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Question: An uncharged capacitor is having the capacitance \(100\mu F\) which has been connected to a battery ...

An uncharged capacitor is having the capacitance 100μF100\mu F which has been connected to a battery of emf 20V20V at t=0t=0 through a resistance of 10Ω10\Omega , then what will be the time at which the stored energy rate in capacitor will be maximum?
A.4W(Js−1) B.2W(Js−1) C.0.2W(Js−1) D.3W(Js−1) \begin{aligned} & 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{aligned}

Explanation

Solution

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=100μFC=100\mu F
Emf of the battery can be mentioned as,
V=20VV=20V
Resistance of the resistor can be written as,
R=10ΩR=10\Omega
Maximum power can be found by taking the ratio of the stored energy in the capacitor to the minimum time taken.
maximum power=stored energy in capacitorminimum 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,
minimum time taken=5×constant of RC circuit\text{minimum time taken=5}\times \text{constant of RC circuit}
That is,
minimum time taken=5×RC \text{minimum time taken=5}\times R\text{C }
Substituting the values in it,
minimum time taken=5×10×100×10−6 minimum time taken=5×103s \begin{aligned} & \text{minimum time taken=5}\times 10\times 100\times {{10}^{-6}} \\\ & \text{minimum time taken}=5\times {{10}^{3}}s \\\ \end{aligned}
The maximum energy stored in the capacitor can be written as,
Emax⁡=12CV2{{E}_{\max }}=\dfrac{1}{2}C{{V}^{2}}
Substituting the values in it will give,
Emax⁡=12×100×10−6×20×20 Emax⁡=0.02J \begin{aligned} & {{E}_{\max }}=\dfrac{1}{2}\times 100\times {{10}^{-6}}\times 20\times 20 \\\ & {{E}_{\max }}=0.02J \\\ \end{aligned}
The maximum power of the circuit can be found by taking the ratio of the maximum energy to the minimum time taken. That is,
Pmax⁡=0.025×10−3=0.04×10−3=4W(Js−1){{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.