A capacitor having capacitance of 100 µF is charged with a potential difference of&nbsp;12 V is connected to an inductor of inductance 10 mH. Find the maximum current through the inductor.

1.2 A

2 A

1.6 A

2.4 A

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Alternating Current Physics Ncert Solutions Class 12th Physics Class 12th

A capacitor having capacitance of 100 µF is charged with a potential difference of 12 V is connected to an inductor of inductance 10 mH. Find the maximum current through the inductor.

Option 1 -

2 A

Option 2 -

1.6 A

Option 3 -

2.4 A

Option 4 -

1.2 A

2 Views | Posted 2 days ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 days ago

Correct Option - 4

Detailed Solution:

$I = Q_{0 . ω}$

$= \frac{C V}{\sqrt[]{L C}} = V \sqrt[]{\frac{C}{L}}$

$= 12 \sqrt[]{\frac{100 \times 1 0^{- 6}}{10 \times 1 0^{- 3}}}$

= 1.2 A

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