Question

An uncharged capacitor of capacitance $4\mu F$, a battery of emf $12volt$ and a resistor of $2.5M\Omega$ are connected in series. The time after which $V_C=3V_R$ is (take $ln2=0.693$)

• A. $6.93seconds$
• B. $13.86seconds$
• C. $7seconds$
• D. $14seconds$

Answer 1

Answer: (B)

$13.86seconds$

Given,

$C=4\mu F$

$V=12V$

$R=2.5M\Omega$

Condition,

$V_C=3V_R$

$V_C=3(V-V_C)=3V-3V_C$

$4V_C=3V$

$V_C=\frac{3V}{4}$. . . . . . . . . .(1)

The voltage across capacitor in $RC$ circuit is given by

$V_C=V(1-e^{-t/RC})$

$\frac{3V}{4}=V(1-e^{-t/RC})$

$\frac{3}{4}=1-e^{-t/RC}$

$e^{-t/RC}=1-\frac{3}{4}=\frac{1}{4}$

Take log both side,

$-t/RC=ln\left(1/4\right)=ln\left(1\right)-ln\left(4\right)$

$-t/RC=0-ln(2{)}^2=-2ln2$

$t=2RCln2$

$t=2\times 2.5\times {10}^6\times 4\times {10}^{-6}\times 0.693$

$t=13.86sec$

The correct option is B.