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Physics LibreTexts

14.12: Another Example

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Figure 14.12.1.png
FIGURE XIV.1

The circuit in Figure XIV.1 contains two equal resistances, two equal capacitances, and a battery. The battery is connected at time t=0. Find the charges held by the capacitors after time t.

Apply Kirchhoff’s second rule to each half:

(˙Q1+˙Q2)RC+Q2=CV,

and ˙Q1RC+Q1Q2=0.

Eliminate Q2:

R2C2¨Q1+3RCQ1+Q1=CV.

Transform, with Q1 and ˙Q1 initially zero:

(R2C2s2+3RCs+1)¯Q1=CVs.

I.e. R2C¯Q1=1s(s2+3as+a2)V,

where a=1/(RC).

That is R2C¯Q1=1s(s+2.618a)(s+0.382a)V.

Partial fractions: R2C¯Q1=[1s+0.1708s+2.618a1.1708s+0.382a]Va2.

That is, ¯Q1=[1s+0.1708s+2.618a1.1708s+0.382a]CV.

Inverse transform: Q1=[1+0.1708e2.618t/(RC)1.1708e0.382t/(RC)].

The current can be found by differentiation.

I leave it to the reader to eliminate Q1 from equations 14.12.1 and 2 and hence to show that

Q2=[10.2764e2.618t/(RC)0.7236e0.382t/(RC)].


This page titled 14.12: Another Example is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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