A real battery can be modeled as an ideal emf source in series with an internal resistance

PHYSICS · Multi-part · CBSE · Physics · Current Electricity · Current Electricity

Problem

A real battery can be modeled as an ideal emf source E=12.0 V\mathcal E = 12.0\,\text{V} in series with an internal resistance r=0.50 Ωr = 0.50\,\Omega. The battery is connected to the external circuit described below. • Resistors R1=2.0 ΩR_1 = 2.0\,\Omega and R2=3.0 ΩR_2 = 3.0\,\Omega are connected in parallel. • That parallel combination is connected in series with resistor R3=4.0 ΩR_3 = 4.0\,\Omega. Treat all connecting wires as having negligible resistance. (a) Determine the equivalent resistance of the external circuit (the combination of R1, R2,R_1,\,R_2, and R3R_3). (b) Calculate the current leaving the battery. (c) Find the terminal potential difference (the potential difference measured across the battery’s terminals). (d) Determine the potential difference across R1R_1 and the current through R1R_1. (e) Express the percentage of the battery’s emf that is lost across its internal resistance when the circuit is operating, and briefly comment on the practical significance of that energy loss.

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