MHT CET · Physics · Electromagnetic Induction
A circuit has self-inductance ' \(L\) ' and carries a current ' \(\mathrm{I}\) '. To prevent sparking when the circuit is switched off, a capacitor which can withstand potential difference ' \(\mathrm{V}\) ' is used. The least capacitance is
- A \(\frac{\mathrm{IV}}{\mathrm{L}}\)
- B \(\mathrm{L}\left(\frac{\mathrm{V}}{\mathrm{L}}\right)^2\)
- C \(\mathrm{L}\left(\frac{\mathrm{I}}{\mathrm{V}}\right)^2\)
- D \(\frac{\mathrm{LI}}{\mathrm{V}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{L}\left(\frac{\mathrm{I}}{\mathrm{V}}\right)^2\)
Step-by-step Solution
Detailed explanation
The energy stored in the inductance is given by \(\mathrm{U}=\frac{1}{2} \mathrm{LI}^2\) Energy stored by the capacitor is given by \(\mathrm{W}=\frac{1}{2} \mathrm{CV}^2\) This energy must be transferred to the capacitor.
\(\therefore \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}^2\)
\(\mathrm{C}=\mathrm{L}\left(\frac{\mathrm{I}}{\mathrm{V}}\right)^2\)
\(\therefore \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}^2\)
\(\mathrm{C}=\mathrm{L}\left(\frac{\mathrm{I}}{\mathrm{V}}\right)^2\)
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