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UNIT 4About 11 min + practice

Electric Circuits

Electricity & Magnetism track · College Board Unit 11. Build the model, derive its consequences, and test the assumptions.

What you’ll learn

  • Use current conservation and energy conservation in circuits.
  • Connect resistivity to dimensions.
  • Derive and interpret charging and discharging RC behavior.
01

Current is charge flow, not charge consumption

Current I=dq/dt is the rate of charge crossing a section. Conventional current points in the direction positive charge would move; electrons in metal drift oppositely. In a steady circuit, charge does not accumulate at a junction, so the sum of currents entering equals the sum leaving.

A resistor transfers energy from the electrical system to internal energy; it does not consume charge. Current is the same through series elements because there is no sustained charge accumulation between them. The energy transferred per charge, or voltage drop, can differ.

PAUSE & TRY IT

Does current decrease as it passes through one series resistor?

Reveal answer

No. In steady state the current entering equals the current leaving.

02

Material and geometry determine resistance

For a uniform wire, R=. Here ρ denotes resistivity, not charge density. For a nonuniform wire, add differential series resistance: dR=ρ(x)dx/A(x). Resistivity can depend on temperature, so a heated component may not remain ohmic.

An ohmic resistor obeys V=IR over the tested range. On a V-versus-I graph the slope is R; on an I-versus-V graph it is . Power dissipated is IV=I2R=. State whether current or voltage is fixed before deciding how changing R changes power.

R=∫ρ(x)dx/A(x)

PAUSE & TRY IT

At fixed voltage, doubling R changes power how?

Reveal answer

P= halves.

03

Kirchhoff’s rules express conservation laws

The junction rule expresses charge conservation. The loop rule expresses energy conservation per charge around a closed path in a lumped circuit: sum voltage changes to zero. Traversing a resistor with its assumed current gives −IR; opposite gives +IR. Crossing an ideal source from negative to positive gives +ε.

Assign current directions, write independent junction and loop equations, and solve. A negative current means the actual direction is opposite your assumption. Do not silently reverse signs halfway through a solution. An ideal voltmeter has very large resistance and is placed in parallel; an ideal ammeter has negligible resistance and is placed in series.

04

Series, parallel, and real batteries

Series resistances add because the same current experiences successive drops. Parallel branches share voltage, so conductances add: _eq=i. A branch short can redirect current; a broken series path stops current through that path. Draw the actual nodes rather than judging by visual placement.

A battery with emf ε and internal resistance r has terminal voltage ε−Ir while delivering current. More current can lower terminal voltage. The maximum current of this simplified model is limited by internal and external resistance; an ideal zero-resistance short is not a physically safe experiment.

05

Derive the RC transient

For a series RC circuit charging from an ideal source ε with initially uncharged capacitor, Kirchhoff gives ε−IR−=0 with I=dq/dt. The solution is q=Cε(1−e−t/RC) and I=()e−t/RC. The time constant τ=RC controls how quickly the transient evolves.

At t=0 an uncharged ideal capacitor has zero voltage; long after connection it behaves as an open circuit for steady DC. Capacitor voltage cannot jump instantaneously in an ordinary finite-current circuit. For discharge through R, q=q0e−t/RC. Current sign depends on the chosen reference direction, but its magnitude decays exponentially.

τ=RC
q_charge=Cε(1−e)
Series RC charging circuit

The battery, resistor, and capacitor form one closed loop. The capacitor’s voltage grows while current decreases.

Series RC charging circuitRCεICharging: ε − IR − q/C = 0 · I = dq/dt
Original ScienceHub diagram · Schematic, not to scale.
An RC charging current decays as capacitor voltage risesIllustrative model, not collected experimental data. Normalized time t/(RC). Voltage and current are divided by their own initial/final reference values, so these curves are dimensionless comparisons.
An RC charging current decays as capacitor voltage rises00.250.50.75101.252.53.755 Time / RCNormalized valueV capacitor / source vo…Current / initial curre…
Read figure values as text

