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Two cells, each of emf 1.5 V and internal resistance 0.5 Ω, are connected in series across a 2 Ω resistor. The current through the resistor is:
A1.0 A
B1.5 A
C3.0 A
D0.75 A
Answer & Solution
Correct answer: A. 1.0 A
Series cells: ε_eq = 1.5 + 1.5 = 3 V and r_eq = 0.5 + 0.5 = 1 Ω, so I = ε_eq/(R + r_eq) = 3/(2+1) = 1.0 A.
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