MHT CET · Physics · Capacitance
A \(1 \mu \mathrm{~F}\) capacitor is charged to 50 V and is then discharged through 10 mH inductor of negligible resistance. The maximum current in the inductor is
- A 0.5 A
- B 1.5 A
- C 1 A
- D 0.15 A
Answer & Solution
Correct Answer
(A) 0.5 A
Step-by-step Solution
Detailed explanation
Energy stored in capacitor \(=\frac{1}{2} \mathrm{CV}^2\)
For maximum current energy stored in inductor, \(\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_0^2\).
\(\mathrm{I}_0=\) Maximum current
\(\mathrm{I}_0^2=\frac{\mathrm{CV}^2}{\mathrm{~L}}=\frac{10^{-6} \times 50 \times 50}{10 \times 10^{-3}}\)
\(=25 \times 10^{-2}\)
\(\mathrm{I}_0=\sqrt{25 \times 10^{-2}}\)
\(\mathrm{I}_0=0.5 \mathrm{~A}\)
For maximum current energy stored in inductor, \(\frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_0^2\).
\(\mathrm{I}_0=\) Maximum current
\(\mathrm{I}_0^2=\frac{\mathrm{CV}^2}{\mathrm{~L}}=\frac{10^{-6} \times 50 \times 50}{10 \times 10^{-3}}\)
\(=25 \times 10^{-2}\)
\(\mathrm{I}_0=\sqrt{25 \times 10^{-2}}\)
\(\mathrm{I}_0=0.5 \mathrm{~A}\)
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