Question

In the circuit shown in figure heat developed across $2\Omega$, $4\Omega$ and $3\Omega$ resistances are in the ratio of

• A. $2:4:3$
• B. $8:4:12$
• C. $4:8:27$
• D. $8:4:27$

Answer 1

Answer: (D)

Current through $2\Omega ,I_1=\frac{2I}{3}$
Heat produced per second,
$H_1=I_1^2\times 2={\left(\frac{2I}{3}\right)}^2\times 2$
$=\frac{8I^2}{9}$
Current through $4\Omega ,I_2=\frac{I}{3}$
Heat produced per second
$H_2=I_2^2\times 4={\left(\frac{I}{3}\right)}^2\times 4$
$=\frac{4I^2}{9}$
Current through $3\Omega$, $=I$
Heat produced $H_3=I^2\times 3$
$=3I^2=\frac{27I^2}{9}$
$\therefore H_1:H_2:H_3=8:4:27$.