/0.003 = 1000 Ω.
  2. p = vi = 3.0 × 0.003 = 9 mW, far below a ¼ W resistor rating.

Safety

  • Use only 1.5–3 V battery packs; never connect to a wall outlet.
  • Never connect the meter on its current range directly across a battery: it is a short circuit.

Home or virtual alternative

  • The breadboard version is the home lab.

PhET: Ohm’s Law · PhET: Circuit Construction Kit: DC

Planet link: The same v·i measurement is what a plug-in meter or smart meter does to show where energy goes; DOE says LEDs use at least 75% less than incandescents. U.S. DOE: Guide to energy-efficient lighting (PDF)

Lab B: Kirchhoff’s laws, series and parallel, and dividers

Week 3

Goal: Verify KVL and KCL by measurement, check series and parallel equivalent resistance, and see a voltage divider sag under load.

∑v=0,∑i=0,v2=vR2R1+R2\sum v=0,\quad \sum i=0,\quad v_2=v\frac{R_2}{R_1+R_2}
KVL, KCL and the divider

Procedure

  1. Build 1 kΩ and 2 kΩ in series across 3 V (PhET or breadboard). Measure the current and each voltage; check that the voltages add to the source.
  2. Rebuild two 1 kΩ resistors in parallel across 3 V. Measure the current in each branch and in the source lead; check KCL.
  3. Treat the series pair as a divider and measure v across the 2 kΩ resistor.
  4. Connect a 2 kΩ “load” across the 2 kΩ resistor and measure again; explain the drop with the parallel combination.

Data table

CircuitCurrentVoltages
1 kΩ + 2 kΩ series1.0 mA1.0 V and 2.0 V
1 kΩ ∥ 1 kΩ3.0 mA each, 6.0 mA total3.0 V
Divider loaded by 2 kΩ1.5 mA from source1.5 V out

Ideal values for a 3 V source.

Sample calculation

  1. Series: i = 3/3000 = 1 mA; 1 V + 2 V = 3 V (KVL).
  2. Loaded divider: 2 kΩ ∥ 2 kΩ = 1 kΩ, so v_out = 3 × 1/(1 + 1) = 1.5 V instead of 2 V.

Safety

  • Low-voltage battery packs only; remove the battery before rewiring.

Home or virtual alternative

  • The breadboard version is the home lab.

PhET: Circuit Construction Kit: DC

Planet link: EIA explains that PV cells are wired into panels and arrays; series strings add voltage and parallel strings add current, as here. U.S. EIA: Photovoltaics and electricity

Lab C: A battery’s Thevenin equivalent and maximum power

Week 6

Goal: Model a real battery as V_Th in series with R_Th by measuring its open-circuit and loaded voltage, then predict the load for maximum power.

RTh=voc−vLiL,Pmax⁡=VTh24RThR_{Th}=\frac{v_{oc}-v_L}{i_L},\quad P_{\max}=\frac{V_{Th}^2}{4R_{Th}}
Thevenin from two measurements

Procedure

  1. In PhET Circuit Construction Kit: DC, give the battery an internal resistance and measure its voltage with no load and with a load resistor.
  2. Home: measure a 2 × AA pack’s open-circuit voltage. Then connect a 22 Ω resistor rated at least 1 W for a few seconds only and read the voltage.
  3. Compute i_L = v_L/22 Ω and R_Th, and predict P_max.
  4. In PhET, sweep the load resistance and confirm the power peaks when R_L = R_Th.

Data table

QuantitySample value
v_oc3.20 V
v_L (22 Ω)3.10 V
i_L0.141 A
R_Th0.71 Ω

Sample values for fresh alkaline cells; worn cells show a much larger R_Th.

Sample calculation

  1. i_L = 3.10/22 = 0.1409 A; R_Th = 0.10/0.1409 = 0.71 Ω.
  2. Load power = 3.10 × 0.1409 = 0.437 W, so use a 1 W resistor. Matching R_L = 0.71 Ω would be close to a short and heat the cells: do not try it at home, only in PhET.

Safety

  • Connect the 22 Ω load only for a few seconds; the resistor and cells get warm.
  • Never short the battery terminals.

Home or virtual alternative

  • The two-measurement step is the home lab; the power sweep is PhET only.

