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

30

$I_1==\frac{I \times 15}{5+6+9}=\frac{3}{4} I $               (i)

and $I_2=\frac{I \times 5}{5+6+9}=\frac{1}{4} I$              (ii)

Heat development in $5 \Omega$ per second

$=\mathrm{I}^2 \times 5=42 \mathrm{Js}^{-1}$     (Given)

∴  $\mathrm{I}_1^2=\frac{42}{5}$                (iii)

From (i) and (iii), we get

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

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

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