Electrostatics Ques 156
- At time $t=0$, a battery of $10 \mathrm{~V}$ is connected across points $A$ and $B$ in the given circuit. If the capacitors have no charge initially, at what time (in second) does the voltage across them become $4 \mathrm{~V}$ ? [Take : $\ln 5=1.6, \ln 3=1.1$ ]
(2010)
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Answer:
Correct Answer: 156.2
Solution:
- Voltage across the capacitors will increase from 0 to $10 \mathrm{~V}$ exponentially. The voltage at time $t$ will be given by
$$ \begin{array}{rlrl} & V & =10\left(1-e^{-t / \tau} \right) \\ \text { Here, } \quad \tau_{c} & =C_{\text {net }} R_{\text {net }} \\ & =\left(1 \times 10^{6}\right)\left(4 \times 10^{-6}\right)=4 \mathrm{~N} \ \therefore \quad V & =10\left(1-e^{-t / 4}\right) \end{array} $$
Substituting $V=4$ volt, we have
$$ \begin{aligned} 4 & =10\left(1-e^{-t / 4}\right) \\ \text { or } \quad e^{-t / 4} & =0.6=\frac{3}{5} \end{aligned} $$
Taking log on both sides, we have,
$$ \begin{alignedat} -\frac{t}{4} & =\ln 3-\ln 5 \\ \text { or } \quad t & =4(\ln 5-\ln 3)=2 \mathrm{~s} \end{aligned} $$
Hence, the answer is 2 .
BETA
