Question

The friends are ready to try a problem. A battery has an emf of 12.0 V and an internal resistance of 0.05 Ω. Its terminals are connected to a load resistance of 1.1 Ω. Find the terminal voltage of the battery. V Calculate the power delivered to the load resistor, the power delivered to the internal resistance of the battery, and the power delivered by the battery. PR = W Or = W Delivered by battery = W

Answer 1

Answer:

Terminal voltage = 11.5 V

Power across load = 119.7 W

Power across internal resistance = 5.4 W

Power delivered by battery = 125.2 W

Explanation:

The current in the circuit is given by

$I=\dfrac{E}{R+r}$

where E is the emf of the battery, R us the load resistance and r is the internal resistance.

$I=\dfrac{12}{1.1+0.05} = 10.43 \text{ A}$

The terminal voltage is the potential difference across the load resistor.

V = I × R = 10.43 A × 1.1 Ω = 11.5 V

The power across any resistance is given by $I^2R$

For the load resistor,

$P_L=10.43^2\times1.1 = 119.7 \text{ W}$

For the internal resistance,

$P_r=10.43^2\times0.05= 5.4\text{ W}$

The power delivered by the battery is

$P = P_L + P_r = 119.7 + 5.4 = 124.1 \text{ W}$

This could also be found by

$P = IE = 10.34\times12 = 125.2 \text{ W}$

The discrepancy in both answers is due to the approximations. The second answer is better.