A 2.58 ? F capacitor that is initially uncharged is connected in series with a 5

Question

A 2.58?F capacitor that is initially uncharged is connected in series with a 5.56k? resistor and an emf source with 82.8V and negligible internal resistance. The circuit is completed at t = 0.

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

Answer is in the form P= something

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?

Answer in the form t=something

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

Answer in the form P=something

Please if you could show your steps it would be greatly appreciated.

Explanation / Answer

a)

Just after switch is closed ,capacitor acts as short circuit ,so power dissipated in resistor is

P=V2/R ==82.82/5560

P=1.233 watts

b)

The Energy stored in capacitor is

E=(1/2)(Q2/C)

The ELectrical energy being dissipated in the capacitor is

Pc=dE/dt =(Q/C)(dQ/dt)

Pc=Q*I/C

Given

PR=PC

I2R=(Q*I/C)

=>I=Q/RC =Q/T

Since Time Constant

T=RC=2.58*10-6*5.56*103

T=0.01434 seconds =14.34 ms

Maximum charge

Qo=CV =2.58*82.8 =213.6 uC

maximum current

Io=Vo/R =82.8/5560=0.0149 A

charge on the capacitor is

Q=Qo[1-e-t/T]

Q=(213.6u)[1-e-t/14.34m]

current in the circuit is

I=Ioe-t/T

Q/T =(0.0149)*e-t/14.34m

(213.6*10-6)[1-e-t/14.34m]/14.34*10-3 =(0.0149)*e-t/14.34m

1-e-t/14.34m=e-t/14.34m

e-t/14.34=(1/2)

-t/14.34m =ln(0.5)

t=9.94 ms or 0.00994 s

c)

Current at t=9.94 ms

I=Ioe-t/T =0.0149*e-9.94/14.34

I=7.45*10-3 A

The energy dissipated in resistor is

P=I2R=(7.45*10-3)2*5560

P=0.3086 Watts

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