MHT CET · Physics · Electromagnetic Induction
A circuit having a self inductance of 1 henry carries a current of 1 A . To prevent the sparking when the circuit is broken, a capacitor which can withstand 500 V is connected across the switch. What is the minimum value of the capacitance of the capacitor?
- A \(2 \mu \mathrm{~F}\)
- B \(4 \mu \mathrm{~F}\)
- C \(6 \mu \mathrm{~F}\)
- D \(8 \mu \mathrm{~F}\)
Answer & Solution
Correct Answer
(B) \(4 \mu \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
The energy stored in the inductance is \(\mathrm{U}=\frac{1}{2} \mathrm{LI}^2\)
\(\mathrm{U}=\frac{1}{2} \times 1 \times 1^2=0.5 \mathrm{~J}\)
This energy must be transferred to the capacitor. Energy stored by a capacitor \(\mathrm{U}=\frac{1}{2} \mathrm{CV}^2\)
\(\begin{array}{ll}\therefore & \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\ \therefore & C=L\left(\frac{I}{V}\right)^2=1\left(\frac{1}{500}\right)^2=4 \mu F\end{array}\)
\(\mathrm{U}=\frac{1}{2} \times 1 \times 1^2=0.5 \mathrm{~J}\)
This energy must be transferred to the capacitor. Energy stored by a capacitor \(\mathrm{U}=\frac{1}{2} \mathrm{CV}^2\)
\(\begin{array}{ll}\therefore & \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\ \therefore & C=L\left(\frac{I}{V}\right)^2=1\left(\frac{1}{500}\right)^2=4 \mu F\end{array}\)
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