V capacitor / source voltage: 0: 0; 0.10416666666666667: 0.09892489427870943; 0.20833333333333334: 0.18806365384936508; 0.3125: 0.2683843710533582; 0.4166666666666667: 0.34075936979955623; 0.5208333333333334: 0.405974679446365; 0.625: 0.4647385714810097; 0.7291666666666666: 0.5176892517087218; 0.8333333333333334: 0.5654017914929218; 0.9375: 0.608394373323201; 1.0416666666666667: 0.6471339185411511; 1.1458333333333333: 0.6820411583440102; 1.25: 0.7134952031398099; 1.3541666666666667: 0.7418376598795473; 1.4583333333333333: 0.7673763420827073; 1.5625: 0.7903886128489022; 1.6666666666666667: 0.8111243971624382; 1.7708333333333333: 0.8298088962049714; 1.875: 0.8466450331550716; 1.9791666666666667: 0.8618156570373211; 2.0833333333333335: 0.875485528555877; 2.1875: 0.8878031094796562; 2.2916666666666665: 0.8989021750127812; 2.3958333333333335: 0.9089032666614493; 2.5: 0.9179150013761012; 2.6041666666666665: 0.9260352511868384; 2.7083333333333335: 0.9333522061435318; 2.8125: 0.939945332104692; 2.9166666666666665: 0.9458862337771784; 3.0208333333333335: 0.9512394323797938; 3.125: 0.9560630663765926; 3.2291666666666665: 0.9604095228902189; 3.3333333333333335: 0.9643260066527476; 3.4375: 0.9678550526731239; 3.5416666666666665: 0.9710349881890298; 3.6458333333333335: 0.9739003489202116; 3.75: 0.9764822541439909; 3.8541666666666665: 0.9788087446664702; 3.9583333333333335: 0.9809050873599727; 4.0625: 0.9827940495741486; 4.166666666666667: 0.9844961464009907; 4.270833333333333: 0.9860298634791853; 4.375: 0.987411857757566; 4.479166666666667: 0.9886571383980641; 4.583333333333333: 0.9897792297828537; 4.6875: 0.9907903183960318; 4.791666666666667: 0.991701385175045; 4.895833333333333: 0.9925223247692634; 5: 0.9932620530009145 • Current / initial current: 0: 1; 0.10416666666666667: 0.9010751057212906; 0.20833333333333334: 0.8119363461506349; 0.3125: 0.7316156289466418; 0.4166666666666667: 0.6592406302004438; 0.5208333333333334: 0.594025320553635; 0.625: 0.5352614285189903; 0.7291666666666666: 0.4823107482912782; 0.8333333333333334: 0.4345982085070782; 0.9375: 0.391605626676799; 1.0416666666666667: 0.3528660814588489; 1.1458333333333333: 0.3179588416559898; 1.25: 0.2865047968601901; 1.3541666666666667: 0.25816234012045264; 1.4583333333333333: 0.2326236579172927; 1.5625: 0.2096113871510978; 1.6666666666666667: 0.18887560283756183; 1.7708333333333333: 0.17019110379502853; 1.875: 0.15335496684492847; 1.9791666666666667: 0.13818434296267892; 2.0833333333333335: 0.12451447144412296; 2.1875: 0.11219689052034373; 2.2916666666666665: 0.10109782498721881; 2.3958333333333335: 0.0910967333385507; 2.5: 0.0820849986238988; 2.6041666666666665: 0.0739647488131616; 2.7083333333333335: 0.06664779385646828; 2.8125: 0.060054667895307945; 2.9166666666666665: 0.05411376622282161; 3.0208333333333335: 0.04876056762020617; 3.125: 0.04393693362340742; 3.2291666666666665: 0.03959047710978117; 3.3333333333333335: 0.035673993347252395; 3.4375: 0.03214494732687607; 3.5416666666666665: 0.028965011810970175; 3.6458333333333335: 0.026099651079788375; 3.75: 0.023517745856009107; 3.8541666666666665: 0.021191255333529852; 3.9583333333333335: 0.01909491264002727; 4.0625: 0.017205950425851383; 4.166666666666667: 0.015503853599009314; 4.270833333333333: 0.013970136520814738; 4.375: 0.012588142242433998; 4.479166666666667: 0.011342861601935855; 4.583333333333333: 0.010220770217146324; 4.6875: 0.00920968160396814; 4.791666666666667: 0.008298614824955013; 4.895833333333333: 0.007477675230736613; 5: 0.006737946999085467

PAUSE & TRY IT

After one charging time constant, what fraction of final charge is present?

