Electrical Power and Watt's Law Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

A direct-current (DC) communication antenna heater operates from a $48\text{ V}$ power source and dissipates $96\text{ W}$ of electrical power. What is the electrical resistance of the heater in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 24

Answer

The electrical resistance of the antenna heater is

24\ \Omega

.

According to Watt's Law, power is related to voltage and resistance by

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

. Rearranging this formula to solve for resistance gives

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

. Substituting

48\text{ V}

for voltage and

96\text{ W}

for power yields

R = \frac{48^2}{96} = \frac{2304}{96} = 24\ \Omega

.

Step-by-Step Solution

1

Select the appropriate form of Watt's Law connecting power (

P

), voltage (

V

), and resistance (

R

).

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

The circuit parameters given are voltage and power, with resistance as the unknown variable.

2

Rearrange the formula to isolate resistance (

R

).

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

Algebraic isolation of

R

allows direct calculation from the known values.

3

Substitute

V = 48\text{ V}

and

P = 96\text{ W}

into the rearranged formula and compute the result.

R = \frac{48^2}{96} = \frac{2304}{96} = 24\ \Omega

Simplifying the arithmetic yields the exact resistance value.

Key Concept

Watt's Law relates power, voltage, and resistance through the formula

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

, which rearranges to

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

.

Alternative Method

First find the current drawn by the heater using

I = \frac{P}{V} = \frac{96\text{ W}}{48\text{ V}} = 2\text{ A}

. Then apply Ohm's Law to determine the resistance:

R = \frac{V}{I} = \frac{48\text{ V}}{2\text{ A}} = 24\ \Omega

.

Estimated Time:1m 0s

Question 2Question →

An electronic circuit contains a $20\text{ k}\Omega$ precision resistor that dissipates $180\text{ mW}$ of electrical power. What is the DC voltage drop across this resistor?

9\text{ V}

60\text{ V}

6\text{ V}

3\text{ V}

Show answer & explanation

Answer: $60\text{ V}$

Answer

The DC voltage drop across the resistor is

60\text{ V}

.

According to Watt's Law, electrical power dissipated in a resistive circuit can be expressed as

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

. Solving for voltage yields

V = \sqrt{P \times R}

. Converting the prefix values gives

180\text{ mW} = 0.18\text{ W}

and

20\text{ k}\Omega = 20{}000\ \Omega

. Multiplying these values gives

V^2 = 0.18 \times 20{}000 = 3600\text{ V}^2

, and taking the square root gives

\sqrt{3600} = 60\text{ V}

.

Step-by-Step Solution

1

Convert given values to standard base units (Watts and Ohms)

P = 180\text{ mW} = 180 \times 10^{-3}\text{ W} = 0.18\text{ W}

;

R = 20\text{ k}\Omega = 20 \times 10^{3}\ \Omega = 20{}000\ \Omega

Standard electrical formulas require base SI units to prevent magnitude errors.

2

Select and rearrange the appropriate Watt's Law formula relating power, voltage, and resistance

P = \frac{V^2}{R} \implies V^2 = P \times R \implies V = \sqrt{P \times R}

The known variables are power (

P

) and resistance (

R

), and the unknown variable is voltage (

V

).

3

Calculate the product of power and resistance

V^2 = (180 \times 10^{-3}\text{ W}) \times (20 \times 10^{3}\ \Omega) = 180 \times 20 = 3600\text{ V}^2

The metric prefix multipliers

10^{-3}

and

10^{3}

cancel each other out (

10^{-3} \times 10^{3} = 1

).

4

Take the square root to determine the voltage

V = \sqrt{3600\text{ V}^2} = 60\text{ V}

Taking the square root of

V^2

yields the electromotive force across the component.

Key Concept

Watt's Law power calculation using

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

and rearranging to solve for voltage with metric prefix conversion.

Alternative Method

First, find the current using

P = I^2 \times R \implies I = \sqrt{\frac{P}{R}} = \sqrt{\frac{0.18\text{ W}}{20{}000\ \Omega}} = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3}\text{ A} = 3\text{ mA}

. Then apply Ohm's Law to find voltage:

V = I \times R = (3 \times 10^{-3}\text{ A}) \times (20 \times 10^{3}\ \Omega) = 60\text{ V}

.

