Question

In: Physics

An RC circuit, hooked up to a battery as shown in the figure,starts with an...

An RC circuit, hooked up to a battery as shown in the figure, starts with an uncharged capacitor. The resistance in the circuit is R = 784.0 Ω the capacitor has capacitance of C = 68.0 μF and the battery maintains the emf of ε = 32.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.

What is the time constant for this circuit?


Tries 0/5

What is the charge on the capacitor after the switch has been closed for t = 2.56×10-2 s?


Tries 0/5

What is the current through the circuit after the switch has been closed for t = 2.56×10-2 s?


Tries 0/5

What is the voltage across the capacitor after the switch has been closed for t = 2.56×10-2 s?

Solutions

Expert Solution

A)

Time constant, T = RC = 784 x 68 x 10^-6 = 0.05 sec

B)

V = Vo(1 - e^-t/T)

V = 32(1 - e^-(0.0256/0.053)) = 12.20 V

Charge, Q = 12.2 x 68 x 10^-6 = 8.30 x 10^-4 C

C)

Current, I = (Vo - V) /R = 0.025 A

D)

V = 32(1 - e^-(0.0256/0.053)) = 12.20 V

Comment in case any doubt please rate my answer ....


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