Reveal answer

1−e-1≈0.632.

PAUSE & TRY IT

An ideal capacitor is fully charged in a DC series RC circuit. What is I?

Reveal answer

Zero in the long-time limit.

06

Use data to test the RC model

For discharge, ln(V_C/V0)=. A plot of ln V_C against t has slope , provided the voltage stays positive and the model applies. Include meter input resistance if it is not much larger than the discharge resistance, since it creates a parallel discharge path.

Estimate time constants from several readings rather than one noisy point. The capacitor is not fully discharged at one time constant: its remaining voltage is e-1≈0.368 of its initial value. A constant sensor offset becomes especially important late in the decay, when the signal is small.

Capacitor discharge, τ=2 s
Capacitor discharge, τ=2 s02.557.51002.557.510 t (s)V_C (V)Capacitor discharge, τ=…
Read figure values as text

Capacitor discharge, τ=2 s: 0: 10; 0.16666666666666666: 9.200444146293233; 0.3333333333333333: 8.464817248906142; 0.5: 7.788007830714049; 0.6666666666666666: 7.1653131057378925; 0.8333333333333334: 6.592406302004438; 1: 6.065306597126334; 1.1666666666666667: 5.580351457700471; 1.3333333333333333: 5.134171190325921; 1.5: 4.723665527410147; 1.6666666666666667: 4.345982085070782; 1.8333333333333333: 3.9984965434484736; 2: 3.6787944117144233; 2.1666666666666665: 3.3846542510674222; 2.3333333333333335: 3.1140322391459767; 2.5: 2.865047968601901; 2.6666666666666665: 2.6359713811572676; 2.8333333333333335: 2.4252107463564867; 3: 2.231301601484298; 3.1666666666666665: 2.052896575799093; 3.3333333333333335: 1.8887560283756182; 3.5: 1.7377394345044515; 3.6666666666666665: 1.598797460796939; 3.8333333333333335: 1.470964673929768; 4: 1.353352832366127; 4.166666666666667: 1.2451447144412295; 4.333333333333333: 1.1455884399268772; 4.5: 1.0539922456186432; 4.666666666666667: 0.9697196786440505; 4.833333333333333: 0.8921851740926011; 5: 0.820849986238988; 5.166666666666667: 0.7552184450877376; 5.333333333333333: 0.6948345122280154; 5.5: 0.6392786120670757; 5.666666666666667: 0.5881647164242988; 5.833333333333333: 0.5411376622282161; 6: 0.49787068367863946; 6.166666666666667: 0.45806314172621476; 6.333333333333333: 0.4214384350927641; 6.5: 0.3877420783172201; 6.666666666666667: 0.356739933472524; 6.833333333333333: 0.32821658326663206; 7: 0.301973834223185; 7.166666666666667: 0.2778293395412425; 7.333333333333333: 0.25561533206507403; 7.5: 0.23517745856009106; 7.666666666666667: 0.2163737071949309; 7.833333333333333: 0.19907342077733686; 8: 0.1831563888873418; 8.166666666666666: 0.16851201259947513; 8.333333333333334: 0.15503853599009315; 8.5: 0.14264233908999255; 8.666666666666666: 0.13123728736940968; 8.833333333333334: 0.12074413323532873; 9: 0.11108996538242306; 9.166666666666666: 0.10220770217146324; 9.333333333333334: 0.09403562551495206; 9.5: 0.08651695203120634; 9.666666666666666: 0.0795994384870645; 9.833333333333334: 0.07323501878765404; 10: 0.06737946999085467

07

Connect microscopic transport to circuit quantities

Current is the rate of charge flow through a cross section. Conventional current follows positive-charge motion; electron drift in a metal is opposite. Charge carriers need not travel around the entire circuit rapidly for an electrical response to occur. Distinguish drift speed from the propagation of the electromagnetic influence.

Resistance relates voltage and current for an ohmic element under appropriate conditions. R= shows how material resistivity and geometry contribute for a uniform wire. Resistivity can depend on temperature, so a changing-temperature device need not keep a constant resistance.

