Question

An $1\mu F$ capacitor $C$ is connected to a battery of $10V$ through a resistance $1M\Omega$. The voltage across $C$ after $1$ sec is approximately

• A. $5.6V$
• B. $7.8V$
• C. $6.3V$
• D. $10V$

Answer 1

Answer: (C)

$6.3V$

Time constant$=\tau =CR=1\times {10}^{-6}\times {10}^6=1s$

So, in 1 time constant, $63\%$ charging is done.

Therefore, $\frac{63}{100}\times q_{max}=\frac{63}{100}\times 1\times 10=6.4\mu C$

$⟹V=\frac{q}{C}=\frac{6.3\mu C}{1\mu F}=6.3V$