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Difficulty: MediumElectrical Power and Watt's Law

An industrial direct-current (DC) indicator lamp dissipates 1.8 W1.8\text{ W} of power when a current of 150 mA150\text{ mA} flows through it. What is the voltage drop across the lamp, in volts?

Answer: 12 V

Answer

The voltage drop across the indicator lamp is 12 V12\text{ V}.
According to Watt's Law, electrical power is the product of voltage and current (P=V×IP = V \times I). Rearranging the formula to solve for voltage gives V=PIV = \frac{P}{I}. Converting 150 mA150\text{ mA} to 0.15 A0.15\text{ A} and evaluating 1.8 W0.15 A\frac{1.8\text{ W}}{0.15\text{ A}} yields 12 V12\text{ V}.

Step-by-Step Solution

1
Convert the current from milliamperes to amperes.
I=150 mA=0.15 AI = 150\text{ mA} = 0.15\text{ A}
Standard electrical formulas require current in base units of amperes (A).
2
Rearrange Watt's Law to solve for voltage.
V=PIV = \frac{P}{I}
Watt's Law defines power as P=V×IP = V \times I, so isolating voltage yields V=PIV = \frac{P}{I}.
3
Substitute the known values into the equation to find the voltage.
V=1.8 W0.15 A=12 VV = \frac{1.8\text{ W}}{0.15\text{ A}} = 12\text{ V}
Dividing the power dissipation (1.8 W1.8\text{ W}) by the current (0.15 A0.15\text{ A}) gives the operating voltage.

Key Concept

Watt's Law describes the mathematical relationship between electrical power, voltage, and current: P=V×IP = V \times I. When power and current are known, voltage is calculated using V=PIV = \frac{P}{I}, provided current is expressed in amperes.
Estimated Time:1m 0s
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Electrical Power and Watt's Law

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