Question

What will be the time required to half the amplitude of charge stored by a capacitor and time for the energy stored to be halved in a series LRC circuit with L=40H, R=0.03 ohms and C=4.8F?

Answer 1

We know that

Charge

$q=q_0(1-e^{\frac{-t}{RC}})$

Gives

$q=\frac{q_0}{2}$

$\frac{q_0}{2}=q_0(1-e^{\frac{-t}{RC}})$

$\frac{1}{2}=(1-e^{\frac{-t}{RC}})$

$-\frac{1}{2}=e^\frac{-t}{RC}\\2=e^\frac{t}{RC}$

$t=RCln2$

Put value

$$t=0.03\times4.8\times ln2=0.0998Sec$$