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2. Electric Potential and Capacitance
hard
Find the ratio of energy stored in $5\,\mu F$ and $4\,\mu F$ capacitor in the given circuit in steady state

A
$0.6$
B
$0.7$
C
$0.8$
D
$0.9$
Solution

$I=\frac{2}{5+4} a m p \quad \Rightarrow \quad I=\frac{2}{9} A m p$
$V_{A C}=5 \times \frac{2}{9}$ volt $=\frac{10}{9}$ volt
$V_{D B}=4 \times \frac{2}{9}$ volt $=\frac{8}{9}$ volt
$\frac{U_{5 \mu F}}{U_{4 \mu F}}=\frac{\frac{1}{2} \times 5\left(\frac{8}{9}\right)^{2}}{\frac{1}{2} \times 4 \times\left(\frac{10}{9}\right)^{2}}=0.8$
Standard 12
Physics