JEE Main 2023 · 25 January · evening · Numerical answer · +4 / −0
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3Ω. Cell 2 has emf of 6V and internal resistance of 6Ω. An external resistor R of 4Ω is connected across A and B. The current flowing through R will be ____________ A.
Answer and solution
Answer 1
Solution
Physics sir · 4 steps
Check the plates first: the two cells face opposite ways, so their emfs fight instead of adding.
- Read the polarity. The longer plate is the positive terminal. Cell 1 has its positive side towards A, while cell 2 has its positive side towards B. So the two cells are in parallel but opposing each other.
- Replace by one cell. Two cells in parallel act like a single cell. Take the potential of A over B as positive, so cell 1 counts as +12 V and cell 2 as -6 V.
E_(eq) = (E₁/r₁ - E₂/r₂)/(1/r₁ + 1/r₂) = (12/3 - 6/6)/(1/3 + 1/6) = 3/0.5 = 6 V
- Internal resistance. The internal resistances combine in parallel: r_(eq) = (3 × 6)/(3 + 6) = 2 Ω.
- So. With R = 4 Ω across A and B, I = E_(eq)/(r_(eq) + R) = 6/(2 + 4) = 1 A.
The common slip: 1.67. Treating both cells as facing the same way gives E_(eq) = 10 V and a current of about 1.67 A, not 1 A.
The idea to keep. For cells in parallel, E_(eq) = Σ (E/r) ÷ Σ (1/r). Give each E a sign from which terminal faces the same point, and combine the internal resistances in parallel.
Answer from the NTA final answer key. Solution by PrepWiser.