A 16 wire is bend to form a square loop. A 9 ~V battery with inte… - Vidyadip

MCQ

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
$42$
✓
$81$
D
$12$

✓

Answer

Correct option: C.

$81$

c
$ 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$

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