Question

An LC circuit has a capacitor with C = 0.2 uF and a solenoid with inductance L = 3 mH. Initially the capacitor is fully charged and has a voltage across its plates of 2 V. The capacitor then begins to discharge. What is the magnitude of the current in the circuit when the voltage across the capacitor is 1 V?

Answer 1

The magnitude of the current in the circuit is 0.01159 A.

Step-by-step explanation:

Given that,

Capacitor
`C = 0.2\mu F`

Inductance
`L=3 mH`

Voltage V = 2 V

We need to calculate the energy when the capacitor is charged

Using formula of energy

`E=(1)/(2)(q^2)/(C)`

`E=(1)/(2)*(C^2V^2)/(C)`

`E=(1)/(2)*CV^2`

Put the value into the formula

`E=(1)/(2)*0.2*10^(-6)*4`

`E=4*10^(-7)`

When the capacitor is discharged

Using formula of energy

`E=(1)/(2)LI^2`

`4*10^(-7)=(1)/(2)*3*10^(-3)* I^2`

`I^2=(2*4*10^(-7))/(3*10^(-3))`

`I=√(0.00027)`

`I=0.0164\ A`

rms value of current,

`I_(rms)=(I_(0))/(√(2))`

`I_(rms)=(0.0164)/(√(2))`

`I_(rms)=0.01159\ A`

Hence, The magnitude of the current in the circuit is 0.01159 A.