JEE Main 2021 · 24 Feb · Shift 1 · Physics Q12

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PhysicsSingle correct+4 / -1Question 12 of 90

A cell E $_1$ of emf 6 V and internal resistance 2 Ω is connected with another cell E $_2$ of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is :

Cells E1 (6 V, 2 Ω) between P and X and E2 (4 V, 8 Ω) between X and Y in a single loop, with the two cells in opposition.

2.0 V

3.6 V

5.6 V

10.0 V

Answer and worked solution

Official answer: C

Opposing cells. Net emf 2 V over 10 Ω gives 0.2 A, driven by E₁ and charging E₂.

Across E₂ (X to Y): $4+0.2\times8=5.6$ V.

(C)

Topics:EMF and internal resistance Kirchhoff's laws

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JEE Main 2021 · 24 Feb · Shift 1, Physics question 12