Stem-Master
Physics / Chapter 27: Circuits / Section 27-8

RC Circuits — Question 6

Section 27-8 · Physics pages 4462–4463

Question 6 Reversing the source makes a charged capacitor cross zero after a finite delay A capacitor initially has upper-plate voltage +8.00 V relative to its lower plate. At t=0 it is connected through R to a source whose maintained upper terminal potential is -4.00 V. Define i positive into the upper capacitor plate and tau=RC. + − E = −4 V R C + − Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find vC(t), i(t) and the time when the capacitor voltage first becomes zero. (b) Find the current at that zero crossing and explain why zero stored charge does not imply zero current.
Show solutionHide solution
Question 6 - Solution Reversing the source makes a charged capacitor cross zero after a finite delay + − E = −4 V R C + − (a) Method and calculation vC(t) = Vf + (Vi −Vf)e−t/τ = −4 + 12e−t/τ V. i(t) = −12 V R e−t/τ, t0 = τ ln 3. (b) Interpretation and further calculation vC(t0) = 0 ⇒q(t0) = 0, i(t0) = −4 V −0 R = −4 V R . ˙q(t0) = i(t0)̸ = 0. Charge passes through zero while the source continues to drive the opposite polarity. Physical check The marked source polarity defines a signed emf; a negative value reverses its actual terminal polarity.

Question 6

Reversing the source makes a charged capacitor cross zero after a finite delay

A capacitor initially has upper-plate voltage +8.00 V relative to its lower plate. At t=0 it is connected through R to a source whose maintained upper terminal potential is -4.00 V. Define i positive into the upper capacitor plate and tau=RC.

Diagram shown in the original SVG worksheet.

  1. Find vC(t), i(t) and the time when the capacitor voltage first becomes zero.

  2. Find the current at that zero crossing and explain why zero stored charge does not imply zero current.

Question 6 - Solution

Reversing the source makes a charged capacitor cross zero after a finite delay

Diagram shown in the original SVG worksheet.

(a) Method and calculation

vC(t)=Vf+(Vi−Vf)e−t/τ=−4+12e−t/τV.v_C(t)=V_f+(V_i-V_f)e^{-t/\tau}=-4+12e^{-t/\tau}\ \mathrm V.i(t)=−12VRe−t/τ,t0=τln⁡3.i(t)=-\frac{12\,\mathrm V}{R}e^{-t/\tau},\quad t_0=\tau\ln3.

(b) Interpretation and further calculation

vC(t0)=0⇒q(t0)=0,i(t0)=−4V−0R=−4VR.v_C(t_0)=0\Rightarrow q(t_0)=0,\quad i(t_0)=\frac{-4\,\mathrm V-0}{R}=-\frac{4\,\mathrm V}{R}.q̇(t0)=i(t0)≠0.\dot q(t_0)=i(t_0)\ne0.Charge passes through zero while the source continues to drive the opposite polarity.

Physical check

The marked source polarity defines a signed emf; a negative value reverses its actual terminal polarity.

← Section questionsChapter contents