Question

In the circuit shown in figure heat developed across 2$15\Omega$, $15.18\Omega$ and $81.15\Omega$ 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)

8 : 4 : 27

Answer 2

: 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$H_{3} = I^{2} \times 3 = 3I^{2} = \frac{27I^{2}}{9}$

$\therefore H_{1}:H_{2}:H_{3} = 8:4:27.$