Question

In the given circuit, $E_1=3V,E_2=2V,E_3=6V,R_1=2\Omega ,R_4=6\Omega ,R_3=2\Omega ,R_2=4\Omega ,C=5\mu F$. The energy stored in the capacitor is :

• A. $1.44\times {10}^{-5}J$
• B. $1\times {10}^{-5}J$
• C. $0.72\times {10}^{-5}J$
• D. $2.88\times {10}^{-5}J$

Answer 1

The correct option is A $1.44\times {10}^{-5}J$

Applying Kirchhoff's law to loop GABFG and EDCBFE, we will get
$4i_1+4i_2=6...\left(1\right)$
$2i_2+3i_2+4i_1+4i_2=3+2\Rightarrow 4i_1+9i_2=5..\left(2\right)$
$\left(1\right)-\left(2\right),-5i_2=1$ in magnitude $i_2=0.2A$
Potential difference E and F is $V=E_2+i_2{R}_2=2+\left(0.2\right)2=2.4V$
As no current passes through $R_1$ so the potential at H and F are same. Thus the potential difference E and H is $V=2.4V$.
Energy stored in capacitor is $U=\frac{1}{2}C{V}^2=0.5\times 5\times {10}^{-6}\times {2.4}^2=1.44\times {10}^{-5}J$