3. In the circuit diagram below, Epsilon is the external voltage applied to the

Question

3. In the circuit diagram below, Epsilon is the external voltage applied to the circuit, C is the capacitance, R is the resistance. and Vc(t) is the time dependent voltage across the capacitor. It is known that the voltage across the capacitor is a solution to the differential equation dVv/dt = -1/CR(Vc - Epsilon) (a) Suppose R = 1 ohm, C = 1/2 F and Epsilon = 2 V. At t = 0, the voltage across the capacitor is zero. Set up the initial value problem. (b) Solve the initial value problem. Hint: use the separable equation method.

Explanation / Answer

dV/dt =(-1/CR)(V-E)

dV/dt =(-1/(1/2)*1)(V-2)

dV/dt =(-2)(V-2)

dV/(V-2) =-2dt

integrate on both sides

integral dV/(V-2) =integral-2dt

ln(V-2)=-2t+c

(V-2)=e-2t+c

V=2+e-2t+c

V=2+Ce-2t

V(0) =0

0=2+Ce0

C=-2

V=2-2e-2t

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