Question

Find the total energy of the capacitors present in the circuit given below

• A. $80.5\mu J$
• B. $100\mu J$
• C. $120.25\mu J$
• D. $90.7\mu J$

Answer 1

The correct option is A $80.5\mu J$

When steady-state condition is achieved it means that now capacitor is fully charged, no current will flow and will start acting as an open circuit and in beginning, it acts as a wire without showing any resistance when it is not charged
Energy stored in each capacitor is given as $E=\frac{1}{2}\frac{Q^2}{C}=\frac{1}{2}C{V}^2$


Now to find the potential difference between E and C
$R_{eq}=3\Omega +3\Omega +2\Omega +2\Omega =10\Omega$
$I=\frac{10V}{10\Omega }=1A$

The potential difference between E and C is,
$V=1(2+3)=5V$
Energy stored, $\begin{array}{}{U}_{EC}=\frac{1}{2}\times 1\times {10}^{-6}\times 5^2& \text{(1)}\end{array}$
Potential difference betweenn A and E is same as in between E and B,
$V_{AE}=10-(2\times 1)=8V$
Energy stored, $\begin{array}{}{U}_{AE}=\frac{1}{2}\times 2\times {10}^{-6}\times 8^2& \text{(2)}\end{array}$
Potential difference between E and D is $2V$
$\begin{array}{}{U}_{ED}=\frac{1}{2}\times 2\times {10}^{-6}\times 2^2& \text{(3)}\end{array}$
Total energy stored,
$U_{total}=\frac{1}{2}\times {10}^{-6}[25+128+8]$
$U_{total}=\frac{161}{2}\times {10}^{-6}=80.5\mu J$