Current Electricity

JEE Main level, numerical answer, medium

203 of 380

Question 203 of 380

Refer to the circuit diagram given below. The heat generated across the 6 Ω6\,\Omega resistance in 100100 second is α100\dfrac{\alpha}{100} J. The value of α\alpha is _______. (Nearest integer)

Show answer and solution

Answer: 3477

Taking the right-hand wire as the reference node at 0 V0\text{ V}, let the left junction (where the 3 Ω3\,\Omega, 4 Ω4\,\Omega and 6 Ω6\,\Omega resistors meet) be Node A at potential VAV_A.

Bottom branch: The positive terminal of the 2 V2\text{ V} battery is at the reference, so its negative terminal (and hence the left end of the 3 Ω3\,\Omega resistor) is at −2 V-2\text{ V}.
Top branch: The positive terminal of the 3 V3\text{ V} battery is at the reference, so its negative terminal (and hence the node between the 6 Ω6\,\Omega resistor and the 3 V3\text{ V} battery) is at −3 V-3\text{ V}.

Applying KCL at Node A (sum of currents leaving the node is zero):
VA−04+VA−(−2)3+VA−(−3)6=0\dfrac{V_A - 0}{4} + \dfrac{V_A - (-2)}{3} + \dfrac{V_A - (-3)}{6} = 0

VA4+VA+23+VA+36=0\dfrac{V_A}{4} + \dfrac{V_A + 2}{3} + \dfrac{V_A + 3}{6} = 0

Multiplying throughout by 1212:
3VA+4(VA+2)+2(VA+3)=03V_A + 4(V_A + 2) + 2(V_A + 3) = 0

3VA+4VA+8+2VA+6=03V_A + 4V_A + 8 + 2V_A + 6 = 0

9VA+14=0⇒VA=−149 V9V_A + 14 = 0 \Rightarrow V_A = -\dfrac{14}{9}\text{ V}

Current through the 6 Ω6\,\Omega resistor (from Node A towards the −3 V-3\text{ V} point):
I6=VA−(−3)6=−149+36=1396=1354 AI_6 = \dfrac{V_A - (-3)}{6} = \dfrac{-\dfrac{14}{9} + 3}{6} = \dfrac{\dfrac{13}{9}}{6} = \dfrac{13}{54}\text{ A}

Heat generated in the 6 Ω6\,\Omega resistor in t=100 st = 100\text{ s}:
H=I62⋅R⋅t=(1354)2⋅6⋅100H = I_6^2 \cdot R \cdot t = \left(\dfrac{13}{54}\right)^2 \cdot 6 \cdot 100

H=1692916⋅600=169⋅100486=16900486 JH = \dfrac{169}{2916} \cdot 600 = \dfrac{169 \cdot 100}{486} = \dfrac{16900}{486}\text{ J}

Given H=α100 JH = \dfrac{\alpha}{100}\text{ J}:
α100=16900486\dfrac{\alpha}{100} = \dfrac{16900}{486}

α=16900⋅100486=1690000486≈3477.37\alpha = \dfrac{16900 \cdot 100}{486} = \dfrac{1690000}{486} \approx 3477.37

Rounding to the nearest integer:
α=3477\alpha = 3477

Hence, the answer is 34773477.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Circuit builder simulationDrag the sliders and watch the numbers in this question change.Revise Current Electricity firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Current Electricity (Kirchoffs Laws), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.