Answer to Question #266229 in Electric Circuits for Jo blue

Question

For a series LRC circuit, if L=30H, R= 0.05 ohms and C= 5.8F, what is the time required for the amplitude of the charge stored by the capacitor to be halved?

Answer 1

We know that

Charge

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

$q=q_0(1-e^{\frac{-t}{0.05\times5.8}})$

$\frac{q_0}{2}=q_0(1-e^{\frac{-t}{0.05\times5.8}})$

$\frac{1}{2}=e^{-\frac{t}{0.05\times5.8}}$

$ln2=\frac{t}{0.05\times5.8}$

$t=0.29\times ln2=0.202sec$