Final answer:
A surge suppressor should be rated at a minimum of 200 joules for basic protection, but preferably 600 to 1000 joules for sensitive equipment. The correct voltage across a capacitor storing 10 J of energy, with a capacitance of 100 µF, is 447.21 V or 4.5 × 10^2 V.
Step-by-step explanation:
A surge suppressor should be rated to handle a specific amount of energy, which is expressed in joules. The energy rating communicates the device's ability to protect against power surges. For basic home electronics, a surge suppressor should be rated at a minimum of 200 joules.
However, for more sensitive and expensive devices, choosing a suppressor with a higher joule rating, such as 600 to 1000 joules or more, is recommended for better protection. In relation to the reference question provided, to find the voltage across a capacitor given the energy stored, you can use the formula for the energy stored in a capacitor, which is E = 1/2 C V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage across the capacitor.
Reorganizing the formula to solve for V, and substituting the given values: E = 1/2 C V^2, 2E = C V^2, V^2 = 2E / C, V = sqrt(2E / C). Using the given values: E = 10 J (joules), C = 100 µF = 100 x 10^-6 F (farads), V = sqrt(2 x 10 J / (100 x 10^-6 F)), V = sqrt(200000 V^2), V = 447.21 V. Therefore, the correct answer is: b. 4.5 × 10^2 V.