CWEA Maintenance Technologist Practice Test

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Prepare for the CWEA Maintenance Technologist Test. Enhance your skills with flashcards and multiple choice questions. Understand each topic with detailed explanations. Get ready for success!

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If 17.32 volts are applied across a resistance of 50 ohms, what is the power dissipated in the resistor?

4.5 watts
3.5 watts
6.0 watts
5.0 watts

The correct answer is: 6.0 watts

To determine the power dissipated in the resistor, you can use the formula for electrical power, which is given by $ P = \frac{V^2}{R} $, where $ P $ is the power in watts, $ V $ is the voltage in volts, and $ R $ is the resistance in ohms. In this scenario, you are given the voltage $ V = 17.32 $ volts and the resistance $ R = 50 $ ohms. Plugging in these values into the formula: 1. First, square the voltage: \[ V^2 = (17.32)^2 = 300.8624 \] 2. Next, divide this value by the resistance: \[ P = \frac{300.8624}{50} = 6.017248 \] Rounding this to a suitable degree of precision gives the power dissipated in the resistor as approximately 6.0 watts. Therefore, the correct answer is 6.0 watts. This shows how power dissipation in a resistive component can be calculated directly from voltage and resistance, highlighting the fundamental relationship between these electrical parameters.