Question

You want to do an experiment with a series of resistors at a constant voltage. The resistors have vales of 187, 210, 347, 458, & 577 Ω. Each can handle a maximum of 0.350 watts. What maximum constant voltage can you use? (Note that in practice you would probably use a little lower voltage t o be safe.)

Answer 1

To find the maximum constant voltage that can be used in the experiment, you can calculate the power dissipated by each resistor using the formula P = V^2 / R. Then, rearrange the formula to solve for V. By plugging in the maximum power of 0.350 watts for each resistor, you can calculate the corresponding voltage. The maximum constant voltage that can be used is approximately 12.80 volts.

Step-by-step explanation:

To find the maximum constant voltage that can be used, we need to determine the power dissipated by each resistor. The power can be calculated using the formula P = V^2 / R, where P is the power, V is the voltage, and R is the resistance. We want to find the maximum voltage that will keep the power below 0.350 watts for all resistors.

1. For the first resistor with a resistance of 187 Ω, P = (V^2) / 187 = 0.350. Rearranging the equation to solve for V, we get V = sqrt(0.350 * 187) = 6.38 V.
2. For the second resistor with a resistance of 210 Ω, P = (V^2) / 210 = 0.350. Rearranging the equation to solve for V, we get V = sqrt(0.350 * 210) = 7.03 V.
3. For the third resistor with a resistance of 347 Ω, P = (V^2) / 347 = 0.350. Rearranging the equation to solve for V, we get V = sqrt(0.350 * 347) = 9.30 V.
4. For the fourth resistor with a resistance of 458 Ω, P = (V^2) / 458 = 0.350. Rearranging the equation to solve for V, we get V = sqrt(0.350 * 458) = 10.92 V.
5. For the fifth resistor with a resistance of 577 Ω, P = (V^2) / 577 = 0.350. Rearranging the equation to solve for V, we get V = sqrt(0.350 * 577) = 12.80 V.

Therefore, the maximum constant voltage that can be used is approximately 12.80 volts.