14.12: Another Example
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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+Q1−Q2=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+2.618a−1.1708s+0.382a]Va2.
That is, ¯Q1=[1s+0.1708s+2.618a−1.1708s+0.382a]CV.
Inverse transform: Q1=[1+0.1708e−2.618t/(RC)−1.1708e−0.382t/(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=[1−0.2764e−2.618t/(RC)−0.7236e−0.382t/(RC)].