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Difficulty: MediumElectrical Power

In a naval shipboard direct-current monitoring circuit, a 75 Ω75\text{ }\Omega dummy load resistor dissipates electrical energy during transmitter testing. If a steady current of 4 A4\text{ A} flows through this resistor, what is the total electrical power dissipated by the resistor, in watts?

Answer: 1200 W

Answer

The total electrical power dissipated by the resistor is 1200 W1200\text{ W}.
According to Watt's Law, power dissipated in a resistor when current and resistance are known is calculated using the formula P=I2RP = I^2 R. Squaring the current (4 A4\text{ A}) gives 16 A216\text{ A}^2, and multiplying by the resistance of 75 Ω75\text{ }\Omega produces an electrical power dissipation of 1200 W1200\text{ W}.

Step-by-Step Solution

1
Identify the known circuit parameters
Current I=4 AI = 4\text{ A} and resistance R=75 ΩR = 75\text{ }\Omega
These are the two independent circuit variables required to calculate electrical power using Watt's Law.
2
Apply Watt's power formula relating current and resistance
P=I2×RP = I^2 \times R
When voltage is not directly given, electrical power can be found directly from the square of the current multiplied by the resistance.
3
Compute the numerical power dissipation
P=(4 A)2×75 Ω=16×75=1200 WP = (4\text{ A})^2 \times 75\text{ }\Omega = 16 \times 75 = 1200\text{ W}
Evaluating (4)2=16(4)^2 = 16 and multiplying by 7575 yields the total power in watts.

Key Concept

Electrical power dissipation in a resistive component using Watt's Law (P=I2RP = I^2 R)
Estimated Time:1m 0s
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Topics

Electrical Power

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