Question

A capacitor of capacity 100 $\mu$F is discharged through a resistance of 1000 ohm:

• A. its time constant is $1\times {10}^{-1}s$
• B. during time = 0.693 $\times {10}^{-1}s$, the charge left on the capacitor is half of its maximum charge
• C. during time = its time constant $\left(\tau \right)$, charge left on the capacitor is 0.63 times the maximum charge
• D. during time = its time constant $\left(\tau \right)$, charge left on the capacitor is 0.37 times the maximum charge

Answer 1

Answer: (A), (B), (D)

The correct options are
A its time constant is $1\times {10}^{-1}s$
B during time = 0.693 $\times {10}^{-1}s$, the charge left on the capacitor is half of its maximum charge
D during time = its time constant $\left(\tau \right)$, charge left on the capacitor is 0.37 times the maximum charge

We know that the formula of charge on a capacitor is -

$q=q_o\left(e^{-t/\tau }\right)$ while discharging

Where time constant $\tau =RC=100\times {10}^6\times 1000=0.1s={10}^{-1}s$

and $q_o$ is maximum charge.

When charge on capacitor is half of maximum charge-

$\frac{q_o}{2}=q_o\left(e^{-t/\tau }\right)$

$⟹e^{-t/\tau }=\frac{1}{2}$

$⟹t=\tau \ln 2=0.693\times {10}^{-1}s$

For time $t=\tau$, charge on capacitor is-

$q=q_o\left(e^{-t/\tau }\right)=q_o{e}^{-1}$

$⟹q=0.37q_o$

That is charge on capacitor is 0.37 times maximum charge.

Answer-(A),(B),(D)