An aircraft pitot tube heating element has an internal electrical resistance of 40\text{ }… — Armed Services Vocational Aptitude Battery (ASVAB)

An aircraft pitot tube heating element has an internal electrical resistance of $40\text{ }\Omega$ and connects directly to a $120\text{ V}$ direct-current (DC) power system. What is the electrical power dissipated by the heater, in watts ( $\text{W}$ )?

Answer: 360 W

Answer

The heating element dissipates

360\text{ W}

of electrical power.

Electrical power (

P

) dissipated by a resistive element when voltage (

V

) and resistance (

R

) are known is given by Watt's Law:

P = \frac{V^2}{R}

. Substituting

V = 120\text{ V}

and

R = 40\text{ }\Omega

gives

P = \frac{120^2}{40} = \frac{14{}400}{40} = 360\text{ W}

Step-by-Step Solution

State the given circuit parameters

V = 120\text{ V}

R = 40\text{ }\Omega

These are the known electrical quantities provided in the problem stem.

Select the appropriate power formula derived from Ohm's Law and Watt's Law

P = \frac{V^2}{R}

Power can be directly calculated from voltage and resistance without needing to solve for current first.

Substitute the values into the formula and solve for power

P = \frac{(120)^2}{40} = \frac{14{}400}{40} = 360\text{ W}

Carrying out the arithmetic yields the total electrical power converted into thermal energy.

Key Concept

Calculating electrical power from voltage and resistance using Watt's Law (

P = \frac{V^2}{R}

)

Estimated Time:1m 0s

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