Physics

Physics

Problem

A car battery that is initially fully charged has an emf of 12.0 V and an internal resistance of 0.30 Ω. After prolonged use, its emf falls to 11.0 V and its internal resistance rises to 0.60 Ω. The battery supplies a headlamp rated at 24 Ω. a. (i) For the fully charged battery, calculate the current drawn by the lamp and the power delivered to the lamp. b. (ii) For the discharged battery, calculate the current and lamp power. c. (iii) For each of the two cases, determine the rate at which energy is dissipated as heat inside the battery. d. (iv) Compare the ratios of useful lamp power to total power supplied by the battery in the two cases and comment briefly on the efficiency change as the battery discharges.
Intuitive explanation

The key here is understanding how a battery's terminal voltage depends on both its emf and internal resistance. When the battery ages, both its emf and internal resistance change—and both effects matter for how bright the headlamp actually gets. You'll need to use Ohm's law for the complete circuit (Vterminal=E−IrV_{\text{terminal}} = \mathcal{E} - Ir) to find the current, then determine how the lamp's brightness changes between the fresh and aged states.

Intuitive explanation

The key here is understanding how a battery's terminal voltage depends on both its emf and internal resistance. When the battery ages, both its emf and internal resistance change—and both effects matter for how bright the headlamp actually gets. You'll need to use Ohm's law for the complete circuit (Vterminal=E−IrV_{\text{terminal}} = \mathcal{E} - Ir) to find the current, then determine how the lamp's brightness changes between the fresh and aged states.

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