PhET: Circuit Construction Kit: DC · OpenStax UP2, 10.1 Electromotive force

Planet link: EIA reports PV module efficiency rising from under 10% in the mid-1980s to nearly 25% now; operating each panel at its best load is part of using that output. U.S. EIA: Photovoltaics and electricity

Lab D: RC charging and the time constant

Week 10

Goal: Watch a capacitor charge through a resistor, measure the time constant, and compare with τ = RC.

vC(t)=V(1−e−t/RC)v_C(t)=V\left(1-e^{-t/RC}\right)
RC charging

Procedure

  1. In PhET Circuit Construction Kit: AC, build a battery, resistor, capacitor and switch; use the voltage chart to watch v_C.
  2. Home: put a 1000 µF capacitor (rated 6.3 V or more, negative lead to battery −) in series with 10 kΩ across 3 V. Read v_C with the multimeter every 10 s.
  3. Find the time at which v_C reaches 63.2% of the final voltage; that is τ.
  4. Plot ln(1 − v_C/V) against t; the slope is −1/τ. Discharge the capacitor through the resistor before storing it.

Data table

t (s)010203050
v_C (V)01.902.592.852.98

Ideal values for R = 10 kΩ, C = 1000 µF and V = 3 V. Electrolytic capacitors are often within ±20%, and the meter itself loads the circuit slightly.

Sample calculation

  1. τ = 10⁴ × 10⁻³ = 10 s.
  2. v_C(10 s) = 3(1 − e⁻¹) = 1.90 V; v_C(50 s) = 3(1 − e⁻⁵) = 2.98 V.

Safety

  • Insert electrolytic capacitors with the correct polarity and below their voltage rating; reversed ones can burst. Wear glasses.

Home or virtual alternative

  • The breadboard version is the home lab.

PhET: Circuit Construction Kit: AC · PhET: Capacitor Lab: Basics

Planet link: Berkeley Lab reports data centers used about 4.4% of U.S. electricity in 2023; every logic switch inside them is a tiny RC charge like this one. Berkeley Lab: Report evaluates increase in electricity demand from data centers (Jan. 2025)

Lab E: RLC resonance

Week 13

Goal: Drive a series RLC circuit with an AC source at different frequencies, find the resonant peak, and compare with ω₀ = 1/√(LC).

∣I∣=VR2+(ωL−1/ωC)2|I|=\frac{V}{\sqrt{R^2+(\omega L-1/\omega C)^2}}
Series RLC current

Procedure

  1. In PhET Circuit Construction Kit: AC (or its Virtual Lab), build an AC source, resistor, inductor and capacitor in series with an ammeter.
  2. Set R = 10 Ω, L = 10 mH and C = 10 µF if available, or note the values PhET uses, and compute f₀.
  3. Step the source frequency through f₀ and record the current amplitude.
  4. Plot |I| against f, find the peak and the half-power points, and compare the bandwidth with ω₀/Q.

Data table

f (Hz)400503600
|I| (A) for 10 V0.5631.0000.667

Calculated values for R = 10 Ω, L = 10 mH, C = 10 µF and a 10 V amplitude source.

Sample calculation

  1. f₀ = 1/(2π√(10⁻² × 10⁻⁵)) = 503.3 Hz; at f₀, |Z| = R = 10 Ω, so |I| = 1 A.
  2. At 400 Hz: X_L = 25.13 Ω, X_C = 39.79 Ω, |Z| = 17.76 Ω, |I| = 0.563 A.

Safety

  • Simulation only: do not build AC circuits from wall power.

Home or virtual alternative

  • This lab is PhET-only for safety; no home wiring.

PhET: Circuit Construction Kit: AC · PhET: Circuit Construction Kit: AC – Virtual Lab

Planet link: AFDC says inductive EV chargers transfer power through an electromagnetic field; tuned LC circuits like this one make that transfer efficient. U.S. DOE Alternative Fuels Data Center: Electric vehicle charging stations

Lab F: Measuring real power at home and transformers in PhET

Week 14

Goal: Use a plug-in power meter to compare the real power of light bulbs and appliances, estimate annual energy, and explore transformer voltage ratios in PhET.

w=Pt,V2V1=N2N1w=Pt,\quad \frac{V_2}{V_1}=\frac{N_2}{N_1}
Energy and the ideal transformer

Procedure

  1. Plug a power meter into a wall outlet and plug a lamp into the meter (never open the meter or the lamp). Record watts with an incandescent or halogen bulb and then with an LED of similar brightness.
  2. If the meter shows VA and power factor, record them for a phone charger and a fan too.
  3. Estimate annual kWh at 3 hours a day and compare.
  4. In PhET Faraday’s Law, change the number of coil turns and the magnet speed, and relate the induced voltage to N and the rate of change of flux.

