Question

A $1$$\mu F$ capacitor is connected in parallel with a battery and a resistor. When the battery is disconnected at $t=0$, the charge on the capacitor falls to $25$ percent of its initial charge in $2ms$. The resistance connected then is [Take ln$2=0.69$]

• A. $14.5k\Omega$
• B. $2.9k\Omega$
• C. $29k\Omega$
• D. $1.45k\Omega$

Answer 1

Answer: (D)

$1.45k\Omega$

When S is opened at $t=0$ s,

$q_0=VC$ and hence $q=q_0{e}^{-\frac{t}{RC}},t>0$

During $t=2ms,q=\frac{q_0}{4}$

$\therefore 4=e^{t/RC}\Rightarrow ln4=\frac{t}{RC}$ or $R=\frac{t}{Cln4}$

Hence, $R=\frac{2\times {10}^{-3}}{2\times 0.69\times 1\times {10}^{-6}}=\frac{2\times {10}^3}{1.38}$

$=1.45k\Omega$