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In the circuit diagram, heat produces in $R, 2 R$ and $1.5 R$ are in the ratio of
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Verified Answer
The correct answer is:
$8: 4: 27$
We have, $l_{1}=\frac{I \times 2 R}{3 R}=\frac{2 I}{3}$
$\begin{array}{ll}
\therefore & H_{1}=l_{1}^{2} R=\frac{4 l^{2}}{9} \times R...(i) \\
\text { Also, } & I_{2}=\frac{I \times R}{3 R}=\frac{l}{3} \\
\therefore & H_{2}=I_{2}^{2}(2 R)=\frac{l^{2}}{9} \times 2 R...(ii) \\
\text { and } & H_{3}=I^{2}(1.5 R)...(iii)
\end{array}$
From Eqs. (i), (ii) and (iii),
$\begin{aligned}
H_{1}: H_{2}: H_{3} &=\frac{4 l^{2}}{9} \times R: \frac{I^{2}}{9} \times 2 R: I^{2} \times 1.5 R \\
&=\frac{4}{9}: \frac{2}{9}: 1.5=4: 2: 13.5 \\
&=8: 4: 27
\end{aligned}$
$\begin{array}{ll}
\therefore & H_{1}=l_{1}^{2} R=\frac{4 l^{2}}{9} \times R...(i) \\
\text { Also, } & I_{2}=\frac{I \times R}{3 R}=\frac{l}{3} \\
\therefore & H_{2}=I_{2}^{2}(2 R)=\frac{l^{2}}{9} \times 2 R...(ii) \\
\text { and } & H_{3}=I^{2}(1.5 R)...(iii)
\end{array}$
From Eqs. (i), (ii) and (iii),
$\begin{aligned}
H_{1}: H_{2}: H_{3} &=\frac{4 l^{2}}{9} \times R: \frac{I^{2}}{9} \times 2 R: I^{2} \times 1.5 R \\
&=\frac{4}{9}: \frac{2}{9}: 1.5=4: 2: 13.5 \\
&=8: 4: 27
\end{aligned}$
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