Question

A $20\mu F$ capacitor is connected to $45V$ battery through a circuit whose resistance is $2000\Omega$. What is the final charge on the capacitor?

• A. $9\times {10}^{-4}C$
• B. $9.154\times {10}^{-4}C$
• C. $9.8\times {10}^{-4}C$
• D. None of these

Answer 1

The correct option is A $9\times {10}^{-4}C$

We know that in steady state the capacitor behaves like as an open circuit i.e., capacitor will not pass the current.
So, the potential difference across the capacitor $=45V$
Hence, the final charge on the capacitor is
$q=CV$
Here, $C=20\mu F,V=45V$
$\therefore q=20\times {10}^{-6}\times 45$
or $q=900\times {10}^{-6}$
or $q=9\times {10}^{-4}C$