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An uncharged capacitor is having the capacitance $100\mu F$ which has been connected to a battery of emf $20V$ at $t=0$ through a resistance of $10\Omega $, then what will be the time at which the stored energy rate in capacitor will be maximum?
$\begin{align}
  & A.4W\left( J{{s}^{-1}} \right) \\
 & B.2W\left( J{{s}^{-1}} \right) \\
 & C.0.2W\left( J{{s}^{-1}} \right) \\
 & D.3W\left( J{{s}^{-1}} \right) \\
\end{align}$

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Last updated date: 17th Jun 2024
Total views: 393.6k
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Answer
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Hint: Maximum power can be found by taking the ratio of the stored energy in the capacitor to the minimum time taken. The minimum time taken can be found by taking five times the constant of RC circuits. The maximum energy in the capacitor will be half the product of the capacitance and the square of the potential. This will help you in answering this question.

Complete answer:
It has been given that the capacitance of the uncharged capacitor will be,
$C=100\mu F$
Emf of the battery can be mentioned as,
$V=20V$
Resistance of the resistor can be written as,
$R=10\Omega $
Maximum power can be found by taking the ratio of the stored energy in the capacitor to the minimum time taken.
$\text{maximum power}=\dfrac{\text{stored energy in capacitor}}{\text{minimum time taken}}$
The minimum time taken can be found by taking five times the constant of RC circuits. This can be written as,
$\text{minimum time taken=5}\times \text{constant of RC circuit}$
That is,
$\text{minimum time taken=5}\times R\text{C }$
Substituting the values in it,
$\begin{align}
  & \text{minimum time taken=5}\times 10\times 100\times {{10}^{-6}} \\
 & \text{minimum time taken}=5\times {{10}^{3}}s \\
\end{align}$
The maximum energy stored in the capacitor can be written as,
${{E}_{\max }}=\dfrac{1}{2}C{{V}^{2}}$
Substituting the values in it will give,
$\begin{align}
  & {{E}_{\max }}=\dfrac{1}{2}\times 100\times {{10}^{-6}}\times 20\times 20 \\
 & {{E}_{\max }}=0.02J \\
\end{align}$
The maximum power of the circuit can be found by taking the ratio of the maximum energy to the minimum time taken. That is,
${{P}_{\max }}=\dfrac{0.02}{5\times {{10}^{-3}}}=0.04\times {{10}^{-3}}=4W\left( J{{s}^{-1}} \right)$
Therefore, the maximum power of the circuit has been found.

It has been mentioned as the option A.

Note:
A capacitor is an electrical device which is useful in storing the electrical field. The electrical field has been stored in between two metallic plates which are separated by a dielectric medium. The unit of capacitance is given as farad.