Skip to main content
GIOPHYSICS
Menu

AP · AP C:E&M · free response

AP Physics C: Electricity and Magnetism · Question 8

AP Physics C: Electricity and Magnetism · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
15
Topics
2
Answer
Complete
A resistor and capacitor in series with a battery and a switchswitch, closed at t = 0RCε
Fig. 8.1A single-loop circuit drawn as a rectangle. A cell of e.m.f. epsilon sits in the left-hand side of the loop, with its long positive plate uppermost. Along the top wire, reading from the left, come an open switch labelled S with the note that it is closed at t = 0, and then a rectangular resistor labelled R. A capacitor labelled C, drawn as two equal parallel plates, sits in the right-hand side of the loop. Everything is in series round the one loop: there are no branches and no junction dots anywhere in the circuit.
free response15 marks

A capacitor of capacitance C, initially uncharged, is connected in series with a resistor of resistance R and a battery of e.m.f. ε. The switch is closed at t = 0.

  1. (a)

    Derive Write the loop equation for the circuit and derive an expression for the charge on the capacitor as a function of time.

    4 marks
  2. (b)

    Sketch On separate axes, sketch the charge on the capacitor and the current in the circuit as functions of time. Label the initial value and the asymptotic value on each.

    4 marks
  3. (c)

    Determine Determine the time, as a multiple of RC, at which the capacitor holds half of its final charge.

    3 marks
  4. (d)

    Explain The total energy delivered by the battery over the whole charging process is Cε², but only ½Cε² is stored in the capacitor. Explain where the remaining energy goes, and explain why this fraction does not depend on the value of R.

    4 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: 1 point: ε − IR − q/C = 0 with I = dq/dt 1 point: separates variables, dq/(εC − q) = dt/(RC) 1 point: integrates with q = 0 at t = 0
01

(a)

4 marks

Derive Write the loop equation for the circuit and derive an expression for the charge on the capacitor as a function of time.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: ε − IR − q/C = 0 with I = dq/dt

  2. 2

    1 point: separates variables, dq/(εC − q) = dt/(RC)

  3. 3

    1 point: integrates with q = 0 at t = 0

  4. 4

    1 point: q(t) = εC(1 − e^(−t/RC))

02

(b)

4 marks

Sketch On separate axes, sketch the charge on the capacitor and the current in the circuit as functions of time. Label the initial value and the asymptotic value on each.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: charge rises from zero, concave down, approaching the asymptote q = εC

  2. 2

    1 point: εC labelled on the charge graph

  3. 3

    1 point: current starts at ε/R and decays, concave up, approaching zero

  4. 4

    1 point: ε/R labelled on the current graph

03

(c)

3 marks

Determine Determine the time, as a multiple of RC, at which the capacitor holds half of its final charge.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: sets εC(1 − e^(−t/RC)) = ½εC

  2. 2

    1 point: e^(−t/RC) = ½, so t/RC = ln 2

  3. 3

    1 point: t = 0.69RC

04

(d)

4 marks

Explain The total energy delivered by the battery over the whole charging process is Cε², but only ½Cε² is stored in the capacitor. Explain where the remaining energy goes, and explain why this fraction does not depend on the value of R.

How to approach it

State the outcome first, then link cause to effect with the relevant physical principle. Each link in the reasoning should be explicit enough to earn its own marking point.

  1. 1

    1 point: the difference, ½Cε², is dissipated as thermal energy in the resistor

  2. 2

    1 point: the battery moves the full charge εC through the full potential difference ε, while the capacitor's stored energy is the integral of q/C dq, which averages to half that

  3. 3

    1 point: a larger R makes the current smaller but the process correspondingly longer

  4. 4

    1 point: the two effects cancel exactly in ∫I²R dt, so the dissipated energy is ½Cε² whatever the resistance

Private on this device

Did your answer earn the marks?

Compare the reasoning point by point, then choose what happens next.