A resistor with 760 is connected to the plates of a charged capacitor with capac

Question

A resistor with 760 is connected to the plates of a charged capacitor with capacitance 5.10 F . Just before the connection is made, the charge on the capacitor is 7.90 mC .

Part A

What is the energy initially stored in the capacitor?

Part B

At what rate is electrical energy being dissipated in the resistor just after the connection is made?

Part C

At what rate is electrical energy being dissipated in the resistor at the instant in time when the remaining capacitor charge is half of the initial charge on the capacitor?

Part D

At what time does the situation described in Part C occur?

Give your answer as a percentage of RC, to three significant figures

Part E

At what time does the situation described in Part C occur?

Give your answer in milliseconds.

Part F

At what rate is electrical energy being dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in Part A?

Part G

At what time does the situation described in Part F occur?

Give your answer as a percentage of RC, to three significant figures.

Explanation / Answer

a)

Energy stored in capacitor, U = 0.5*Q^2/C

So, U = 0.5*(7.9*10^-3)^2/(5.1*10^-6) = 6.12 J <------answer

b)

Just after connection is made, the voltage across capacitor and resistor, V = Q/C

So, V = 7.9*10-3/(5.1*10^-6) = 1549 V

So, current across resistor, I = V/R = 1549/760 = 2.04 A <-------answer

c)

Charge on the capcitor, Q = Qo*e^(-t/RC)

when Q = Qo/2

So, 1/2 = e^(-t/RC)

So, t = -ln(0.5)*RC = -ln(0.5)*760*5.1*10^-6 = 2.69*10^-3 s

So, Current, I = Io(e^(-t/RC))

where Io = 2.04 A

So, I = 2.04*(1/2) = 1.02 A <-------answer

d)

t = -ln(0.5)*RC = 0.693*RC = 69.3 percent of RC<-------answer

e)

t = 2.69*10^-3 s <-------answer

f)

Energy stored, U = 0.5*Q^2/C = 0.5*( Qo*e^(-t/RC))^2/C

= (0.5*Qo^2/C)*e^(-2t/RC) = Uo*e^(-2t/RC)

So, U = Uo/2

then , 1/2 = e^(-2t/RC)

So, t = 1.34 ms

So, at that time, current , I = 1.02*e^(-1.34*10^-3/(760*5.1*10^-6)) = 0.722 A

So, rate of dissipation of energy, P = I^2*R = 0.722^2*760 = 396.2 W <--------answer

g)

t = (ln(0.5)/2)*RC = 0.347*RC = 34.7 percent of RC<--------answer

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