A 2.27 F capacitor that is initially uncharged is connected in series with a 7.1

Question

A 2.27 F capacitor that is initially uncharged is connected in series with a 7.19 k resistor and an emf source with 52.1 V and negligible internal resistance. The circuit is completed at t = 0. Just after the circuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? 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? At the time calculated in part B, what is the rate at which electrical energy is being dissipated in the resistor?

Explanation / Answer

(a)

Firstly it acts as a short circuit,

so the current(i) is 52.1V/7.19k = 7.25 mA,

power is 52.1V x 7.25mA = 377.73 mW or 0.378 watts

(b)

current into a cap, charging
v = v[1–e (–t/)] ;
v = battery voltage; i = current after time t ;R =resistance ;C = capacitance ;t = time
RC = = time constant

Vc = 52.1[1–e(–t/(2.2µ•7.19m)]
Vc = 52.1[1–e(–t/(0.016)]
Vr = 52.1 – Vc

= 52.1 – 52.1[1–e(–t/(0.016)]
Vr = 52.1[e(–t/(0.016) ]
since the currents in the C and the R are the same, the power (and therefore energy per unit time) is proportional to voltage. So we can set the voltages equal and solve for t

52.1[1–e(–t/(0.016)] = 52.1[e(–t/(0.016)]
1–e (–t/(0.016)) = e(–t/(0.016))
2e(–t/(0.016)) = 1
e(–t/(0.016)) = 1/2
ln both sides
–t/(0.016) = –0.693
t = 0.0111 seconds
(C)

Vr = 52.1[e(–t/(0.016)) ]
Vr = 52.1[e(– 0.0111/(0.016)) ]
Vr = 52.1x (0.366) = 19.069 volts
P = E²/R = (52.1)2/ (7.19x103)

P=0.3775 watts

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