Question

A capacitor of capacitance C=1 $\mu$F is suddenly connected to a battery of 100 volt through a resistor R = 100 $\Omega$. The time taken for the capacitor to be charged to get 50 V is: [Take ln 2 = 0.69] 

 

• A. $0.69 \times 10^{-4}$ s
• B. $0.30 \times 10^{-4}$ s
• C. $1.44 \times 10^{-4}$ s
• D. $3.33 \times 10^{-4}$ s

Hint:

The time taken to reach half the maximum voltage ($V_0/2$) during charging is always equal to $\tau \ln(2)$. This is a useful shortcut for RC circuit problems.

Answer 1

The Correct Option is A

The voltage $V(t)$ across a charging capacitor in a series RC circuit at time t is given by:
$V(t) = V_0 (1 - e^{-t/\tau})$
where $V_0$ is the battery voltage and $\tau$ is the time constant, $\tau = RC$.
First, calculate the time constant $\tau$:
$R = 100 \, \Omega$
$C = 1 \, \mu F = 1 \times 10^{-6} \, F$
$\tau = RC = (100 \, \Omega)(1 \times 10^{-6} \, F) = 10^{-4} \, s$
Now, we find the time t when the voltage across the capacitor $V(t)$ is 50 V. The battery voltage $V_0$ is 100 V.
$50 = 100 (1 - e^{-t/10^{-4}})$
Divide by 100:
$0.5 = 1 - e^{-t/10^{-4}}$
Rearrange the equation:
$e^{-t/10^{-4}} = 1 - 0.5 = 0.5 = \frac{1}{2}$
Take the natural logarithm (ln) of both sides:
$\ln(e^{-t/10^{-4}}) = \ln(\frac{1}{2}) = -\ln(2)$
$-\frac{t}{10^{-4}} = -\ln(2)$
$t = 10^{-4} \times \ln(2)$
Using the given value $\ln(2) = 0.69$:
$t = 10^{-4} \times 0.69 = 0.69 \times 10^{-4} \, s$