Estimated Time:1m 30s

Question 3Question →

A DC circuit consists of a $20\text{ V}$ power source connected in series with a $4\text{ }\Omega$ resistor ( $R_1$ ). After passing through $R_1$ , the circuit splits into two parallel branches containing an $8\text{ }\Omega$ resistor ( $R_2$ ) and a $24\text{ }\Omega$ resistor ( $R_3$ ). The branches then recombine and connect back to the negative terminal of the power source. What is the electrical power dissipated by the $8\text{ }\Omega$ resistor ( $R_2$ )?

Show answer & explanation

Answer: 18

Answer

18 Watts

To find the power dissipated by a single component in a compound circuit, you must isolate the parameters for that specific component. By finding the total resistance (

10\text{ }\Omega

) and total current (

2\text{ A}

), we can determine that the voltage dropping specifically across the parallel section is

12\text{ V}

. Because components in parallel share the same voltage, the

8\text{ }\Omega

resistor receives the full

12\text{ V}

. Using Watt's Law,

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

, we find

\frac{12^2}{8} = 18\text{ W}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the parallel section (

R_p

).

R_p = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\text{ }\Omega

Before analyzing the whole circuit, parallel branches must be simplified into a single equivalent resistance.

2

Calculate the total resistance of the circuit (

R_t

).

R_t = 4\text{ }\Omega + 6\text{ }\Omega = 10\text{ }\Omega

The equivalent parallel resistance is in series with the

4\text{ }\Omega

resistor.

3

Calculate the total current (

I_t

) leaving the power source.

I_t = \frac{20\text{ V}}{10\text{ }\Omega} = 2\text{ A}

Ohm's law (

I = \frac{V}{R}

) relates the total source voltage to the total circuit resistance.

4

Calculate the voltage drop across the parallel section (

V_p

).

V_p = 2\text{ A} \times 6\text{ }\Omega = 12\text{ V}

The total current flows through the equivalent parallel resistance. This gives the voltage available specifically to the parallel branches.

5

Calculate the power dissipated by the

8\text{ }\Omega

resistor (

R_2

).

P = \frac{12^2}{8} = \frac{144}{8} = 18\text{ W}

Using Watt's Law (

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

), we apply the voltage specific to the parallel branch (

12\text{ V}

) and the individual resistance (

8\text{ }\Omega

).

Key Concept

Calculating power dissipation in a specific component of a series-parallel circuit requires finding the exact voltage or current acting on that specific component, not the entire circuit.

Question 4Question →

Match each circuit modification on the left with its corresponding effect on power dissipation relative to the baseline power ( $P_0$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Doubling voltage with constant resistance matches

4P_0

; Halving current with constant resistance matches

\frac{1}{4}P_0

; Doubling resistance with constant voltage matches

\frac{1}{2}P_0

; Tripling current with one-third resistance matches

3P_0

.

Each modification is correctly determined by applying Watt's Law formulas (

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

and

P = I^2 R

). Doubling voltage quadruples power (

2^2 = 4

), halving current reduces power to one-fourth (

(\frac{1}{2})^2 = \frac{1}{4}

), doubling resistance at constant voltage reduces power by half (

\frac{1}{2}

), and tripling current while reducing resistance to one-third increases power threefold (

3^2 \times \frac{1}{3} = 3

).

Step-by-Step Solution

1

Select the appropriate form of Watt's Law for each scenario depending on which parameters are held constant and which are modified.

Use

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

when voltage and resistance are given, and

P = I^2 R

when current and resistance are given.

Using the formula directly containing the relevant independent variables simplifies scaling calculations.

2

Calculate the scaling factor for doubling voltage with constant resistance.

P' = \frac{(2V)^2}{R} = 4\frac{V^2}{R} = 4P_0

.

Power is directly proportional to the square of the voltage across a fixed load.

3

Calculate the scaling factor for halving current with constant resistance.

