Electrical Power Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

A direct current (DC) automotive lamp connected to a $12\text{ V}$ power source draws a current of $2\text{ A}$ . What is the total electrical power consumed by the lamp?

6\text{ W}

14\text{ W}

24\text{ W}

48\text{ W}

Show answer & explanation

Answer: $24\text{ W}$

Answer

24\text{ W}

According to Watt's Law, the electrical power (

P

) consumed by a circuit load is equal to the voltage (

V

) across the component multiplied by the current (

I

) flowing through it:

P = V \times I

. Substituting the given values yields

P = 12\text{ V} \times 2\text{ A} = 24\text{ W}

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage (

V

) =

12\text{ V}

, Current (

I

) =

2\text{ A}

These are the two electrical quantities necessary to calculate power directly.

2

Select the appropriate electrical power formula (Watt's Law)

P = V \times I

When voltage and current are known, power is equal to their product.

3

Substitute the given values into the formula and calculate

P = 12\text{ V} \times 2\text{ A} = 24\text{ W}

Multiplying electromotive force by current gives total power in watts.

Key Concept

Electrical power in a DC circuit is calculated using Watt's Law (

P = V \times I

), where power in watts is the product of voltage in volts and current in amperes.

Estimated Time:45s

Question 2Question →

A $24\text{ V}$ direct current (DC) power supply delivers a steady current of $3\text{ A}$ to a resistive heating element. What is the total electrical power consumed by the heating element in watts ( $\text{W}$ )?

Show answer & explanation

Answer: 72

Answer

The total electrical power consumed is

72\text{ W}

.

Electrical power in a direct current circuit is calculated using the formula

P = V \times I

. Substituting the given voltage of

24\text{ V}

and current of

3\text{ A}

yields

P = 24 \times 3 = 72\text{ W}

.

Step-by-Step Solution

1

Identify the given values from the circuit problem

Voltage

V = 24\text{ V}

and current

I = 3\text{ A}

Electrical power calculations require knowing at least two circuit parameters.

2

Select the appropriate electrical power formula

P = V \times I

Watt's Law defines electrical power as the product of voltage across a component and the current flowing through it.

3

Substitute the values and calculate the final result

P = 24\text{ V} \times 3\text{ A} = 72\text{ W}

Multiplying volts by amperes gives power directly in watts.

Key Concept

Electrical power is the rate at which electrical energy is transferred in a circuit, calculated using Watt's Law (

P = V \times I

).

Estimated Time:45s

Question 3Question →

An aircraft landing light is connected to a $28\text{ V}$ DC bus and draws a current of $4\text{ A}$ . What is the total electrical power consumed by the light?

7\text{ W}

32\text{ W}

112\text{ W}

1120\text{ W}

Show answer & explanation

Answer: $112\text{ W}$

Answer

112\text{ W}

According to Watt's Law, electrical power (

P

) is equal to the product of voltage (

V

) and current (

I

). Multiplying

28\text{ V}

by

4\text{ A}

yields

112\text{ W}

.

Step-by-Step Solution

1

Identify the given electrical values from the circuit

Voltage

V = 28\text{ V}

, Current

I = 4\text{ A}

These are the two known parameters required to calculate power directly.

2

Select Watt's Law power formula that relates voltage, current, and power

P = V \times I

Electrical power in watts is equal to electromotive force in volts multiplied by current in amperes.

3

Calculate the total power dissipation

P = 28\text{ V} \times 4\text{ A} = 112\text{ W}

Multiplying

28

by

4

gives the power consumption in watts.

Key Concept

Electrical Power Calculation using Watt's Law (

P = V \times I

)

Estimated Time:45s

Question 4Question →

In electrical circuits, the power consumed or dissipated can be calculated using different formulas depending on which circuit parameters are known. Match each set of known circuit parameters on the left with the correct electrical power formula on the right. How should each parameter combination be paired with its corresponding power formula?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Voltage and Current pairs with

P = V \times I

; Current and Resistance pairs with

P = I^2 R

; Voltage and Resistance pairs with

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

; Electrical Energy and Time pairs with

P = \frac{E}{t}

.