Data table

BulbP (W)kWh/yr at 3 h/day
Incandescent6065.7
LED (similar light)99.9

Sample readings; DOE’s guide says LEDs use at least 75% less energy than incandescents. Your bulbs set your numbers.

Sample calculation

  1. 60 W × 3 h × 365 = 65.7 kWh; 9 W gives 9.86 kWh.
  2. Savings = 85% for this sample LED, consistent with “at least 75%”.

Safety

  • Use only a commercial plug-in meter and plug-in appliances; never touch outlet contacts or open devices. Let hot bulbs cool before handling.

Home or virtual alternative

  • The plug-in meter step is the home lab; transformers are explored in PhET.

PhET: Faraday’s Law · OpenStax UP2, 15.6 Transformers

Planet link: EIA estimates transmission and distribution losses at about 5%; every watt saved at the bulb also avoids those losses upstream. U.S. DOE: Guide to energy-efficient lighting (PDF) · U.S. EIA: How much electricity is lost in transmission and distribution?

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Exam prep practice sets

Time yourself: about 10 minutes per problem. Then compare with the full solutions.

Midterm practice set (Units 1–5)

  1. Find the equivalent resistance of 12 Ω in parallel with (4 Ω + 2 Ω).

    Show full solution
    1. 12 × 6/(12 + 6).

    Answer: 4 Ω.

  2. A 24 V source feeds 3 kΩ and 1 kΩ in series. Find the voltage across 1 kΩ.

    Show full solution
    1. Divider: 24 × 1/4.

    Answer: 6 V.

  3. A node connects to 12 V through 4 Ω, to ground through 12 Ω, and receives 2 A from a current source. Find its voltage.

    Show full solution
    1. (v − 12)/4 + v/12 = 2, so 4v = 60.

    Answer: 15 V.

  4. A network has v_oc = 18 V and i_sc = 6 A. Find R_Th and the maximum power it can deliver.

    Show full solution
    1. R_Th = 3 Ω; P = 18²/(4 × 3).

    Answer: 3 Ω and 27 W.

  5. An inverting amplifier has R_f = 33 kΩ and R_in = 3.3 kΩ. Find v_o for v_in = −0.4 V.

    Show full solution
    1. Gain = −10.

    Answer: +4 V.

Final practice set (all units)

  1. A 20 µF capacitor charges from 0 V through 5 kΩ toward 10 V. Find τ and v_C at 0.2 s.

    Show full solution
    1. τ = 0.1 s; 10(1 − e⁻²).

    Answer: τ = 0.1 s, v_C = 8.65 V.

  2. A series RLC has R = 40 Ω, L = 0.1 H and C = 100 µF. Classify the response and find ω_d if it rings.

    Show full solution
    1. α = 200 s⁻¹; ω₀ = 316.2 rad/s.

    Answer: Underdamped, ω_d = 244.9 rad/s.

  3. Find the impedance of 20 Ω in series with 100 µF at 60 Hz.

    Show full solution
    1. Z_C = −j26.53 Ω.

    Answer: Z = 20 − j26.53 Ω = 33.22∠−52.99° Ω.

  4. A load draws 12 A RMS from 120 V RMS at pf 0.85 lagging. Find P and Q.

    Show full solution
    1. S = 1440 VA; sin θ = 0.5268.

    Answer: P = 1224 W, Q = 758.6 VAR.

  5. A 20:1 transformer has 2400 V on its primary and a 10 A secondary load. Find V₂ and I₁.

    Show full solution
    1. V₂ = 2400/20; I₁ = 10/20.

    Answer: 120 V and 0.5 A.

  6. Find the power of a balanced 208 V three-phase load drawing 50 A at unity power factor.

    Show full solution
    1. √3 × 208 × 50.

    Answer: 18.01 kW.