P' = \left(\frac{1}{2}I\right)^2 R = \frac{1}{4}I^2 R = \frac{1}{4}P_0

.

Power is directly proportional to the square of current through a fixed load.

4

Calculate the scaling factor for doubling resistance at constant voltage.

P' = \frac{V^2}{2R} = \frac{1}{2}\frac{V^2}{R} = \frac{1}{2}P_0

.

At constant voltage, power is inversely proportional to resistance.

5

Calculate the scaling factor for tripling current while reducing resistance to one-third.

P' = (3I)^2 \cdot \left(\frac{1}{3}R\right) = 9I^2 \cdot \frac{1}{3}R = 3I^2 R = 3P_0

.

The squared current factor of 9 multiplied by the fractional resistance factor of 1/3 results in an overall factor of 3.

Key Concept

Watt's Law Power Scaling Relationships:

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

Estimated Time:1m 30s

Question 5Question →

An electric heating element is rated to dissipate $1.8\text{ kW}$ of power when connected across a standard $120\text{ V}$ DC supply line. If a control circuit reduces the direct current flowing through the heating element to $5\text{ A}$ while its internal resistance remains constant, what is the new power dissipated by the element?

600\text{ W}

200\text{ W}

40\text{ W}

0.2\text{ W}

Show answer & explanation

Answer: $200\text{ W}$

Answer

200\text{ W}

The correct answer of

200\text{ W}

is found by first converting

1.8\text{ kW}

to

1800\text{ W}

. Using the initial ratings, the element's resistance is

R = \frac{V^2}{P} = \frac{120^2}{1800} = 8\text{ }\Omega

. With the current controlled to

5\text{ A}

, the dissipated power is

P = I^2 R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

.

Step-by-Step Solution

1

Convert the rated power from kilowatts to watts

P_1 = 1.8\text{ kW} = 1800\text{ W}

Metric prefixes must be converted to base SI units before performing Watt's Law calculations.

2

Calculate the constant internal resistance (

R

) of the heating element using the initial rating

R = \frac{V_1^2}{P_1} = \frac{(120\text{ V})^2}{1800\text{ W}} = \frac{14400}{1800} = 8\text{ }\Omega

The heating element's resistance remains unchanged and serves as the bridge to calculate power under the new operating current.

3

Calculate the new power dissipation with the reduced current (

I_2 = 5\text{ A}

)

P_2 = I_2^2 \times R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

Watt's Law specifies that power dissipation across a known resistance is directly proportional to the square of the current.

Key Concept

Watt's Law Power and Resistance Relationships (

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

and

P = I^2 R

)

Alternative Method

Determine the initial current

I_1 = \frac{P_1}{V_1} = \frac{1800\text{ W}}{120\text{ V}} = 15\text{ A}

. Since

P \propto I^2

for a constant resistance, scaling the power by the ratio of the currents squared gives

P_2 = P_1 \times \left(\frac{I_2}{I_1}\right)^2 = 1800\text{ W} \times \left(\frac{5\text{ A}}{15\text{ A}}\right)^2 = 1800 \times \left(\frac{1}{3}\right)^2 = \frac{1800}{9} = 200\text{ W}

.

Estimated Time:1m 30s

Question 6Question →

An automotive spotlight operates on a $12\text{ V}$ direct current (DC) electrical system and draws a steady current of $4\text{ A}$ . How much electrical power is consumed by the spotlight?

3\text{ W}

48\text{ W}

16\text{ W}

192\text{ W}

Show answer & explanation

Answer: $48\text{ W}$

Answer

The power consumed by the spotlight is

48\text{ W}

.

According to Watt's Law, electrical power (

P

) is determined by multiplying voltage (

V

) by current (

I

), using the formula

P = V \times I

. Substituting

12\text{ V}

and

4\text{ A}

gives

P = 12\text{ V} \times 4\text{ A} = 48\text{ W}

.

Step-by-Step Solution

1

Identify the given electrical values from the problem statement.

Voltage (

V

) =

12\text{ V}

, Current (

I

) =

4\text{ A}

.

Establishing the known quantities determines which form of Watt's Law to apply.

