Physics-(general)

NEET Physics (general) MCQ Question

Type: MCQ-numerical-Easy-Class 12

Two cells with emfs ε1=1.5 V\varepsilon_1 = 1.5 \text{ V} and ε2=1.0 V\varepsilon_2 = 1.0 \text{ V} and internal resistances r1=0.5Ωr_1 = 0.5 \Omega and r2=0.3Ωr_2 = 0.3 \Omega are connected in series. What is the equivalent emf of the combination?

A

5 V

B

0 V

C

5 V

D

0 V

Correct Answer

Option A

Detailed Explanation

The equivalent emf for cells in series is the sum of the individual emfs: εeq=ε1+ε2=1.5+1.0=2.5 V\varepsilon_{eq} = \varepsilon_1 + \varepsilon_2 = 1.5 + 1.0 = 2.5 \text{ V}.

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