Physics-Kirchhoff's Laws

NEET Physics Kirchhoff's Laws MCQ Question

Type: MCQ-numerical-Medium-Class 12

In a circuit, two cells with emfs ε1=1.5 V\varepsilon_1 = 1.5 \text{ V} and ε2=2.0 V\varepsilon_2 = 2.0 \text{ V} and internal resistances r1=0.5Ωr_1 = 0.5 \Omega and r2=0.5Ωr_2 = 0.5 \Omega are connected in parallel. What is the equivalent emf (εeq\varepsilon_{eq}) of the combination?

A

75 V

B

00 V

C

50 V

D

50 V

Correct Answer

Option A

Detailed Explanation

Using the formula for parallel cells from the NCERT context, εeq=(ε1r2+ε2r1)/(r1+r2)\varepsilon_{eq} = (\varepsilon_1 r_2 + \varepsilon_2 r_1) / (r_1 + r_2), we calculate εeq=(1.5×0.5+2.0×0.5)/(0.5+0.5)=1.75 V\varepsilon_{eq} = (1.5 \times 0.5 + 2.0 \times 0.5) / (0.5 + 0.5) = 1.75 \text{ V}.

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