2

Select the appropriate Watt's Law power formula relating power, voltage, and current.

P = V \times I

Electrical power in watts equals the product of potential difference in volts and current in amperes.

3

Substitute the values into the formula and calculate.

P = 12\text{ V} \times 4\text{ A} = 48\text{ W}

Carrying out the multiplication yields the total power dissipated.

Key Concept

Watt's Law relates electrical power (

P

) to voltage (

V

) and current (

I

) through the formula

P = V \times I

.

Estimated Time:45s

Question 7Question →

Match each Watt's Law power formula or electrical power unit on the left with its corresponding definition or application condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

P = V \times I

matches power from voltage and current;

P = I^2 R

matches power from current and resistance;

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

matches power from voltage and resistance;

1\text{ Watt (W)}

matches the unit equivalent to one joule per second.

Each formula represents a direct algebraic variation of Watt's Law derived by combining

P = V \times I

with Ohm's Law (

V = IR

), and the watt is defined as one joule per second.

Step-by-Step Solution

1

Identify the base relationship of Watt's Law

P = V \times I

Electrical power is defined as the rate at which electrical energy is transferred, calculated by multiplying voltage across a load by the current flowing through it.

2

Derive the formula relating power to current and resistance

P = I^2 R

By substituting

V = I \times R

from Ohm's Law into

P = V \times I

, power is expressed in terms of current squared times resistance.

3

Derive the formula relating power to voltage and resistance

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

By substituting

I = \frac{V}{R}

from Ohm's Law into

P = V \times I

, power is expressed in terms of voltage squared divided by resistance.

4

Identify the standard unit and definition of power

1\text{ W} = 1\text{ J/s}

Power measures the rate of doing work or dissipating energy, where one watt equals one joule per second.

Key Concept

Watt's Law and Electrical Power Formulas

Estimated Time:45s

Question 8Question →

A direct-current (DC) ventilation blower operates on a $28\text{ V}$ aircraft electrical bus and draws a steady current of $5\text{ A}$ . How many watts ( $\text{W}$ ) of electrical power are consumed by the blower?

Show answer & explanation

Answer: 140

Answer

The power consumed by the blower is

140\text{ W}

.

According to Watt's Law, electrical power (

P

) dissipated by a direct-current device equals the product of the applied voltage (

V

) and the current (

I

). Multiplying

28\text{ V}

by

5\text{ A}

gives

140\text{ W}

.

Step-by-Step Solution

1

Identify the known electrical quantities given in the scenario.

Voltage

V = 28\text{ V}

and Current

I = 5\text{ A}

.

Identifying the known parameters determines which form of Watt's Law to use.

2

Apply the fundamental Watt's Law formula for electrical power (

P = V \times I

).

P = 28\text{ V} \times 5\text{ A} = 140\text{ W}

.

Electrical power in a direct-current circuit is calculated by multiplying the potential difference across the load by the current flowing through it.

Key Concept

Watt's Law for DC Power (

P = V \times I

)

Estimated Time:45s

Question 9Question →

A direct current (DC) cooling fan in an electronics enclosure operates at a voltage of $24\text{ V}$ and draws a current of $1.5\text{ A}$ . What is the total electrical power consumed by the fan?

16\text{ W}

25.5\text{ W}

36\text{ W}

54\text{ W}

Show answer & explanation

Answer: $36\text{ W}$

Answer

36\text{ W}

According to Watt's Law, electrical power in a DC circuit is determined by multiplying the voltage by the current (

P = V \times I

). Multiplying

24\text{ V}

by

1.5\text{ A}

results in

36\text{ W}

of power dissipation.

Step-by-Step Solution

1

Identify the given electrical quantities and the appropriate Watt's Law formula.

Voltage

V = 24\text{ V}

, Current

I = 1.5\text{ A}

, Formula:

P = V \times I

Electrical power in a DC circuit is directly defined as the product of electromotive force (voltage) and current.

2

Substitute the values into the formula and calculate the power.

P = 24\text{ V} \times 1.5\text{ A} = 36\text{ W}

Multiplying

24

by

1.5

gives

36

, with the standard unit of power being watts (W).

