Refer to the circuit diagram given below. The heat generated across the 6 Ω resistance in 100 second is (α)/(100) J. The value of α is ______. (Nearest integer) Numerical answer type. Work it out, then open the answer.

Answer: 3477. Take the right node as 0 V. The left end of 3 Ω is at 2 V and the 3 V cell puts the end of 6 Ω at 3 V. KCL at the junction X: (V - 2)/(3) + V/4 + (V - 3)/(6) = 0 ⇒ 9V = 14, V = 14/9 V. Current in 6 Ω: (3 - 14/9)/(6) = 13/54 A. H = I² R t = (13/54)² × 6 × 100 ≈ 34.77 J = (3477)/(100) J. So α = 3477.

Refer to the circuit diagram given below. The heat generated across… | JEE Main 2026 (6 Apr, Shift 1)

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Electrical Power & Heating Effect

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

Numerical answer type. Work it out, then open the answer.

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Answer: 3477

Take the right node as 0 V. The left end of $3\ \Omega$ is at 2 V and the $3\text{ V}$ cell puts the end of $6\ \Omega$ at 3 V.
KCL at the junction $X$ : $\dfrac{V - 2}{3} + \dfrac{V}{4} + \dfrac{V - 3}{6} = 0 \Rightarrow 9V = 14,\ V = \dfrac{14}{9}\text{ V}$ .
Current in $6\ \Omega$ : $\dfrac{3 - 14/9}{6} = \dfrac{13}{54}\text{ A}$ .
$H = I^2 R t = \left(\dfrac{13}{54}\right)^2 \times 6 \times 100 \approx 34.77\text{ J} = \dfrac{3477}{100}\text{ J}$ .
So $\alpha = 3477$ .

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