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In the circuit shown in figure, if the resistance $5\Omega$ develops a heat $42\mathrm{J}$ per second, the heat developed in $2\Omega$ must be about $\mathrm{(in\: Js^{-1})}$ :

Option: 1
$25$

Option: 2
$20$

Option: 3
$30$

Option: 4
$35$

Answers (1)

best_answer

$\mathrm{I}_1=\frac{\mathrm{I} \times 15}{5+6+9}=\frac{3}{4} \mathrm{I}$ (i)

and $\mathrm{I}_2=\frac{\mathrm{I} \times 5}{5+6+9}=\frac{1}{4} \mathrm{I}$ (ii)

Heat development in $5 \Omega$ per second

$\begin{aligned} & =\mathrm{I}_1^2 \times 5=42 \mathrm{Js}^{-1} \\ \end{aligned}$ (Given)

$\begin{aligned} \therefore \quad \mathrm{I}_1^2=\frac{42}{5} \end{aligned}$ (iii)

From (i) and (iii), we get

$\mathrm{I}^2=\frac{16 \times 42}{9 \times 5}$

Heat developed in $2 \Omega$ per second $=\mathrm{I}^2 \times 2$

$=\frac{16 \times 42 \times 2}{9 \times 5} \approx 30 \mathrm{Js}^{-1}$

Hence option 3 is correct.

Posted by

Divya Prakash Singh

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