Question 106: Electrostatics & Capacitance

Question

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 $$4 \mu F$$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $$\frac{x}{2} \mu J$$, where $$x=$$ _________

Accepted Answer

$$\begin{aligned}
& I=\frac{V}{R_{	ext {eq }}} I=\frac{V}{R_{	ext {eq }}}=\frac{9}{1+\frac{12 	imes 4}{12+4}}=\frac{9}{4} \
& I_1=\frac{9}{4} 	imes \frac{4}{16}=\frac{9}{16} \
& V_A-V_B=I_1 	imes 8=\frac{9}{16} 	imes 8=\frac{9}{2} V \
& 	herefore U=\frac{1}{2} 	imes 4 	imes \frac{81}{4} \mu J \
& 	herefore U=\frac{81}{2} \mu J \
& 	herefore x=81
\end{aligned}$$

Electrostatics & Capacitance

A $16 \Omega wire is bend to form a square loop. A 9 \mathrm{~V} battery with internal resistance 1 \Omega is connected across one of its sides. If a 4 \mu F capacitor is connected across one of its diagonals, the energy stored by the capacitor will be \frac{x}{2} \mu J, where x=$ _________

Integer / Numeric Answer Type