Refer to the circuit diagram given below. The heat generated across the 6 Ohm resistance in 100 second is alpha/100 J. The value of alpha is _______. (Nearest integer)

The correct answer is Option 3477. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q46 – JEE-MAIN 2026

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Home Physics Current ElectricityQuestion #53

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)

Enter Calculated Numerical Value:

Clue 1 of 3: Circuit Analysis

To find the heat generated in a resistor, you first need to determine the current flowing through it.

🧠 Full Solution Path

Step 1: Apply Nodal Analysis to find the potential at the common node.✦ Active

Let the potential of the bottom wire be $0 V$ . The potential at the node connected to the positive terminal of the $2 V$ battery (left of $3 Ω$ ) is $2 V$ . The potential at the node connected to the positive terminal of the $3 V$ battery (right of $6 Ω$ ) is $3 V$ . Let the potential of the central node (where $3 Ω$ , $6 Ω$ , and $4 Ω$ meet) be $V_{A}$ . Applying Kirchhoff's Current Law (KCL) at node $V_{A}$ :

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

Multiplying by 12 (LCM of 3, 6, 4):

4 (V_{A} - 2) + 2 (V_{A} - 3) + 3 V_{A} = 0

Simplifying the equation:

4 V_{A} - 8 + 2 V_{A} - 6 + 3 V_{A} = 0

⟹ 9 V_{A} - 14 = 0

⟹ V_{A} = \frac{14}{9} V

Step 2: Calculate the current through the

6 Ω

resistor.○ Expand

The current $I_{6 Ω}$ flowing from node $V_{A}$ towards the $3 V$ battery is:

I_{6 Ω} = \frac{V_{A} - 3}{6} = \frac{\frac{14}{9} - 3}{6} = \frac{\frac{14 - 27}{9}}{6} = \frac{- 13 / 9}{6} = - \frac{13}{54} A

The magnitude of the current through the $6 Ω$ resistor is $| I_{6 Ω} | = \frac{13}{54} A$ .

💡 Teacher's Secret Hint

Remember that the direction of current does not affect the heat generated, only its magnitude squared.

Step 3: Calculate the heat generated and the value of

α

.○ Expand

The heat generated $H$ in the $6 Ω$ resistor over time $t = 100$ seconds is given by Joule's Law: $H = I^{2} R t$ .

H = {(\frac{13}{54})}^{2} \times 6 \times 100 = \frac{169}{2916} \times 600 = \frac{101400}{2916} = \frac{8450}{243} J

The problem states that the heat generated is $\frac{α}{100}$ J. Therefore:

\frac{α}{100} = \frac{8450}{243}

α = \frac{8450 \times 100}{243} = \frac{845000}{243} \approx 3477.366

Rounding to the nearest integer, the value of $α$ is $3477$ .

💡 Teacher's Secret Hint

Pay close attention to the units and the form in which $α$ is requested.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 46 – JEE-MAIN 2026

Clue 1 of 3: Circuit Analysis

🧠 Full Solution Path