The circuit shown in the figure below contains a 12.0-V battery, a 2.00 mF capac

Question

The circuit shown in the figure below contains a 12.0-V battery, a 2.00 mF capacitor and a resistor. The capacitor is initially without charge. a) If the switch is thrown so as to include the battery in the circuit, the capacitor will begin to charge. It takes 3.60 s for the capacitor to reach 50.0% of maximum charge. With this information, find the resistance in the resistor. b) What is the time constant for this circuit? c) What is the maximum charge the capacitor can have? d) After reaching maximum charge, the switch is moved back to its original position (so that the battery is not included in the circuit). What is the current through the resistor after one time constant has passed?

Explanation / Answer

a) V = VO[ 1 - e^(-t/RC)]

0.50Vo = Vo [ 1 - e^(-3.60 /RC)]

-3.60 / RC = ln(0.50)

RC = 5.19

R = 5.19 / 2 x 10^-3 = 2596.85 Ohm

b) Time Constant = RC = 5.19 s

C) Q = CV

Qmax = 2 x 10^-3 x 12 = 0.024 C

D) V = V0(e^(-t/RC))

V = 12(e^-1) = 4.41 V

I = V/R = 4.41 / 2596.85 = 1.70 x 10^-3 A

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