An RC circuit, hooked up to a battery as shown in the figure, starts with an unc
ID: 1422284 • Letter: A
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 = 307.0 ? the capacitor has capacitance of C = 96.0 ?F and the battery maintains the emf of ? = 17.0 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 = 9.43×10-3 s?
C) What is the current through the circuit after the switch has been closed for t = 9.43×10-3 s?
D) What is the voltage across the capacitor after the switch has been closed for t = 9.43×10-3 s?
Explanation / Answer
a) What is the time constant for this circuit?
t = RC
t = (307)(96 * 10^(-6))
t = 0.029472 seconds
b) What is the charge on the capacitor after the switch has been closed for t = 9.43 × 10^-3 seconds?
charge on the capacitor = C * E * [1 - e^(-t/(RC))]
charge on the capacitor = (96 * 10^(-6)) * 17 * [1 - e^(-(9.43 * 10^(-3))/0.029472)]
charge on the capacitor = 1632 * 10^(-6) * [1 - 0.726]
charge on the capacitor = 1632 * 10^(-6) * [0.274]
charge on the capacitor = 447.168 * 10^(-6)
c) What is the current through the circuit after the switch has been closed for t = 9.43 × 10^-3 seconds?
Maximum current = E / R
Maximum current = 17 / 307
Maximum current = 0.055374 amps
current through circuit = Maximum current * e^(-t/(RC))
current through circuit = 0.05537 * e^(-(9.43 * 10^(-3))/0.029472)
current through circuit = 0.05537 * 0.726
current through circuit = 0.0401 amps
current through circuit = 40.1 milliamps
d) What is the voltage across the capacitor after the switch has been closed for t =9.43 × 10^-3 seconds ?
voltage across capacitor = E * [1 - e^(-t/(RC))]
voltage across capacitor = 17 * [1 - e^(-(9.43 * 10^(-3))/0.029472)]
voltage across capacitor = 17 * [1 - 0.726]
voltage across capacitor = 17 * [0.274]
voltage across capacitor = 4.658 v
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