A 16 μ F capacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is

Question

A 16 μ F capacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is (A) 0.2 A (B) 40 mA (C) 2 A (D) 0.4 A

Tardigrade Admin · Accepted Answer

Energy of charged capacitor = energy generated across inductor coil
E =

\frac{1}{2} C V^{2} = \frac{1}{2} L i^{2}

Given,

C = 16 μ F = 16 \times 1 0^{- 6} F, V = 20

volt
L = 40 m H = 40

\times 1 0^{- 3}

H

∴ \frac{1}{2} \times 16 \times 1 0^{- 6} \times (20)^{2} = \frac{1}{2} \times 40 \times 1 0^{- 3} \times i^{2}

\Rightarrow i^{2} = \frac{16}{100}

\Rightarrow i = 0.4

A

Q. A $16 μ F$ capacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is

0.2 A

40 mA

2 A

0.4 A

0.8 A

Q. A 16μFcapacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is

0.2 A

40 mA

2 A

0.4 A

0.8 A

Energy of charged capacitor = energy generated across inductor coil E = 21​CV2=21​Li2 Given, C=16μF=16×10−6F,V=20 volt L = 40 m H = 40 ×10−3 H ∴21​×16×10−6×(20)2=21​×40×10−3×i2 ⇒i2=10016​ ⇒i=0.4 A

Q. A $16 μ F$ capacitor is charged to a 20 V potential. The battery is then disconnected and pure 40 mH coil is connected across the capacitor, so that L-C oscillations are setup. The maximum current in the coil is