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Formula sheet

i=dq/dt, p=vi, v=iRi=dq/dt,\ p=vi,\ v=iR
R=ρL/A, p=i2R=v2/RR=\rho L/A,\ p=i^2R=v^2/R
∑i=0, ∑v=0\sum i=0,\ \sum v=0
Rs=∑R, 1/Rp=∑1/RR_s=\sum R,\ 1/R_p=\sum 1/R
v2=vR2/(R1+R2)v_2=vR_2/(R_1+R_2)
VTh=voc, RTh=voc/isc, Pmax⁡=VTh2/(4RTh)V_{Th}=v_{oc},\ R_{Th}=v_{oc}/i_{sc},\ P_{\max}=V_{Th}^2/(4R_{Th})
vo=−(Rf/Rin)vin, vo=(1+Rf/Rg)vinv_o=-(R_f/R_{in})v_{in},\ v_o=(1+R_f/R_g)v_{in}
i=C dv/dt, v=L di/dt, w=12Cv2, 12Li2i=C\,dv/dt,\ v=L\,di/dt,\ w=\tfrac12Cv^2,\ \tfrac12Li^2
τ=RC, L/R; x(t)=x(∞)+[x(0)−x(∞)]e−t/τ\tau=RC,\ L/R;\ x(t)=x(\infty)+[x(0)-x(\infty)]e^{-t/\tau}
α=R/2L, ω0=1/LC, ωd=ω02−α2\alpha=R/2L,\ \omega_0=1/\sqrt{LC},\ \omega_d=\sqrt{\omega_0^2-\alpha^2}
ZL=jωL, ZC=1/(jωC), Vrms=Vm/2Z_L=j\omega L,\ Z_C=1/(j\omega C),\ V_{rms}=V_m/\sqrt2
P=VIcos⁡θ, Q=VIsin⁡θ, S2=P2+Q2P=VI\cos\theta,\ Q=VI\sin\theta,\ S^2=P^2+Q^2
V2/V1=N2/N1, P3ϕ=3VLILcos⁡θV_2/V_1=N_2/N_1,\ P_{3\phi}=\sqrt3V_LI_L\cos\theta

Constants

ConstantValue
1 kilowatt-hour (kWh)3.6 × 10⁶ J
1 horsepower (electric)746 W
Elementary charge e1.602 × 10⁻¹⁹ C
U.S. household service120 V and 240 V at 60 Hz
RMS of a sine waveV_rms = V_peak/√2 ≈ 0.7071 V_peak
1 ampere-hour (Ah)3600 C

Source: NIST Guide to the SI, Appendix B.9: conversion factors

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Free texts and simulations

Use these free textbooks, lectures and simulations alongside the course.

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Glossary

Apparent power
S = V_rms I_rms, in VA.
Capacitor
Element with i = C dv/dt that stores energy in an electric field.
Impedance
Complex ratio V/I in sinusoidal steady state.
Inductor
Element with v = L di/dt that stores energy in a magnetic field.
KCL
Currents leaving a node sum to zero.
KVL
Voltages around a loop sum to zero.
Mesh
A loop that contains no other loop.
Node
A point where two or more elements connect.
Norton equivalent
Current source I_N in parallel with R_N.
Op amp
High-gain differential amplifier used with feedback.
Passive sign convention
Current enters the + terminal of an element that absorbs power.
Phasor
Complex number giving a sinusoid’s amplitude and phase.
Power factor
cos θ = P/S.
Quality factor
Q = ω₀L/R, sharpness of resonance.
RMS
Root-mean-square value; V_m/√2 for a sinusoid.
Reactive power
Q = V_rms I_rms sin θ, in VAR.
Supernode
Two nodes joined by a voltage source, treated together in KCL.
Thevenin equivalent
Voltage source V_Th in series with R_Th.
Time constant
τ = RC or L/R.
Transformer
Coupled coils that change AC voltage by the turns ratio.

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Self-assessment rubric

No grade or credit is given. Use this to judge honestly where you are in each unit.

LevelWhat it looks like
4 · MasteryLabels circuits fully, chooses an efficient method (dividers, nodal, mesh, Thevenin, phasors), checks power balance and limiting cases, and works safely in the lab.
3 · ProficientSolves the practice problems with occasional sign or unit slips.
2 · DevelopingFollows worked examples but mixes up series and parallel, or the sign convention.
1 · BeginningPlugs numbers into formulas without a labeled diagram; needs to revisit the unit.

Suggested weighting if you want a score

PartWeight
Practice problems20%
Labs25%
Midterm practice set25%
Final practice set30%

Aim for level 3 or better in every unit before moving to the next course.

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