Current is not consumed by a resistor. Energy is transferred as charges move through potential differences, while charge conservation constrains the circuit. Power can be expressed as IV, I2R or for the suitable resistive model. State whether current or voltage is held fixed before comparing heating after a resistance change.

08

Translate the circuit topology before simplifying

Elements are in series when the same current must pass through them without a branching junction between. They are in parallel when both terminals connect to the same two nodes and therefore share voltage. Physical closeness in a drawing does not determine series or parallel status.

Kirchhoff’s junction rule expresses charge conservation. The loop rule expresses energy conservation around a circuit under the relevant lumped-circuit assumptions. Choose current directions; a negative solution means the actual direction is opposite your choice. A guessed direction is not a reason to alter the algebraic signs midway.

An ideal voltmeter draws negligible current and is connected across an element. An ideal ammeter has negligible resistance and is placed in series with the branch being measured. Real instruments can load a circuit, so use their stated resistance when the problem includes it.

09

Derive the structure of an RC transient

During charging of an initially uncharged capacitor through a resistor, the loop equation is E−IR−=0 with I=dq/dt. Initially q=0, so current is . As capacitor voltage rises, resistor voltage and current fall. At long time, the ideal capacitor carries no steady DC current through its dielectric.

The time constant τ=RC sets the time scale. After one τ, a charging capacitor has reached about 63% of its final charge and the current has fallen to about 37% of its initial value. One time constant is not “fully charged.” Exponential approach is asymptotic in the ideal model.

For discharge through a resistor, charge and voltage decay exponentially while stored field energy becomes thermal energy. Energy decays with the square of voltage, so its exponential time dependence differs from that of charge. Initial capacitor voltage cannot jump instantaneously in the ordinary ideal finite-current model.

PAUSE & TRY IT

At long time in a simple series RC charging circuit, is the resistor voltage equal to the battery voltage?

Reveal answer

No. Current approaches zero, so IR approaches zero. The capacitor voltage approaches the battery voltage.

10

Read switching conditions and graphs

Before solving a switched circuit, determine the capacitor’s voltage just before switching from the prior state. Use continuity of capacitor voltage for the immediate state after the switch. Then identify the new long-time state and resistance governing the transition.

A graph of ln|V−Vfinal| against time is linear for a single ideal exponential transient, with slope . A nonzero offset must be removed before taking the logarithm. Experimental departures can arise from multiple time constants, leakage, source resistance or measurement loading.

Use limiting behavior as a check: immediately after connecting an uncharged capacitor, its voltage is zero; after a long charging interval, current approaches zero in the simple series circuit. A proposed expression that violates either endpoint cannot represent that experiment.

11

Connect microscopic transport with circuit quantities

Current is charge crossing a section per time: I=dQ/dt. Conventional current follows positive-charge motion; electron drift in a metal is opposite. The electric signal establishing a circuit and the slow average drift of individual electrons are different phenomena.

For a uniform conductor with constant resistivity, R=. A longer conductor has greater resistance; a larger cross-sectional area has less. Current density J= relates to field through the material model J=σE for an ohmic conductor. Do not confuse resistivity, a material property under specified conditions, with resistance, which also depends on geometry.

Electrical power can be IV; for an ohmic resistor it can also be I2R or . The choice of fixed quantity matters when comparing resistors. Increasing R at fixed I increases power, while increasing R at fixed V decreases power. State the circuit constraint before predicting heating.

12

Derive an RC transient from a loop equation

For a charging series RC circuit, Kirchhoff’s loop rule gives V source−IR−=0 with I=dQ/dt under the chosen direction. Rearranging gives a first-order differential equation. For an initially uncharged capacitor, Q=CV source(1−e−t/RC) and I=(V source/R)e−t/RC.

At the initial instant, capacitor voltage cannot jump in the ideal finite-current model, so it is zero for an initially uncharged capacitor. Long after connection to a DC source, current approaches zero and capacitor voltage approaches the source voltage. These limiting cases provide checks on the exponential solution.

During discharge through R, Q=Q0e−t/RC, with current sign determined by the chosen reference direction. One time constant reduces the remaining difference from the final state to , about 37%. It does not complete the process. Energy initially stored in the capacitor becomes thermal energy in the ideal resistor model.

