To determine the current (\( I \)) when power (\( P \)) and resistance (\( R \)) are known, use the formula:
\[ I = \sqrt{\dfrac{P}{R}} \]
where:
- \( I \) is the current (in amperes, A),
- \( P \) is the power (in watts, W),
- \( R \) is the resistance (in ohms, Ω).
Problem 1: Current in a Light Bulb
Scenario: A light bulb dissipates \( 60 \, \text{W} \) of power and has a resistance of \( 240 \, \Omega \). What is the current through the light bulb?
Calculation:
1. Given:
\[ P = 60 \, \text{W} \]
\[ R = 240 \, \Omega \]
2. Substitute into the Current Formula:
\[ I = \sqrt{\dfrac{P}{R}} \]
\[ I = \sqrt{\dfrac{60}{240}} \]
3. Calculate:
\[ I = \sqrt{0.25} = 0.5 \, \text{A} \]
Answer: The current through the light bulb is \( 0.5 \, \text{A} \).
Problem 2: Current Through a Heater
Scenario: A heater operates at \( 1500 \, \text{W} \) of power and has a resistance of \( 25 \, \Omega \). Determine the current flowing through the heater.
Calculation:
1. Given:
\[ P = 1500 \, \text{W} \]
\[ R = 25 \, \Omega \]
2. Substitute into the Current Formula:
\[ I = \sqrt{\dfrac{P}{R}} \]
\[ I = \sqrt{\dfrac{1500}{25}} \]
3. Calculate:
\[ I = \sqrt{60} \approx 7.75 \, \text{A} \]
Answer: The current flowing through the heater is approximately \( 7.75 \, \text{A} \).
Problem 3: Current in an Electric Fan
Scenario: An electric fan uses \( 80 \, \text{W} \) of power and has a resistance of \( 50 \, \Omega \). What is the current in the fan?
Calculation:
1. Given:
\[ P = 80 \, \text{W} \]
\[ R = 50 \, \Omega \]
2. Substitute into the Current Formula:
\[ I = \sqrt{\dfrac{P}{R}} \]
\[ I = \sqrt{\dfrac{80}{50}} \]
3. Calculate:
\[ I = \sqrt{1.6} \approx 1.26 \, \text{A} \]
Answer: The current in the electric fan is approximately \( 1.26 \, \text{A} \).
