Question 12

A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is:

$$P_{\text{ext}} = V \cdot I \implies 0.5 = 2.5 \cdot I \implies I = \frac{0.5}{2.5} = 0.2\ \text{A}$$

Voltage drop across the internal resistance:

$$V_{\text{int}} = E - V = 3.0 - 2.5 = 0.5\ \text{V}$$

Power dissipated within the internal resistance:

$$P_{\text{int}} = V_{\text{int}} \cdot I = 0.5\ \text{V} \times 0.2\ \text{A} = 0.10\ \text{W}$$

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