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Question: An ideal capacitor of capacitance \(0.2\mu F\) is charged to a potential difference of 10V. The char...

An ideal capacitor of capacitance 0.2μF0.2\mu F is charged to a potential difference of 10V. The charging battery id then disconnected. The capacitor is then connected to an ideal inductor of self-inductance 0.5mH. The current at a time when the potential difference across the capacitor is 5V, is:
A. 0.17A
B. 0.15A
C. 0.34A
D. 0.25A

Explanation

Solution

Capacitor is an electronic device which can store electrical energy. If we charge a capacitor, the energy will be stored in the capacitor. Then if we connect an inductor in the circuit, we can use the stored energy. In the system, the initial energy will be equal to the final output energy.

Complete step by step answer:
We have a capacitor of capacitance C=0.2μF=2×10−7FC=0.2\mu F=2\times {{10}^{-7}}F .
The capacitor is charged to a potential of V0=10V{{V}_{0}}=10V .
The energy stores in a capacitor can be given as,
E=12CV2E=\dfrac{1}{2}C{{V}^{2}}
After that, the capacitor is connected to an inductor of inductance, L=0.5mH=5×10−4HL=0.5mH=5\times {{10}^{-4}}H .
The energy stores in an inductor can be given as,
E=12LI2E=\dfrac{1}{2}L{{I}^{2}}
We need to find the current flowing through the circuit at a time when the potential across the capacitor is 5V.
Here, we can say that the initial energy of the system will be equal to the final energy of the system.
Let, the current flowing through the circuit is I.
So, we can write,
Ei=Efc+EL ⇒12CV02=12CV2+12LI2 ⇒12×2×10−7×102=12×2×10−7×52+12×5×10−4I2 ⇒10−5=2.5×10−6+2.5×10−4I2 ⇒7.5×10−6=2.5×10−4I2 ⇒I2=7.5×10−62.5×10−4 ⇒I2=3×10−2 ∴I=0.17A \begin{aligned} & {{E}_{i}}={{E}_{fc}}+{{E}_{L}} \\\ & \Rightarrow \dfrac{1}{2}CV_{0}^{2}=\dfrac{1}{2}C{{V}^{2}}+\dfrac{1}{2}L{{I}^{2}} \\\ & \Rightarrow \dfrac{1}{2}\times 2\times {{10}^{-7}}\times {{10}^{2}}=\dfrac{1}{2}\times 2\times {{10}^{-7}}\times {{5}^{2}}+\dfrac{1}{2}\times 5\times {{10}^{-4}}{{I}^{2}} \\\ & \Rightarrow {{10}^{-5}}=2.5\times {{10}^{-6}}+2.5\times {{10}^{-4}}{{I}^{2}} \\\ & \Rightarrow 7.5\times {{10}^{-6}}=2.5\times {{10}^{-4}}{{I}^{2}} \\\ & \Rightarrow {{I}^{2}}=\dfrac{7.5\times {{10}^{-6}}}{2.5\times {{10}^{-4}}} \\\ & \Rightarrow {{I}^{2}}=3\times {{10}^{-2}} \\\ & \therefore I=0.17A \\\ \end{aligned}
So, the current flowing through the circuit at a time when the potential across the capacitor will be 5V is 0.17A.

So, the correct answer is “Option A”.

Note: Capacitance of a capacitor can be defined as the ability or the capacity of the capacitor to store electrical energy in it. An inductor also stores energy in it. But the difference between the capacitor and the inductor is that the capacitor stores energy in the form of an electric field and the inductor stores energy in the form of a magnetic field. The unit of capacitance is Farad and the unit of inductance is Henry.