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

Question

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 = 140.0 ? the capacitor has capacitance of C = 31.0 ?F and the battery maintains the emf of ? = 34.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.

a. What is the time constant for this circuit?

b. What is the charge on the capacitor after the switch has been closed for t = 4.30×10-3 s?

c. What is the current through the circuit after the switch has been closed for t = 4.30×10-3 s?

d. What is the voltage across the capacitor after the switch has been closed for t = 4.30×10-3 s?

Explanation / Answer

a)

time constant t = R*C

t = (140)*(31*10^(-6))

t = 0.00434 seconds

b)

charge on the capacitor = C * V0 * [1 - e^(-t/(RC))]

charge on the capacitor = (31*10^(-6))*34*[1 - e^(-(4.3*10^(-3))/0.00434)]

charge on the capacitor = (31*10^(-6))*34*[1 - 0.37128]

charge on the capacitor = 66.26*10^(-5)

c)

Maximum current = V0 / R

Maximum current = 34 / 140

Maximum current = 0.2428 amps

current through circuit = Maximum current * e^(-t/(RC))

current through circuit = 0.2428 * e^(-(4.3*10^(-3))/0.00434)

current through circuit = 0.2428 * 0.37128

current through circuit = 0.09014 amps

d)

voltage across capacitor = V0*[1 - e^(-t/(RC))]

voltage across capacitor = 34*[1 - e^(-(4.3*10^(-3))/0.00434)]

voltage across capacitor = 34*[1 - 0.37128]

voltage across capacitor = 21.37 volt

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