A capacitor of capacitance C = 3 mu F has been charged so that the potential dif

Question

A capacitor of capacitance C = 3 mu F has been charged so that the potential difference between its plates is V_0 = 95V. The capacitor is then connected to a resistor of resistance R = 9.5K Ohm. The switch S is closed, and the capacitor begins to discharge. Randomized Variables C = 3 mu F V_0 = 95 V R = 9.5 k Ohm Calculate the time constant tou of the circuit seconds. Calculate the charge Q_0 on the capacitor before the switch is closed, in coulombs. Calculate the potential difference V_C in volts between the capacitor plates at time t = 5.0 ms after the switch is closed. Calculate the time T in s after which the charge on the capacitor has decreased to one fourth its maximum value.

Explanation / Answer

here,
Capacitance, C = 3 UF = 3*10^-6 F
Potential difference, Vo = 95 V
Resistance, R = 9.5 kOhms = 9500 Ohms

Part a:
Time consant, tao = RC = 9500 * 3*10^-6
tao = 0.029 s

Part B:
Charge , Qo = C*Vo
Qo = 3*10^-6 * 95
Qo = 2.85*10^-4 C

Part C:
From ohms law, Io = Vo/R
Io = 95/9500
Io = 0.01 A

Part D:
time = 5 ms = 0.005 s

Vc = Vs(e^(-t/tao) )
Vc = 95(e^(-0.005/0.029) )
Vc = 79.955 V

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