JEE Main 2026 Apr 6 Shift 1, Physics Q21: Energy, Power And Heating Effect

JEE Main2026Apr 6, Shift 1Physics

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

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Solution

Label the junction between the $3 Ω$ resistor and the parallel combination as node X (voltage $x$ ); ground the opposite rail at $0$ V.

Applying KCL at node X with conventions as in the figure:

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

Multiply through by $12$ : $2 (x - 3) + 4 (x - 2) + 3 x = 0 \Rightarrow 9 x - 14 = 0 \Rightarrow x = \frac{14}{9}$ V.

Current through $6 Ω$ branch: $I_{6} = \frac{3 - x}{6} = \frac{3 - 14/9}{6} = \frac{13/9}{6} = \frac{13}{54}$ A.

Heat dissipated in $100$ s: $H = I_{6}^{2} R t = (\frac{13}{54})^{2} \cdot 6 \cdot 100 = \frac{169}{2916} \cdot 600 \approx 34.77$ J.

$H = \frac{α}{100} \Rightarrow α = 3477$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q21: Energy, Power And Heating Effect

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