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What is the power dissipated in a resistor of 5Ω carrying a current of 2A?
What is the power dissipated in a resistor of 5Ω carrying a current of 2A?
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Q1
What is the power dissipated in a resistor of 5Ω carrying a current of 2A?
5W
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Power is given by P = I^2 * R = (2A)^2 * 5Ω = 4A^2 * 5Ω = 20W.
Questions & Step-by-step Solutions
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Q
Q: What is the power dissipated in a resistor of 5Ω carrying a current of 2A?
Solution:
Power is given by P = I^2 * R = (2A)^2 * 5Ω = 4A^2 * 5Ω = 20W.
Steps: 7
Show Steps
Step 1: Identify the values given in the problem. We have a resistor with a resistance of 5 ohms (Ω) and a current of 2 amperes (A).
Step 2: Recall the formula for power (P) in a resistor, which is P = I^2 * R.
Step 3: Substitute the current (I) and resistance (R) into the formula. Here, I = 2A and R = 5Ω.
Step 4: Calculate I^2. This means multiplying the current by itself: (2A) * (2A) = 4A^2.
Step 5: Now multiply the result from Step 4 by the resistance: 4A^2 * 5Ω.
Step 6: Perform the multiplication: 4 * 5 = 20.
Step 7: The final result is 20 watts (W), which is the power dissipated in the resistor.
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