To determine the electrical power (\( P \)) when voltage (\( V \)) and resistance (\( R \)) are known, use the formula:
\[ P = \dfrac{V^2}{R} \]
where:
- \( P \) is the power (in watts, W),
- \( V \) is the voltage (in volts, V),
- \( R \) is the resistance (in ohms, Ω).
Problem 1: Power of a Light Bulb
Scenario: A light bulb operates at \( 120 \, \text{V} \) and has a resistance of \( 240 \, \Omega \). What is the power consumed by the bulb?
Calculation:
1. Given:
\[ V = 120 \, \text{V} \]
\[ R = 240 \, \Omega \]
2. Substitute into the Power Formula:
\[ P = \dfrac{V^2}{R} \]
\[ P = \dfrac{(120)^2}{240} \]
3. Calculate:
\[ P = \dfrac{14400}{240} = 60 \, \text{W} \]
Answer: The power consumed by the bulb is \( 60 \, \text{W} \).
Problem 2: Power of an Electric Heater
Scenario: An electric heater operates with a voltage of \( 230 \, \text{V} \) and has a resistance of \( 50 \, \Omega \). Determine the power output.
Calculation:
1. Given:
\[ V = 230 \, \text{V} \]
\[ R = 50 \, \Omega \]
2. Substitute into the Power Formula:
\[ P = \dfrac{V^2}{R} \]
\[ P = \dfrac{(230)^2}{50} \]
3. Calculate:
\[ P = \dfrac{52900}{50} = 1058 \, \text{W} \]
Answer: The power output of the heater is \( 1058 \, \text{W} \).
Problem 3: Power of a Motor
Scenario: A motor is powered by \( 440 \, \text{V} \) and has a resistance of \( 10 \, \Omega \). What is the power consumed by the motor?
Calculation:
1. Given:
\[ V = 440 \, \text{V} \]
\[ R = 10 \, \Omega \]
2. Substitute into the Power Formula:
\[ P = \dfrac{V^2}{R} \]
\[ P = \dfrac{(440)^2}{10} \]
3. Calculate:
\[ P = \dfrac{193600}{10} = 19360 \, \text{W} \]
Answer: The power consumed by the motor is \( 19360 \, \text{W} \).
