Question

A $ 20 \, \mu F $ capacitor is connected to $ 45 \,V $ 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

Answer: (A)

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 $= 45 \,V$
Hence, the final charge on the capacitor is
$q=cv$
Here, $C =20\,\mu F, V=45\,V$
$\therefore q=20\times10^{-6}\times45$
or $q=900\times10^{-6}$
or $q=9\times10^{-4}\,C$