Search any question & find its solution
Question:
Answered & Verified by Expert
A $1 \mu \mathrm{F}$ capacitor $C$ is connected to a battery of $10 \mathrm{V}$ through a resistance 1 Mega Ohm. The voltage across $C$ after 1 s is approximately
Options:
Solution:
1954 Upvotes
Verified Answer
The correct answer is:
$6.3 \mathrm{V}$
The time constant of the circuit
Here,
$$
\begin{aligned}
\tau &=C \cdot R \\
C &=1 \mu F=1 \times 10^{-6}
\end{aligned}
$$
$$
\begin{array}{l}
R=1 \mathrm{M\Omega}=1 \times 10^{6} \Omega \\
\tau=1 \times 10^{-6} \times 1 \times 10^{6}=1 \mathrm{s}
\end{array}
$$
We know that in one (1) time constant $63 \%$ charging is done. $\therefore \quad \frac{63}{100} \times q_{\max }$
$=\frac{63}{100} \times 1 \times 10^{-6} \times 10$
$=\frac{63}{100} \times 1 \mu \mathrm{F} \times 10=6.3 \mu \mathrm{C}$
$V=\frac{\mathrm{q}}{\mathrm{C}}=\frac{6.3 \times 10^{-6} \mathrm{C}}{1 \times 10^{-6} \mathrm{F}}=6.3 \mathrm{V}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.