Key Concept

Watt's Law calculates electrical power (

P

) as the product of voltage (

V

) and current (

I

), expressed as

P = V \times I

.

Estimated Time:45s

Question 10Question →

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?

40\text{ W}

80\text{ W}

10\text{ W}

800\text{ W}

Show answer & explanation

Answer: $80\text{ W}$

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

1

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

.

2

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.

3

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

Question 11Question →

In direct-current (DC) circuit analysis, different forms of Watt's Law are used depending on which electrical quantities are known. Match each set of known circuit parameters on the left with the correct formula on the right used to calculate the target value directly.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Each given set of circuit parameters matches the direct algebraic form of Watt's Law designed for those specific variables: Voltage and Current to

P = V \times I

; Current and Resistance to

P = I^2 \times R

; Voltage and Resistance to

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

; and Power and Voltage to

I = \frac{P}{V}

.

Each match correctly pairs the given known variables with the exact Watt's Law equation variant required to solve directly for the unknown quantity without needing extraneous calculation steps.

Step-by-Step Solution

1

Identify the base definition of electrical power in terms of voltage and current.

P = V \times I

Power in watts equals electromotive force in volts multiplied by electron current flow in amperes.

2

Derive the power formula when only current and resistance are given using Ohm's Law (

V = I \times R

).

P = (I \times R) \times I = I^2 \times R

Replacing voltage with the product of current and resistance calculates power without needing voltage explicitly.

3

Derive the power formula when only voltage and resistance are given using Ohm's Law (

I = \frac{V}{R}

).

P = V \times \left(\frac{V}{R}\right) = \frac{V^2}{R}

Replacing current with voltage divided by resistance calculates power without needing current explicitly.

4

Rearrange the primary power formula to solve for current when power and voltage are known.

I = \frac{P}{V}

Dividing both sides of

P = V \times I

by voltage isolates the current parameter.

Key Concept

Watt's Law and its derived variations relate electrical power (

P

), voltage (

V

), current (

I

), and resistance (

R

).

Estimated Time:45s

Question 12Question →

A direct-current (DC) power supply maintains a constant potential difference of $15\text{ V}$ across a circuit resistor that dissipates $4.5\text{ W}$ of electrical power. What is the electrical resistance of the resistor in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 50

Answer

The resistance of the resistor is

50\ \Omega

.

According to Watt's Law, power dissipated by a resistor with a known voltage across it is given by

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

. Solving for resistance gives

R = \frac{V^2}{P} = \frac{15^2}{4.5} = \frac{225}{4.5} = 50\ \Omega

.

Step-by-Step Solution

1

State the known values and identify the target variable.

Voltage

V = 15\text{ V}

, Power

P = 4.5\text{ W}

, Resistance

R = ?

Establishing the known electrical parameters determines the appropriate form of Watt's Law to apply.

2

Select and rearrange the Watt's Law power equation that relates power, voltage, and resistance.

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

Since voltage and power are known, using

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

allows direct solving for resistance.

3

Calculate the squared voltage.

V^2 = 15^2 = 225\text{ V}^2

Evaluating the numerator is the first calculation step in solving for

R

.

4

Divide the squared voltage by the power dissipation.

R = \frac{225}{4.5} = 50\ \Omega

Carrying out the division yields the final resistance value in ohms.

Key Concept

Watt's Law relates electrical power, voltage, and resistance through the formula

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

, which rearranges algebraically to

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

.

Alternative Method

First find the circuit current using

I = \frac{P}{V} = \frac{4.5\text{ W}}{15\text{ V}} = 0.3\text{ A}

. Then apply Ohm's Law to determine the resistance:

R = \frac{V}{I} = \frac{15\text{ V}}{0.3\text{ A}} = 50\ \Omega

.

Estimated Time:1m 0s

Question 13Question →

An electronics technician measures a power dissipation of $20\text{ W}$ across a $500\ \Omega$ test resistor in a direct-current circuit. What is the electrical current flowing through this resistor in milliamperes ( $\text{mA}$ )?

