Refer to the circuit diagram given below. The heat generated across the $6 \Omega$ resistance in 100 second is $\frac{\alpha}{100} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$. (Nearest integer)

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Current Electricity

Refer to the circuit diagram given below. The heat generated across the

6Ω

resistance in 100 second is

\frac{α}{100} J

. The value of

α

____

. (Nearest integer)

Answer: 3477

Let the junction where the

3 Ω

6 Ω

and

4 Ω

branches meet be at potential

x

, with the lower rail as reference. The branch currents leaving the junction sum to zero:

\frac{x - 3}{6} + \frac{x - 2}{3} + \frac{x}{4} = 0

, the

3

V and

2

V sources setting the far ends of the

6 Ω

and

3 Ω

branches. Clearing denominators,

2 (x - 3) + 4 (x - 2) + 3 x = 0 \Rightarrow 9 x = 14

, so

x = \frac{14}{9}

V. The current in the

6 Ω

branch is

i = \frac{3 - x}{6} = \frac{3 - 14/9}{6} = \frac{13}{54}

A. The heat in 100 s is

H = i^{2} R t = (\frac{13}{54})^{2} \times 6 \times 100 = 34.77

= \frac{α}{100}

. Hence

α = 3477

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Integer answer