Each parameter combination matches its standard mathematical expression: Voltage and Current relate directly through Watt's Law (

P = V \times I

), Current and Resistance relate through

P = I^2 R

, Voltage and Resistance relate through

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

, and Energy and Time match the foundational definition of power (

P = \frac{E}{t}

).

Step-by-Step Solution

1

Identify the base relationship of electrical power from fundamental physical definitions

Power is defined as energy divided by time:

P = \frac{E}{t}

.

This establishes the core definition of power in Joules per second (Watts).

2

Identify Watt's Law for circuits when potential difference and flow rate are known

Power is given by

P = V \times I

.

Multiplying potential difference (Volts) by current (Amperes) yields electrical power in Watts.

3

Use Ohm's Law substitutions to find power formulas for resistive circuits

When current and resistance are known:

P = I^2 R

. When voltage and resistance are known:

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

.

Substituting

V = IR

or

I = \frac{V}{R}

allows power to be determined when one electrical variable is omitted.

Key Concept

Electrical Power and Watt's Law Formulations

Estimated Time:45s

Question 5Question →

A heating resistor with a resistance of $20\text{ \Omega}$ has a steady current of $3\text{ A}$ flowing through it. What is the total electrical power dissipated by the resistor?

60\text{ W}

120\text{ W}

180\text{ W}

18\text{ W}

Show answer & explanation

Answer: $180\text{ W}$

Answer

180\text{ W}

Electrical power dissipated by a resistor when current and resistance are known is determined by the formula

P = I^2 \times R

. Substituting the given values gives

P = (3\text{ A})^2 \times 20\text{ \Omega} = 9 \times 20 = 180\text{ W}

.

Step-by-Step Solution

1

Identify the known circuit parameters and the appropriate power formula.

Known values:

I = 3\text{ A}

,

R = 20\text{ \Omega}

. Applicable formula:

P = I^2 \times R

.

When current and resistance are known, Watt's law in the form

P = I^2 R

gives electrical power directly.

2

Square the current value.

I^2 = (3\text{ A})^2 = 9\text{ A}^2

.

Power dissipation is proportional to the square of the current flowing through a resistor.

3

Multiply the squared current by the resistance.

P = 9 \times 20 = 180\text{ W}

.

Multiplying squared current by resistance yields total dissipated power in watts.

Key Concept

Electrical power dissipation in a resistive component using Joule's/Watt's law (

P = I^2 R

).

Estimated Time:45s

Question 6Question →

A workshop bench grinder is connected to a $120\text{ V}$ electrical supply line and draws a steady current of $5\text{ A}$ . What is the total electrical power consumed by the grinder in watts ( $\text{W}$ )?

Show answer & explanation

Answer: 600

Answer

600 W

Electrical power is calculated using the formula

P = V \times I

, where

P

is power in watts,

V

is electromotive force in volts, and

I

is current in amperes. Substituting the given values gives

P = 120\text{ V} \times 5\text{ A} = 600\text{ W}

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage

V = 120\text{ V}

and Current

I = 5\text{ A}

Determining the known values is necessary to select the correct power formula.

2

Apply Watt's Law formula for power

P = V \times I

Watt's Law defines electrical power in watts as the product of potential difference in volts and current in amperes.

3

Calculate the total power

P = 120\text{ V} \times 5\text{ A} = 600\text{ W}

Multiplying voltage by current yields the total power consumed by the electrical device.

Key Concept

Electrical power (

P

) in a circuit is the rate at which electrical energy is converted into another form of energy, calculated using Watt's Law as

P = V \times I

.

Estimated Time:45s

Question 7Question →

In a direct current (DC) avionics test circuit, a $48\text{ V}$ power supply is connected in series with a $30\text{ }\Omega$ current-limiting resistor and a $50\text{ }\Omega$ load resistor. What is the electrical power dissipated by the $50\text{ }\Omega$ load resistor, in watts?

