A ceramic load resistor in a direct-current (DC) circuit has a resistance of 20\ \Omega an… — Armed Services Vocational Aptitude Battery (ASVAB)

A ceramic load resistor in a direct-current (DC) circuit has a resistance of $20\ \Omega$ and carries a steady current of $2\text{ A}$ . What is the total electrical power dissipated by this resistor?

A
$40\text{ W}$
$80\text{ W}$ Answer
C
$10\text{ W}$
D
$800\text{ W}$

Answer

The total electrical power dissipated by the resistor is

80\text{ W}

According to Watt's Law, when current (

I

) and resistance (

R

) are known, electrical power is calculated using the formula

P = I^2 R

. Substituting

I = 2\text{ A}

and

R = 20\ \Omega

gives

P = (2\text{ A})^2 \times 20\ \Omega = 4 \times 20 = 80\text{ W}

Step-by-Step Solution

Identify the given electrical values and select the appropriate form of Watt's Law.

Given: current

I = 2\text{ A}

, resistance

R = 20\ \Omega

. Formula:

P = I^2 R

When current and resistance are known, the power dissipated is calculated using

P = I^2 R

Square the current value.

I^2 = (2\text{ A})^2 = 4\text{ A}^2

Watt's Law requires current to be squared before multiplying by resistance.

Multiply the squared current by the resistance to calculate power in watts.

P = 4\text{ A}^2 \times 20\ \Omega = 80\text{ W}

Multiplying amperes squared by ohms yields power in watts.

Key Concept

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

P = I^2 R

)

Estimated Time:45s

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