40\text{ mA}

200\text{ mA}

0.2\text{ mA}

100\text{ mA}

Show answer & explanation

Answer: $200\text{ mA}$

Answer

200\text{ mA}

According to Watt's Law, power dissipated in a resistive load is

P = I^2 R

. Rearranging to solve for current gives

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}

. Converting amperes to milliamperes gives

0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}

.

Step-by-Step Solution

1

Identify known parameters and select the Watt's Law formula relating power, resistance, and current.

P = 20\text{ W}

,

R = 500\ \Omega

, and

P = I^2 R

The circuit power and resistance are given, and current is the unknown quantity.

2

Rearrange the formula to solve for current (

I

) and substitute the given values.

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{20\text{ W}}{500\ \Omega}} = \sqrt{0.04} = 0.2\text{ A}

Isolating

I

requires dividing power by resistance and taking the square root.

3

Convert the calculated current from amperes to milliamperes.

0.2\text{ A} \times 1{}000\text{ mA/A} = 200\text{ mA}

The question asks for the final answer in milliamperes.

Key Concept

Watt's Law relates power, current, and resistance via

P = I^2 R

, which rearranges to

I = \sqrt{\frac{P}{R}}

.

Alternative Method

First compute the voltage drop across the resistor using

V = \sqrt{P \times R} = \sqrt{20\text{ W} \times 500\ \Omega} = \sqrt{10{}000} = 100\text{ V}

. Then apply Ohm's Law to find current:

I = \frac{V}{R} = \frac{100\text{ V}}{500\ \Omega} = 0.2\text{ A} = 200\text{ mA}

.

Estimated Time:1m 0s

Question 14Question →

A technician evaluates four direct-current (DC) electrical components operating under steady-state conditions. Match each component and operating condition on the left with its corresponding dissipated electrical power on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The automotive lamp (

12\text{ V}, 750\text{ mA}

) matches

9\text{ W}

; the heating element (

50\ \Omega, 0.6\text{ A}

) matches

18\text{ W}

; the instrumentation sensor (

600\ \Omega, 200\text{ mA}

) matches

24\text{ W}

; and the relay coil (

100\ \Omega, 60\text{ V}

) matches

36\text{ W}

.

Each electrical component's power dissipation is determined by selecting the appropriate algebraic form of Watt's Law based on the known parameters:

P = V \times I

for the indicator lamp (

12\text{ V} \times 0.75\text{ A} = 9\text{ W}

),

P = I^2 \times R

for the heating element (

(0.6\text{ A})^2 \times 50\ \Omega = 18\text{ W}

) and sensor (

(0.2\text{ A})^2 \times 600\ \Omega = 24\text{ W}

), and

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

for the relay coil (

(60\text{ V})^2 / 100\ \Omega = 36\text{ W}

).

Step-by-Step Solution

1

Calculate power for the automotive indicator lamp using voltage and current

P = 12\text{ V} \times 0.75\text{ A} = 9\text{ W}

Convert milliamperes to amperes (

750\text{ mA} = 0.75\text{ A}

) and apply the fundamental Watt's Law equation

P = V \times I

.

2

Calculate power for the heating element using current and resistance

P = (0.6\text{ A})^2 \times 50\ \Omega = 0.36 \times 50 = 18\text{ W}

Apply Watt's Law in the form

P = I^2 \times R

when current and resistance are known.

3

Calculate power for the instrumentation sensor using current and resistance

P = (0.2\text{ A})^2 \times 600\ \Omega = 0.04 \times 600 = 24\text{ W}

Convert

200\text{ mA}

to

0.2\text{ A}

and calculate power using

P = I^2 \times R

.

4

Calculate power for the communications relay coil using voltage and resistance

P = \frac{(60\text{ V})^2}{100\ \Omega} = \frac{3600}{100} = 36\text{ W}

Apply Watt's Law in the form

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

when potential difference and resistance are known.

Key Concept

Watt's Law and Electrical Power Dissipation

Estimated Time:1m 15s

Question 15Question →

A direct-current (DC) relay coil has an internal resistance of $80\ \Omega$ and draws a steady current of $250\text{ mA}$ . How much electrical power is dissipated by the coil?

