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 current in the circuit 12.0 after the connections are made.

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Quantrex Academy · Accepted Answer

Correct answer: B. 2.21 A

The time constant of the RC circuit is given by

\tau = RC

.

Substituting the given values:

\tau = 1.00 \times 12.0 \times 10^{-6} = 12.0 \mu\text{s}

The current in the circuit during the charging of a capacitor is given by:

I = I_0 e^{-t/\tau}

The initial current

I_0

is:

I_0 = \dfrac{E}{R} = \dfrac{6.00}{1.00} = 6.00 \text{ A}

At

t = 12.0 \mu\text{s}

, we have

t = \tau

. Substituting this into the current equation:

I = 6.00 \times e^{-1} = \dfrac{6.00}{e}

Using

e \approx 2.718

:

I = \dfrac{6.00}{2.718} \approx 2.21 \text{ A}

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

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