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2. Electric Potential and Capacitance
hard
A $16\ \Omega$ wire is bend to form a square loop. A $9 \mathrm{~V}$ battery with internal resistance $1\ \Omega$ is connected across one of its sides. If a $4\ \mu \mathrm{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{x}{2} \ \mu \mathrm{J}$. where $x=$________.
A
$52$
B
$42$
C
$81$
D
$12$
(JEE MAIN-2024)
Solution
$ I=\frac{V}{R_{e q}} I=\frac{V}{R_{e q}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} $
$ I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} $
$ V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V $
$ \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu \mathrm{J} $
$ \therefore U=\frac{81}{2} \mu \mathrm{J} $
$ \therefore \mathrm{X}=81$
Standard 12
Physics
