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8. Consider the circuit from question 3. Relabel the battery as Va and then change the capacitor on the right across the capaR R

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Answer #1

By KVL across second loop,

q/C - iR - Vb = 0

q/C - R*dq/dt - Vb = 0

dq/dt = q/RC - Vb/R

dq/[q/RC - Vb/R] = dt

after integrating

RC* ln(q/RC - Vb/R) = t

After applying the limit that q at t=0 will be CVa and at some time t it will be CV,

CV/RC - Vb/R)/(CVa/RC - Vb/R) = e^(-t/RC)

(V - Vb)/(Va - Vb) = e^(-t/RC)

V = Vb + (Va - Vb)e^(-t/RC)

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