PAUSE & TRY IT

What is continuous when a capacitor is switched between ordinary finite-current circuits?

Reveal answer

Its voltage cannot change instantaneously in the ideal model because an instantaneous charge change would require an unbounded current.

FROM IDEA TO APPLICATION

Worked examples

EXAMPLE 1

Find a capacitor from a decay

A capacitor discharges through 200 kΩ. A fit of ln(V_ V) against time has slope −0.25 s-1. Find C.

Reveal worked solution
  1. Slope=, so RC=4 s.
  2. C= F.
Result & interpretation

C=20 μF.

EXAMPLE 2

A tapered resistive element

A wire has constant area A and resistivity ρ(x)=ρ0(1+) for 0≤x≤L. Find R.

Reveal worked solution
  1. Elements along the wire are in series.
  2. R=∫0ᴸρ0(1+)dx/A.
Result & interpretation

R=.

EXAMPLE 3

Read an RC charging curve

A 10 V source charges a 20 μF capacitor through 50 kΩ. Find τ and the initial current.

Reveal worked solution
  1. RC=(50×103)(20×10-6)=1.0 s.
  2. Initial capacitor voltage is zero, so I0==2.0×10-4 A.
  3. After 1.0 s, capacitor voltage is approximately 6.32 V and current is about 0.0736 mA.
Result & interpretation

τ=1.0 s and initial current=0.200 mA.

EXAMPLE 4

Check a charging circuit at one time constant

An initially uncharged 10 μF capacitor charges through 100 kΩ from a 9 V source.

Reveal worked solution
  1. τ=RC=(100,000)(10×10-6)=1 s.
  2. At t=τ, VC=9(1−e-1)≈5.69 V.
  3. I=(,000)e-1≈33.1 μA.
Result & interpretation

After 1 s, voltage is about 63% of its final value, not 100%.

EXAMPLE 5

Distinguish fixed-current and fixed-voltage heating

A resistance doubles. How does power change?

Reveal worked solution
  1. At fixed voltage, P= becomes half as large.
  2. At fixed current, P=I2R becomes twice as large.
Result & interpretation

The constraint determines the answer; “more resistance means more heating” is incomplete.

MAKE THE DISTINCTION

Common mistakes, clearer reasoning

The trapA capacitor always acts like a short circuit.

The better explanationIts initial and long-time behavior differ; an initially charged capacitor also has a nonzero starting voltage.

The trapA negative solved current is an invalid answer.

The better explanationIt indicates flow opposite the assumed direction.

RETRIEVE BEFORE YOU REVEAL

Practice checkpoints

Revisit the quick checks from this guide without looking back. Explain why, then reveal the answer.

1. Does current decrease as it passes through one series resistor?

Reveal answer

No. In steady state the current entering equals the current leaving.

2. At fixed voltage, doubling R changes power how?

Reveal answer

P= halves.

3. After one charging time constant, what fraction of final charge is present?

Reveal answer

1−e-1≈0.632.

4. An ideal capacitor is fully charged in a DC series RC circuit. What is I?

Reveal answer

Zero in the long-time limit.

5. At long time in a simple series RC charging circuit, is the resistor voltage equal to the battery voltage?

Reveal answer

No. Current approaches zero, so IR approaches zero. The capacitor voltage approaches the battery voltage.

6. What is continuous when a capacitor is switched between ordinary finite-current circuits?

Reveal answer

Its voltage cannot change instantaneously in the ideal model because an instantaneous charge change would require an unbounded current.

Key language

Current
Rate of charge transfer, dq/dt.
Resistivity
Material property in R=ρL/A.
Emf
Energy supplied per charge by a source.
Time constant
Characteristic transient time RC.
Junction rule
Algebraic sum of currents at a node is zero.
Loop rule
Algebraic sum of voltage changes around a circuit loop is zero.
Connect it to the course

Connect the equations to the physical assumptions. Explain directions, signs, limiting cases, and units before accepting a numerical result.

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Written for ScienceHub · Original instructional material. Course framework reference ↗. These notes are independently authored and are not College Board materials. External photographs retain their credited licenses.

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