Show answer & explanation

Answer: 18

Answer

The power dissipated by the

50\text{ }\Omega

load resistor is

18\text{ W}

.

In a series circuit, the total resistance is the sum of all individual resistances (

30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega

). With a

48\text{ V}

supply, the circuit current is

I = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}

. The electrical power dissipated by the

50\text{ }\Omega

resistor is determined by

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

.

Step-by-Step Solution

1

Find the equivalent total resistance of the series circuit

R_{\text{total}} = 30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega

Resistances in a series circuit are directly additive to determine total opposition to current flow.

2

Calculate the total circuit current using Ohm's Law

I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}

Current is uniform throughout all components connected in a series loop.

3

Calculate the power dissipation in the load resistor using Watt's Law

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

Power dissipated by a resistor is proportional to the square of the current flowing through it multiplied by its resistance.

Key Concept

Series circuit current and resistor power dissipation using combined Ohm's Law and Watt's Law (

P = I^2 R

or

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

)

Alternative Method

Find the voltage drop across the load resistor using the voltage divider rule:

V_{\text{load}} = 48\text{ V} \times \left(\frac{50\text{ }\Omega}{30\text{ }\Omega + 50\text{ }\Omega}\right) = 48 \times \frac{5}{8} = 30\text{ V}

. Then compute power using

P = \frac{V_{\text{load}}^2}{R_{\text{load}}} = \frac{30^2}{50} = \frac{900}{50} = 18\text{ W}

.

Estimated Time:1m 15s

Question 8Question →

A continuous-duty DC servomotor coil is connected to a $50\text{ V}$ power line and draws a steady current of $4\text{ A}$ . During a brownout condition, the source voltage drops to $40\text{ V}$ while the internal resistance of the coil remains constant. What is the electrical power dissipated by the servomotor coil under this reduced voltage condition?

40\text{ W}

128\text{ W}

160\text{ W}

200\text{ W}

Show answer & explanation

Answer: $128\text{ W}$

Answer

128\text{ W}

The coil's resistance is determined first using Ohm's Law:

R = \frac{50\text{ V}}{4\text{ A}} = 12.5\text{ }\Omega

. Under the reduced supply voltage of

40\text{ V}

, power dissipation is calculated using

P = \frac{V^2}{R} = \frac{40^2}{12.5} = \frac{1600}{12.5} = 128\text{ W}

. Alternatively, the new current is

I = \frac{40\text{ V}}{12.5\text{ }\Omega} = 3.2\text{ A}

, giving

P = 40\text{ V} \times 3.2\text{ A} = 128\text{ W}

.

Step-by-Step Solution

1

Calculate the constant internal resistance of the coil from initial operating values

R = \frac{V_1}{I_1} = \frac{50\text{ V}}{4\text{ A}} = 12.5\text{ }\Omega

The coil's physical resistance remains constant despite changes in external supply voltage.

2

Calculate the new current drawn by the coil at the reduced voltage of

40\text{ V}

I_2 = \frac{V_2}{R} = \frac{40\text{ V}}{12.5\text{ }\Omega} = 3.2\text{ A}

By Ohm's Law, reduced voltage across a fixed resistance causes a proportional reduction in current.

3

Calculate the new electrical power dissipated under the brownout condition

P_2 = V_2 \times I_2 = 40\text{ V} \times 3.2\text{ A} = 128\text{ W}

(or

P_2 = \frac{V_2^2}{R} = \frac{40^2}{12.5} = 128\text{ W}

)

Watt's Law defines power as the product of voltage and current or the square of voltage divided by resistance.

Key Concept

Electrical power in a resistive load varies with the square of voltage (

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

) when resistance is held constant.

Alternative Method

Proportional scaling using Watt's Law: Since

P \propto V^2

for constant resistance,

P_2 = P_1 \times \left(\frac{V_2}{V_1}\right)^2 = 200\text{ W} \times \left(\frac{40}{50}\right)^2 = 200 \times 0.64 = 128\text{ W}

.

