A 2.31-microFarad capacitor that is initially uncharged is connected in series with a 5.58−kΩ resistor and an emf source with ε=72.3 V and negligible internal resistance. The circuit is completed at t = 0. Find (a) Just after the circuit 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 being stored in the capacitor?

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Given :-

C=2.31 \: \mu F

R = 5.58 \, K\Omega

and emf(\varepsilon )=72.3\: V


(1) Initially capacitor is uncharged means there is no charge on capacitor i.e

q(t=0)=0.

we know the relation i.e q = CV 

since initially charge on capacitor is zero therefore potential across the capacitor is also zero hence it acts like a wire with no resistance.

it means we are not going to consider capacitor in the circuit at initial stage so there is only emf source and resister in the circuit. the current through this resistor is,

I=\frac{\varepsilon }{R}

I=\frac{72.3}{5.58\times 10^{3}}

I=0.0129\: A

the rate at which electrical energy is being dissipated in the resistor is therefore,

P_{R}=I^{2}R                                         .........................1

P_{R}=I^{2}R=(0.0129\, A)^{2}(5.58\times 10^{3})

P_{R}=0.928 \: W

(2) The energy stored in the capacitor is given by 

U_{c}=\frac{q^{2}}{2C}

the rate at which electrical energy is being stored in the capacitor is therefore,

P_{C}=\frac{\mathrm{d}U_{C}}{\mathrm{d} t}=\frac{q}{C}\frac{\mathrm{d} q}{\mathrm{d} t}

P_{C}=\frac{qI}{C}...............\because \frac{\mathrm{d} q}{\mathrm{d} t}=I.....................2

We have given rate of electrical energy being dissipated in the resistor equal to the rate at which electrical energy being stored in the capacitor

therefore from equation 1 and 2

P_{R}=P_{C}

I^{2}R=\frac{qI}{C}

\therefore I=\frac{q}{RC}                                         .....................3

Charge over the capacitor is given by 

q=Q(1-e^{\frac{-t}{RC}})

q=C\varepsilon (1-e^{\frac{-t}{RC}})\: \: \: \: \because Q=C\varepsilon ..............4

We Know

I=\frac{\mathrm{d} q}{\mathrm{d} t}=\frac{\varepsilon }{R}e^{\frac{-t}{RC}}                           ....................5

putting the value of equation 4 and 5 in the equation 3 we get

\frac{C\varepsilon (1-e^{\frac{-t}{RC}})}{RC}=\frac{\varepsilon }{R}e^{\frac{-t}{RC}}

\therefore 1-e^{\frac{-t}{RC}}=e^{\frac{-t}{RC}}

\therefore e^{\frac{t}{RC}}=2

\therefore t=RCln (2)

Putting the given values 

t=5.58\times 10^{3}\times 2.31\times 10^{-6}ln (2)

t=8.93 \times 10^{-3} \: s

Therefore Part A answer is P_{R}=0.928 \: W and

Part B answer is t=8.93 \times 10^{-3} \: s


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