20\text{ W}

5{}000\text{ W}

5\text{ W}

320\text{ W}

Show answer & explanation

Answer: $5\text{ W}$

Answer

The electrical power dissipated by the coil is

5\text{ W}

.

To find the power dissipated when current and resistance are known, apply Watt's Law in the form

P = I^2 R

. First, convert

250\text{ mA}

to

0.25\text{ A}

. Squaring the current gives

(0.25\text{ A})^2 = 0.0625\text{ A}^2

. Multiplying by the resistance yields

0.0625\text{ A}^2 \times 80\ \Omega = 5\text{ W}

.

Step-by-Step Solution

1

Convert the current from milliamperes (mA) to amperes (A).

I = 250\text{ mA} = 0.25\text{ A}

Standard electrical formulas require current in base units of amperes.

2

Calculate the power dissipation using Watt's Law (

P = I^2 R

).

P = (0.25\text{ A})^2 \times 80\ \Omega = 0.0625\text{ A}^2 \times 80\ \Omega = 5\text{ W}

When current and resistance are known, power equals the square of current multiplied by resistance.

Key Concept

Calculating electrical power using current and resistance (

P = I^2 R

) with metric prefix conversion.

Estimated Time:1m 0s

Question 16Question →

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

Show answer & explanation

Answer: 12

Answer

The voltage drop across the indicator lamp is

12\text{ V}

.

According to Watt's Law, electrical power is the product of voltage and current (

P = V \times I

). Rearranging the formula to solve for voltage gives

V = \frac{P}{I}

. Converting

150\text{ mA}

to

0.15\text{ A}

and evaluating

\frac{1.8\text{ W}}{0.15\text{ A}}

yields

12\text{ V}

.

Step-by-Step Solution

1

Convert the current from milliamperes to amperes.

I = 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 = \frac{P}{I}

Watt's Law defines power as

P = V \times I

, so isolating voltage yields

V = \frac{P}{I}

.

3

Substitute the known values into the equation to find the voltage.

V = \frac{1.8\text{ W}}{0.15\text{ A}} = 12\text{ V}

Dividing the power dissipation (

1.8\text{ W}

) by the current (

0.15\text{ A}

) gives the operating voltage.

Key Concept

Watt's Law describes the mathematical relationship between electrical power, voltage, and current:

P = V \times I

. When power and current are known, voltage is calculated using

V = \frac{P}{I}

, provided current is expressed in amperes.

Estimated Time:1m 0s

Question 17Question →

A direct-current (DC) soldering iron operates on a $24\text{ V}$ power supply and dissipates $72\text{ W}$ of power. What is the electrical resistance of the heating element?

0.33\ \Omega

8\ \Omega

3\ \Omega

24\ \Omega

Show answer & explanation

Answer: $8\ \Omega$

Answer

8\ \Omega

According to Watt's Law, power dissipated by a resistance is

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

. Solving for resistance gives

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

. Substituting

24\text{ V}

and

72\text{ W}

yields

R = \frac{24^2}{72} = \frac{576}{72} = 8\ \Omega

.

Step-by-Step Solution

1

Identify the given values and select the Watt's Law formula that relates voltage, power, and resistance.

V = 24\text{ V}

,

P = 72\text{ W}

, and

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

The circuit voltage and power dissipation are known, and the resistance is the target variable.

2

Rearrange the formula to isolate resistance (

R

).

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

Multiplying both sides by

R

and dividing by

P

provides the direct expression for resistance.

3

Square the voltage and divide by the power to find the resistance in ohms.

R = \frac{24^2}{72} = \frac{576}{72} = 8\ \Omega

Evaluating the numerical expression yields the final resistance value.

Key Concept

Watt's Law relates power, voltage, and resistance through the relationship

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

, which can be rearranged as

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

.

