Determining the electric current when the power and resistance are known is crucial for various electrical applications. This relationship is expressed through a specific formula, which we will explore with practical examples.
The Formula: \( P = I^2 \cdot R \)
To find the current, we rearrange the formula as follows:
\[ I = \sqrt{\dfrac{P}{R}} \]
Where:
- \( P \) is the electric power (measured in watts, W)
- \( I \) is the current (measured in amperes, A)
- \( R \) is the resistance (measured in ohms, \(\Omega\))
Example 1: Current of a Refrigerator
Question: A refrigerator consumes 200 watts of power and has a resistance of 60 ohms. What is the current?
Given:
- \( P = 200 \) W
- \( R = 60 \) \(\Omega\)
Using the formula:
\[ I = \sqrt{\dfrac{P}{R}} = \sqrt{\dfrac{200}{60}} = \sqrt{3.33} \approx 1.83 \, \text{A} \]
Result: The current of the refrigerator is approximately 1.83 amperes.
Example 2: Current of a Washing Machine
Question: A washing machine operates at 1800 watts and has a resistance of 40 ohms. What is the current?
Calculation:
Given:
- \( P = 1800 \) W
- \( R = 40 \) \(\Omega\)
Using the formula:
\[ I = \sqrt{\dfrac{P}{R}} = \sqrt{\dfrac{1800}{40}} = \sqrt{45} \approx 6.71 \, \text{A} \]
Result: The current of the washing machine is approximately 6.71 amperes.
Example 3: Current of a Television
Question: A television consumes 150 watts of power and has a resistance of 120 ohms. What is the current?
Calculation:
Given:
- \( P = 150 \) W
- \( R = 120 \) \(\Omega\)
Using the formula:
\[ I = \sqrt{\dfrac{P}{R}} = \sqrt{\dfrac{150}{120}} = \sqrt{1.25} \approx 1.12 \, \text{A} \]
Result: The current of the television is approximately 1.12 amperes.
