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Q.
In the circuit shown in figure heat developed across $2\,\Omega$, $4\,\Omega$ and $3\,\Omega$ resistances are in the ratio of
Current Electricity
Solution:
Current through $2\,\Omega, I_{1}=\frac{2I}{3}$
Heat produced per second,
$H_{1}=I^{2}_{1} \times 2=\left(\frac{2I}{3}\right)^{2} \times 2$
$=\frac{8I^{2}}{9}$
Current through $4\,\Omega, I_{2}=\frac{I}{3}$
Heat produced per second
$H_{2}=I^{2}_{2} \times4=\left(\frac{I}{3}\right)^{2} \times 4$
$=\frac{4I^{2}}{9}$
Current through $3\,\Omega$, $=I$
Heat produced $H_{3}=I^{2} \times 3$
$=3I^{2}=\frac{27I^{2}}{9}$
$\therefore H_{1} : H_{2} : H_{3}=8 : 4: 27$.
