Through resistanceless leads, a capacitor of capacitance 12.0 is connected to a battery of emf 6.00 V and internal resistance 1.00 . Determine the power dissipated as heat 12.0 after the connections are made.

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Quantrex Academy · Accepted Answer

Correct answer: C. 4.87 W

The time constant of the RC circuit is given by

\tau = RC

.

Substituting the given values,

\tau = (1.00 \Omega) \times (12.0 \mu\text{F}) = 12.0 \mu\text{s}

.

The current in the circuit during charging is given by

I(t) = \dfrac{V}{R} e^{-t/\tau}

.

At

t = 12.0 \mu\text{s}

, we have

t = \tau

.

I = \dfrac{6.00}{1.00} e^{-1} = \dfrac{6.00}{e} \text{ A}

.

The power dissipated as heat in the internal resistance is

P = I^2 R

.

P = \left( \dfrac{6.00}{e} \right)^2 \times 1.00 = \dfrac{36.0}{e^2} \text{ W}

.

Using

e \approx 2.718

, we get

P \approx \dfrac{36.0}{7.389} \approx 4.87 \text{ W}

.

Through resistanceless leads, a capacitor of capacitance 12.0 is...

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