Estimated Time:1m 30s

Question 9Question →

Match each electrical load scenario on the left with its corresponding steady-state power dissipation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The fan motor (

12\text{ V}, 2.5\text{ A}

) matches

30\text{ W}

; the shunt resistor (

0.5\text{ }\Omega, 8\text{ A}

) matches

32\text{ W}

; the heating element (

24\text{ V}, 16\text{ }\Omega

) matches

36\text{ W}

; and the parallel resistors (

50\text{ V}, 2 \times 50\text{ }\Omega

) match

100\text{ W}

.

Each load condition requires selecting the appropriate formula derived from Watt's Law and Ohm's Law:

P = VI

yields

30\text{ W}

for the motor;

P = I^2R

yields

32\text{ W}

for the current shunt;

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

yields

36\text{ W}

for the heating resistor; and parallel reduction with

P = \frac{V^2}{R_{eq}}

yields

100\text{ W}

for the dual load.

Step-by-Step Solution

1

Calculate the power consumed by the cooling fan motor using

P = V \times I

.

P = 12\text{ V} \times 2.5\text{ A} = 30\text{ W}

When voltage and current are explicitly provided, the direct product gives the electrical power.

2

Calculate the power dissipated in the shunt resistor using

P = I^2 \times R

.

P = (8\text{ A})^2 \times 0.5\text{ }\Omega = 64 \times 0.5 = 32\text{ W}

When current and resistance are known, squaring the current and multiplying by resistance gives power dissipation.

3

Calculate the power dissipated in the heating element using

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

.

P = \frac{(24\text{ V})^2}{16\text{ }\Omega} = \frac{576}{16} = 36\text{ W}

When voltage across a component and its resistance are known, dividing the squared voltage by the resistance gives the dissipated power.

4

Determine the equivalent parallel resistance of the two

50\text{ }\Omega

resistors and calculate total circuit power.

R_{eq} = \frac{50\text{ }\Omega \times 50\text{ }\Omega}{50\text{ }\Omega + 50\text{ }\Omega} = 25\text{ }\Omega

;

P = \frac{(50\text{ V})^2}{25\text{ }\Omega} = \frac{2500}{25} = 100\text{ W}

Combining parallel branches determines the total circuit load before applying the power equation.

Key Concept

Calculation of electrical power dissipation across varied circuit configurations using Watt's Law formulas (

P = VI

,

P = I^2R

, and

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

).

Estimated Time:2m 0s

Question 10Question →

A direct current circuit is powered by a $48\text{ V}$ source. The circuit contains a main line with resistor $R_1$ ( $8\text{ }\Omega$ ) which then splits into two parallel branches containing resistor $R_2$ ( $12\text{ }\Omega$ ) and resistor $R_3$ ( $24\text{ }\Omega$ ) respectively. How much electrical power is dissipated specifically by resistor $R_2$ ?

48\text{ W}

72\text{ W}

108\text{ W}

192\text{ W}

Show answer & explanation

Answer: $48\text{ W}$

Answer

The power dissipated specifically by resistor

R_2

is

48\text{ W}

.

To find the power dissipated by

R_2

, we must determine either the specific voltage across it or the specific current through it. First, we find the parallel equivalent resistance of

R_2

and

R_3

:

12\text{ }\Omega

and

24\text{ }\Omega

in parallel give

8\text{ }\Omega

. Adding

R_1

(

8\text{ }\Omega

) gives a total circuit resistance of

16\text{ }\Omega

. The total circuit current is

48\text{ V} / 16\text{ }\Omega = 3\text{ A}

. This

3\text{ A}

current flows through the parallel equivalent resistance (

8\text{ }\Omega

), creating a voltage drop of

24\text{ V}

across the parallel branches. Using this specific voltage for

R_2

, we calculate its power dissipation:

P = V^2 / R = 24^2 / 12 = 576 / 12 = 48\text{ W}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the parallel network containing

R_2

and

R_3

.

R_p = \frac{12 \times 24}{12 + 24} = \frac{288}{36} = 8\text{ }\Omega

This condenses the parallel branches into a single equivalent resistance to find the total circuit resistance.

2

Calculate the total resistance of the circuit.

