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In this circuit, when certain current flows, the heat produced in $5 \Omega$ is $4.05 \mathrm{~J}$ in a time t. The heat produced in $2 \Omega$ coil in the same time interval is
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The correct answer is:
$2.88$
Let, the current distribution is as shown, then by current divider rule,
$H_{1}=\frac{(6+9)}{(6+9)+5} I=\frac{3}{4} I$
Let $\mathrm{H}$ and $\mathrm{H}_{1}$ be heat produced in $2 \Omega$ and $5 \Omega$ resistors respectively in time t, then
$$
\begin{gathered}
\mathrm{H}=\mathrm{I}^{2}(2) \mathrm{t}=2 \mathrm{I}^{2} \mathrm{t} \\
\text { and } \mathrm{H}_{1}=\mathrm{I}_{1}^{2}(5) \mathrm{t}=\mathrm{II}_{1}^{2} \mathrm{t}=\frac{45}{16} \mathrm{I}^{2} \mathrm{t} \\
\therefore \quad \frac{\mathrm{H}}{\mathrm{H}_{\mathrm{I}}}=\frac{32}{45}=\frac{32}{45} \times 4.05 \mathrm{~J}=2.88 \mathrm{~J}
\end{gathered}
$$
$H_{1}=\frac{(6+9)}{(6+9)+5} I=\frac{3}{4} I$
Let $\mathrm{H}$ and $\mathrm{H}_{1}$ be heat produced in $2 \Omega$ and $5 \Omega$ resistors respectively in time t, then
$$
\begin{gathered}
\mathrm{H}=\mathrm{I}^{2}(2) \mathrm{t}=2 \mathrm{I}^{2} \mathrm{t} \\
\text { and } \mathrm{H}_{1}=\mathrm{I}_{1}^{2}(5) \mathrm{t}=\mathrm{II}_{1}^{2} \mathrm{t}=\frac{45}{16} \mathrm{I}^{2} \mathrm{t} \\
\therefore \quad \frac{\mathrm{H}}{\mathrm{H}_{\mathrm{I}}}=\frac{32}{45}=\frac{32}{45} \times 4.05 \mathrm{~J}=2.88 \mathrm{~J}
\end{gathered}
$$
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