Question 3-10B3

In a parallel circuit, the voltage across all components is the same. To determine the maximum voltage for the entire circuit, we must first calculate the maximum voltage each individual resistor can withstand without exceeding its power rating. The relationship between power (P), voltage (V), and resistance (R) is given by $P = V^2 / R$. Rearranging this, we find $V = \sqrt{P \times R}$. 1. **For the 500-ohm, 2-watt resistor:** $V_1 = \sqrt{2 \text{ W} \times 500 \text{ } \Omega} = \sqrt{1000} \approx 31.62 \text{ volts}$. 2. **For the 1500-ohm, 1-watt resistor:** $V_2 = \sqrt{1 \text{ W} \times 1500 \text{ } \Omega} = \sqrt{1500} \approx 38.73 \text{ volts}$. Since both resistors are in parallel, the same voltage is applied across each. To ensure that *neither* resistor exceeds its wattage rating, the applied voltage must be limited to the lowest of the individual maximum voltages. If we apply 38.7 volts (option C), the 500-ohm resistor would dissipate $(38.7^2 / 500) \approx 3 \text{ watts}$, exceeding its 2-watt rating. Therefore, the maximum voltage that can be applied across the parallel circuit is 31.6 volts, as this is the limit for the 500-ohm resistor. At this voltage, the 1500-ohm resistor would dissipate $(31.6^2 / 1500) \approx 0.66 \text{ watts}$, which is well within its 1-watt rating.