RC Circuits — Question 1

Section 27-8 · Physics pages 4452–4453

Question 1 A precharged capacitor approaches a new source voltage from its actual initial value At t=0, a capacitor C=100 microfarad initially at vC=3.00 V is connected through R=2.00 kilohm to an ideal 12.0 V source. Voltage vC is positive on the upper capacitor plate; current i enters that plate. + − 12 V 2 kΩ 100 µF + − Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Derive vC(t) and i(t), including their initial values. (b) Find the first time vC reaches 9.00 V and the current then.
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Question 1 - Solution A precharged capacitor approaches a new source voltage from its actual initial value + − 12 V 2 kΩ 100 µF + − (a) Method and calculation RC dvC dt + vC = E, τ = RC = .200 s. vC = 12 −9e−t/τ V, i = C ˙vC = 4.50e−t/τ mA. (b) Interpretation and further calculation vC(0) = 3.00 V, i(0) = 4.50 mA, e−t/τ = (12 −9)/9 = 1/3. t = τ ln 3 = .21972 s, i = (12 −9)/2000 = 1.50 mA. Physical check Capacitor voltage is continuous through switching with finite resistance; its derivative and current can change abruptly.

Question 1

A precharged capacitor approaches a new source voltage from its actual initial value

At t=0, a capacitor C=100 microfarad initially at vC=3.00 V is connected through R=2.00 kilohm to an ideal 12.0 V source. Voltage vC is positive on the upper capacitor plate; current i enters that plate.

Diagram shown in the original SVG worksheet.

  1. Derive vC(t) and i(t), including their initial values.

  2. Find the first time vC reaches 9.00 V and the current then.

Question 1 - Solution

A precharged capacitor approaches a new source voltage from its actual initial value

Diagram shown in the original SVG worksheet.

(a) Method and calculation

RCdvCdt+vC=E,τ=RC=.200s.RC\frac{dv_C}{dt}+v_C=E,\quad\tau=RC=.200\,\mathrm s.vC=12−9e−t/τV,i=Cv̇C=4.50e−t/τmA.v_C=12-9e^{-t/\tau}\ \mathrm V,\quad i=C\dot v_C=4.50e^{-t/\tau}\ \mathrm{mA}.

(b) Interpretation and further calculation

vC(0)=3.00V,i(0)=4.50mA,e−t/τ=(12−9)/9=1/3.v_C(0)=3.00\,\mathrm V,\quad i(0)=4.50\,\mathrm{mA},\quad e^{-t/\tau}=(12-9)/9=1/3.t=τln⁡3=.21972s,i=(12−9)/2000=1.50mA.t=\tau\ln3=.21972\,\mathrm s,\quad i=(12-9)/2000=1.50\,\mathrm{mA}.

Physical check

Capacitor voltage is continuous through switching with finite resistance; its derivative and current can change abruptly.