A 12.0 V battery with internal resistance 0.50 Ω supplies a 5.50 Ω load. (a) Calculate total resistance and current. [3] (b) Calculate terminal potential difference and show that it matches the load voltage. [3] (c) Calculate power in the load, power dissipated inside the battery and efficiency of transfer to the load. [4] (d) Explain why the battery terminal voltage falls when current increases. [2]

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Total resistance is 6.00 Ω and current is 2.00 A. Terminal/load voltage is 11.0 V. Load power is 22.0 W, internal dissipation is 2.00 W and transfer efficiency is 91.7%. (a) The internal and external resistances are in series for the complete circuit, so R_total = 0.50 + 5.50 = 6.00 Ω. (a) Current is I = emf/R_total = 12.0/6.00 = 2.00 A. (b) Lost voltage inside the battery is Ir = 2.00 × 0.50 = 1.00 V, so terminal potential difference is 12.0 − 1.00 = 11.0 V. (b) Across the load, V = IR = 2.00 × 5.50 = 11.0 V, which independently checks the terminal-voltage calculation. (c) Load power is I²R = 2.00² × 5.50 = 22.0 W; internal power is I²r = 2.00² × 0.50 = 2.00 W. Total 24.0 W also equals emf × current. (c) Efficiency to the load is 22.0/24.0 × 100 = 91.7%. The remainder heats the battery internally. (d) A larger current produces a larger internal drop Ir, so terminal voltage emf − Ir falls even if the emf is approximately unchanged.

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