Question

Figure shows part of a circuit. Energy stored in $2\mu F$ capacitor at steady state is:

• A. $100\mu J$
• B. $50\mu J$
• C. $200\mu J$
• D. $80\mu J$

Answer 1

The correct option is A $100\mu J$

By junction law, we can see the current through $5\Omega$ is $2A$.
So, potential difference across $5\Omega$ or $2\mu F$ is $V=5\times 2=10V$
Hence, energy stored in the capacitor is$\frac{1}{2}C{V}^2=\frac{1}{2}\times 2\times {10}^2=100\mu J$