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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 $: \ln 5=1.6, \ln 3=1.1]$
[Take $: \ln 5=1.6, \ln 3=1.1]$
Solution:
1383 Upvotes
Verified Answer
The correct answer is:
2
Voltage across the capacitors will increase from 0 to $10 \mathrm{~V}$ exponentially. The voltage at time $t$ will be given by
$$
\begin{aligned}
& V=10\left(1-e^{-t / \tau} \mathrm{C}\right) \\
& \text { Here } \tau_c=C_{\text {net }} R_{\text {net }} \\
& =\left(1 \times 10^6\right)\left(4 \times 10^{-6}\right)=4 \mathrm{~s} \\
& \therefore \quad V=10\left(1-e^{-t / 4}\right) \\
&
\end{aligned}
$$
Substituting $V=4$ volt we have,
$$
4=10\left(1-e^{-t / 4}\right)
$$
Substituting $V=4$ volt we have,
$$
4=10\left(1-e^{-t / 4}\right)
$$
or $e^{-t / 4}=0.6=\frac{3}{5}$
Taking $\log$ both sides we have,
$$
-\frac{t}{4}=\ln 3-\ln 5
$$
or $\quad t=4(\ln 5-\ln 3)=2 \mathrm{~s}$.
Hence, the answer is 2 .
$$
\begin{aligned}
& V=10\left(1-e^{-t / \tau} \mathrm{C}\right) \\
& \text { Here } \tau_c=C_{\text {net }} R_{\text {net }} \\
& =\left(1 \times 10^6\right)\left(4 \times 10^{-6}\right)=4 \mathrm{~s} \\
& \therefore \quad V=10\left(1-e^{-t / 4}\right) \\
&
\end{aligned}
$$
Substituting $V=4$ volt we have,
$$
4=10\left(1-e^{-t / 4}\right)
$$
Substituting $V=4$ volt we have,
$$
4=10\left(1-e^{-t / 4}\right)
$$
or $e^{-t / 4}=0.6=\frac{3}{5}$
Taking $\log$ both sides we have,
$$
-\frac{t}{4}=\ln 3-\ln 5
$$
or $\quad t=4(\ln 5-\ln 3)=2 \mathrm{~s}$.
Hence, the answer is 2 .
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