Question

Two capacitors of capacities $1\mu \text{F}$ and $C\mu \text{F}$ are connected in series and this combination is charged to a potential difference of $120\text{V}$. If the charge on the $1\mu \text{F}$ capacitor is $80\mu \text{C}$, then the energy stored in the capacitor $C$ (in micro joule) is

• A. $1600$
• B. $1800$
• C. $2000$
• D. $1400$

Answer 1

The correct option is A $1600$

For first capacitor
$C_1=1\mu \text{F}$
$q_1=80\mu \text{C}$
Potential across $1\mu \text{F}$ capacitor is
$V_1=\frac{q_1}{C_1}=\frac{80}{1}$
$V_1=80\text{V}$

For second capacitor
$C_2=C\mu \text{F}$
$q_2=80\mu \text{C}$
Potential across $C\mu \text{F}$ capacitor is
$V_2=V-V_1$
$=120-80$
$V_2=40\text{V}$

$C=\frac{q_2}{V_2}$
$C=\frac{80\times {10}^{-6}}{40}$
$C=2\times {10}^{-6}F$

Energy stored in capacitor $\left(C\mu \text{F}\right)$ is
$U=\frac{1}{2}CV{{}_2}^2$
$=\frac{1}{2}(2\times {10}^{-6})\times 40\times 40$
$=1600\times {10}^{-6}\text{J}=1600\mu \text{J}$