Problem 26.71 A 2.17 ? F capacitor that is initially uncharged is connected in s

Question

Problem 26.71

A 2.17 ?F capacitor that is initially uncharged is connected in series with a 6.88 k? resistor and an emf source with 88.5 V and negligible internal resistance. The circuit is completed at t = 0.

Part A:

Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor?

P=?

Part B:

At what value of t is the rate at which electrical energy is being dissipated in the resistor equal to the rate at which electrical energy is being stored in the capacitor?

t=?

Part C:

At the time calculated in part B, what is the rate at which electrical energy is being dissipated in the resistor?

P=?

Thank you.

Explanation / Answer

Given,

Capacitance, C = 2.17 uF

Resistance, R = 6.88 kohm

Voltage, V = 88.5 V

A) P = V^2/R = 88.5^2/6880 = 1.138 J/s

B) Time constant, T = RC = 6880 x 2.17 x 10^-6 = 0.015 sec

Hence, time at which rate of dissipation of energy in the resistor,

t = T ln(2) = 0.015 x ln2 = 0.010 sec

C) P = (V^2/R) e^-(2t/T)

P = (88.5^2/6880) e^-(2 x 0.010/0.015) = 0.286 J/s

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