a 2.00*10^-6 F capacitor that is initially uncharged is connected in series with

Question

a 2.00*10^-6 F capacitor that is initially uncharged is connected in series with a 6000 ohm resistor and an emf of 90.0 volts and negligible internal resistance. The circuit is completed at t = 0.(a) Just after the curcuit is completed, what is the rate at which electrical energy is being dissipated in the resistor? (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 bieng stored in the capacitor? (c) At the time calculated in part (b), what is the rate at which electrical energy is being dissipated in the resistor ?

Explanation / Answer

a)

P = V^2 / R

= 90^2 / 6000

= 1.35 W

===========================

b)

Voltage in capacitor = 90*[1–e^(–t/(2*10^-6*6000))]
= 90*[1–e^(–t/(0.012))]
Voltage in resistor + Voltage in capacitor = 90 v

Voltage in resistor = 90 -  Voltage in capacitor

= 90*e^(–t/(0.012))

now equating both Vc and Vr we solve t

90*[1–e^(–t/(0.012))] = 90*e^(–t/(0.012))
1–e^(–t/(0.012)) = e^(–t/(0.012))
2e^(–t/(0.012)) = 1
e^(–t/(0.012)) = 1/2
–t/(0.012) = –0.693
t = 0.00832 s

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c)

V = Vo*[1–e^(–t/)]

= 90*e^(–t/(0.012))

= 90*e^(– 0.00832/(0.012))
= 90(0.5)

= 45 V

P = V^2 / R

= 45^2 / 6000

= 0.338 W

= 0.3375 watts

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