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Physics / Chapter 27: Circuits / Section 27-8

RC Circuits — Question 8

Section 27-8 · Physics pages 4466–4467

Question 8 Leakage changes both the final voltage and the observed charging time constant A capacitor C has leakage resistance RL in parallel. It charges from E=12.0 V through R=1.00 megohm and approaches 8.00 V. After disconnecting the source branch completely, its voltage decays with time constant 4.00 s. + − E R RL C Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Infer RL and C from these two observations. (b) Find the charging time constant and the initial charging current when the capacitor starts uncharged.
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Question 8 - Solution Leakage changes both the final voltage and the observed charging time constant + − E R RL C (a) Method and calculation 8 = 12 RL R + RL ⇒RL = 2R = 2.00 MΩ. τoff= RLC = 4.00 s ⇒C = 2.00 µF. (b) Interpretation and further calculation τon = (R ∥RL)C = 2 3 × 106(2 × 10−6) = 4/3 s. iC(0+) = E/R = 12.0 µA, iL(0+) = 0. Physical check A lower steady voltage and a faster charging decay can both arise from the same leakage path.

Question 8

Leakage changes both the final voltage and the observed charging time constant

A capacitor C has leakage resistance RL in parallel. It charges from E=12.0 V through R=1.00 megohm and approaches 8.00 V. After disconnecting the source branch completely, its voltage decays with time constant 4.00 s.

Diagram shown in the original SVG worksheet.

  1. Infer RL and C from these two observations.

  2. Find the charging time constant and the initial charging current when the capacitor starts uncharged.

Question 8 - Solution

Leakage changes both the final voltage and the observed charging time constant

Diagram shown in the original SVG worksheet.

(a) Method and calculation

8=12RLR+RL⇒RL=2R=2.00MΩ.8=12\frac{R_L}{R+R_L}\Rightarrow R_L=2R=2.00\,\mathrm{M\Omega}.τoff=RLC=4.00s⇒C=2.00μF.\tau_{\mathrm{off}}=R_LC=4.00\,\mathrm s\Rightarrow C=2.00\,\mathrm{\mu F}.

(b) Interpretation and further calculation

τon=(R∥RL)C=23×106(2×10−6)=4/3s.\tau_{\mathrm{on}}=(R\parallel R_L)C=\frac{2}{3}\times10^6(2\times10^{-6})=4/3\,\mathrm s.iC(0+)=E/R=12.0μA,iL(0+)=0.i_C(0^+)=E/R=12.0\,\mathrm{\mu A},\quad i_L(0^+)=0.

Physical check

A lower steady voltage and a faster charging decay can both arise from the same leakage path.

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