Estimated Time:1m 0s

Question 18Question →

An electronics technician evaluates four direct-current (DC) circuit devices under steady-state conditions. Match each electrical device and operating parameter on the left with its correct calculated electrical power dissipation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The backlight (

12\text{ V}, 0.5\text{ A}

) matches

6\text{ W}

; the dummy load (

50\ \Omega, 0.4\text{ A}

) matches

8\text{ W}

; the defroster element (

60\text{ V}, 300\ \Omega

) matches

12\text{ W}

; and the bleeder resistor (

4\text{ k}\Omega, 100\text{ V}

) matches

2.5\text{ W}

.

Each load correctly maps to its power calculation derived from Watt's Law formulas (

P = V \times I

,

P = I^2 \times R

, and

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

) with proper unit conversions.

Step-by-Step Solution

1

Calculate the power of the instrument panel backlight using known voltage and current.

P = V \times I = 12\text{ V} \times 0.5\text{ A} = 6\text{ W}

When voltage (

V

) and current (

I

) are provided, the direct formula for electrical power is

P = V \times I

.

2

Calculate the power of the dummy load resistor using known current and resistance.

P = I^2 \times R = (0.4\text{ A})^2 \times 50\ \Omega = 0.16 \times 50 = 8\text{ W}

When current (

I

) and resistance (

R

) are known, substitute Ohm's Law (

V = I \times R

) into Watt's Law to get

P = I^2 \times R

.

3

Calculate the power of the defroster element using known voltage and resistance.

P = \frac{V^2}{R} = \frac{(60\text{ V})^2}{300\ \Omega} = \frac{3600}{300} = 12\text{ W}

When voltage (

V

) and resistance (

R

) are known, substitute

I = \frac{V}{R}

into Watt's Law to get

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

.

4

Convert the bleeder resistance from kilo-ohms to ohms and calculate its power dissipation.

R = 4\text{ k}\Omega = 4000\ \Omega

, then

P = \frac{V^2}{R} = \frac{(100\text{ V})^2}{4000\ \Omega} = \frac{10000}{4000} = 2.5\text{ W}

Metric prefixes must be converted to base units before applying Watt's Law formulas.

Key Concept

Electrical power in direct-current circuits is calculated using Watt's Law variations:

P = V \times I

,

P = I^2 \times R

, and

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

, depending on which two electrical parameters are known.

Estimated Time:1m 15s

Question 19Question →

A direct-current (DC) avionics heating resistor with a resistance of $200\ \Omega$ dissipates $18\text{ W}$ of electrical power under steady-state operating conditions. What is the magnitude of the current flowing through the resistor?

90\text{ mA}

300\text{ mA}

0.30\text{ mA}

60\text{ mA}

Show answer & explanation

Answer: $300\text{ mA}$

Answer

300\text{ mA}

According to Watt's Law, power dissipation in a resistive load is given by

P = I^2 R

. Rearranging to solve for current gives

I = \sqrt{\frac{P}{R}} = \sqrt{\frac{18\text{ W}}{200\ \Omega}} = \sqrt{0.09\text{ A}^2} = 0.3\text{ A}

. Multiplying by

1{}000

converts the current to

300\text{ mA}

.

Step-by-Step Solution

1

Select the appropriate Watt's Law formula that relates power (

P

), current (

I

), and resistance (

R

).

P = I^2 R

The known quantities are electrical power (

18\text{ W}

) and resistance (

200\ \Omega

), and the target quantity is current (

I

).

2

Rearrange the formula to solve for the squared current (

I^2

).

I^2 = \frac{P}{R} = \frac{18\text{ W}}{200\ \Omega} = 0.09\text{ A}^2

Isolating the variable term allows direct computation of

I^2

.

3

Take the square root of both sides to calculate the current in amperes (

A

).

I = \sqrt{0.09\text{ A}^2} = 0.3\text{ A}

Current is the square root of

I^2

.

4

Convert the current from amperes (

A

) to milliamperes (

\text{mA}

).

0.3\text{ A} \times 1{}000\text{ mA/A} = 300\text{ mA}

Expressing the value in standard sub-unit milliamperes.

Key Concept

Watt's Law relates power, current, and resistance through the formula

P = I^2 R

, which can be rearranged as

I = \sqrt{\frac{P}{R}}

.

Estimated Time:1m 0s

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