Capacitance Question 6
Question 6 - 2024 (29 Jan Shift 1)
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 $\mathrm{x}=$
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Answer: (81)
Solution:
$I=\frac{V}{R_{e q}} I=\frac{V}{R_{e q}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4}$
$I_{1}=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16}$
$V_{A}-V_{B}=I_{1} \times 8=\frac{9}{16} \times 8=\frac{9}{2} V$
$\therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu \mathrm{J}$
$\therefore U=\frac{81}{2} \mu J$
$\therefore x=81$
