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Question:

Two batteries with e.m.f. 12V and 13V are connected in parallel across a load resistor of 10Ω. The internal resistance of the two batteries are 1Ω and 2Ω respectively. The voltage across the load lies between:

11.4V and 11.5V

11.5V and 11.6V

11.7V and 11.8V

11.6V and 11.7V

Solution:

Equivalent resistance of r1 and r2, req = 1 × 2 / (1 + 2) = 0.66Ω
Equivalent emf of E1 and E2, Eeq = [E1r2 + E2r1] / req ⇒ Eeq = [12 × 2 + 13 × 1] × 0.66 = 12.33 V
Potential across R, VR = R / (R + req) × Eeq ⇒ VR = 10 / (10 + 0.66) × 12.33 = 11.57 V