You connect a battery, resistor, and capacitor as in (Figure 1), where epsilon =

Question

You connect a battery, resistor, and capacitor as in (Figure 1), where epsilon = 56.0 V, C = 5.00 mu F, and R = 140 Ohm The switch S is closed at t = 0. When the voltage across the capacitor is 8.00 V, what is the magnitude of the current in the circuit? Express your answer with the appropriate units. I = 0.343 At what time t after the switch is closed is the voltage across the capacitor 8.00 V? Express your answer with the appropriate units. t=108 mu s When the voltage across the capacitor is 8.00 V, at what rate is energy being stored in the capacitor? Express your answer with the appropriate units P = 76.8 W

Explanation / Answer

C)

charge on the capacitor, Q = Qo*( 1 - e^(-t/RC) )

Qo = 5*10^-6*56 = 2.8*10^-4 C

So, Q = 2.8*10^-4*(1 - e^(-t/(140*5*10^-6)))

= 2.8*10^-4*(1 - e^(-t*1428.6))

So, Energy stored in capacitor , U = 0.5*Q^2/C

= 0.5*(2.8*10^-4*(1 - e^(-t*1428.6)) )^2/(5*10^-6)

= (0.5*(2.8*10^-4)^2/(5*10^-6))*(1 - e^(-t*1428.6) )^2

= 7.84*10^-3*( 1 - e^(-t*1428.6) )^2

So, rate of energy being stored = dU/dt = d/dt ( 7.84*10^-3*( 1 - e^(-t*1428.6) )^2)

= 7.84*10^-3*2*(1 - e^(-t*1428.6))*(e^(-t*1428.6))*(1428.6)

= 22.4*(1 - e^(-t*1428.6))*(e^(-t*1428.6))

Voltage across the capcitor, Vc = 56*( 1 - e^(-t/RC))

So, when Vc = 8V,

8 = 56*(1 - e^(-t/RC))

So, e^(-t/RC) = 0.857

So, e^(-t/(140*5*10^-6)) = 0.857

So, t = 1.08*10^-4 s

So, at this time, rate of energy being stored in capacitor = 22.4*(1 - e^(-(1.08*10^-4)*1428.6))*(e^(-(1.08*10-4)*1428.6))

= 2.73 W

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