Question

A 60 uF capacitor is charged to 100 volts. This
charged capacitor is connected across a 1.5 mH
coil, so that LC oscillations occur. The maximum
current in the coil is :-
(1) 1.5 A
(2) 2 A
(3) 15 A
(4) 20 A​

Answer 1

So maximum current will be 20 A

And option (4) will be correct option

Explanation:

Potential difference V = 100 volts

We have given capacitance of the capacitor $C=60\mu F$

And inductance $L=1.3mH=1.3\times 10^{-3}H$

In LC oscillators energy stored by capacitor and inductor is same

So $\frac{1}{2}LI^2=\frac{1}{2}CV^2$

$\frac{1}{2}\times 1.3\times 10^{-3}\times I^2=\frac{1}{2}\times 60\times 10^{-6}\times 100^2$

$I=20A$

So maximum current will be 20 A

And option (4) will be correct option

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Answer 2

The maximum  current in the coil is 20 A.

(4) is correct option

Explanation:

Given that,

Capacitor = 60μF

Voltage = 100 V

Inductance = 1.5 mH

We need to calculate the maximum current in the coil

At resonance condition

$\dfrac{1}{2}LI^2=\dfrac{1}{2}CV^2$

$I^2=\dfrac{CV^2}{L}$

Put the value into the formula

$I^2=\dfrac{60\times10^{-6}\times(100)^2}{1.5\times10^{-3}}$

$I=\sqrt{\dfrac{60\times10^{-6}\times(100)^2}{1.5\times10^{-3}}}$

$I=20\ A$

Hence, The maximum  current in the coil is 20 A.

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Topic : maximum current in the coil

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