R_{total} = R_1 + R_p = 8 + 8 = 16\text{ }\Omega

Adding the series resistor to the equivalent parallel resistance gives the total resistance.

3

Calculate the total current flowing from the source.

I_{total} = \frac{V_{total}}{R_{total}} = \frac{48}{16} = 3\text{ A}

Ohm's Law allows us to find the total current, which flows entirely through

R_1

.

4

Calculate the voltage drop across the parallel network.

V_p = I_{total} \times R_p = 3 \times 8 = 24\text{ V}

Since components in parallel share the same voltage, this is the voltage across both

R_2

and

R_3

.

5

Calculate the power dissipated by

R_2

.

P_{R2} = \frac{V_p^2}{R_2} = \frac{24^2}{12} = \frac{576}{12} = 48\text{ W}

Using Watt's Law with the specific voltage across

R_2

and its specific resistance gives its power dissipation.

Key Concept

Calculating power in a series-parallel circuit requires identifying the specific voltage across or current through the individual component, rather than using total circuit values.

Question 11Question →

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

Show answer & explanation

Answer: 1200

Answer

The total electrical power dissipated by the resistor is

1200\text{ W}

.

According to Watt's Law, power dissipated in a resistor when current and resistance are known is calculated using the formula

P = I^2 R

. Squaring the current (

4\text{ A}

) gives

16\text{ A}^2

, and multiplying by the resistance of

75\text{ }\Omega

produces an electrical power dissipation of

1200\text{ W}

.

Step-by-Step Solution

1

Identify the known circuit parameters

Current

I = 4\text{ A}

and resistance

R = 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 = 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\text{ A})^2 \times 75\text{ }\Omega = 16 \times 75 = 1200\text{ W}

Evaluating

(4)^2 = 16

and multiplying by

75

yields the total power in watts.

Key Concept

Electrical power dissipation in a resistive component using Watt's Law (

P = I^2 R

)

Estimated Time:1m 0s

Question 12Question →

A direct-current ( $24\text{ V}$ DC) cooling fan motor installed in an electronics enclosure has an internal resistance of $16\text{ }\Omega$ . What is the total electrical power consumed by the fan motor during normal operation?

1.5\text{ W}

24\text{ W}

36\text{ W}

384\text{ W}

Show answer & explanation

Answer: $36\text{ W}$

Answer

The total electrical power consumed by the fan motor is

36\text{ W}

.

Electrical power (

P

) delivered to a resistive load is given by

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

. Substituting the given parameters yields

P = \frac{(24\text{ V})^2}{16\text{ }\Omega} = \frac{576}{16} = 36\text{ W}

. Alternatively, by Ohm's Law, the current drawn is

I = \frac{24\text{ V}}{16\text{ }\Omega} = 1.5\text{ A}

, yielding

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

.

Step-by-Step Solution

1

Determine the current flowing through the motor using Ohm's Law (

I = \frac{V}{R}

).

I = \frac{24\text{ V}}{16\text{ }\Omega} = 1.5\text{ A}

Current is needed to calculate power using Watt's Law (

P = V \times I

).

2

Calculate the electrical power using Watt's Law (

P = V \times I

) or directly using

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

.

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

(or

P = \frac{24^2}{16} = \frac{576}{16} = 36\text{ W}

)

Electrical power is the rate of energy consumption and is given by the product of voltage and current.

Key Concept

Electrical power dissipation in a resistive direct-current load calculated from voltage and resistance (

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

).

Estimated Time:1m 0s

Question 13Question →

An electronics technician measures the operating parameters for four direct-current (DC) subsystems on an avionics test bench. Match each electrical subsystem described on the left with its correct operating power consumption on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The

150\text{ }\Omega

solenoid at

0.4\text{ A}

dissipates

24\text{ W}

; the

28\text{ V}

transceiver at

2.5\text{ A}

consumes

70\text{ W}

; the

12\text{ V}

,

24\text{ }\Omega

backlight consumes

6\text{ W}

; and the

120\text{ V}

,

40\text{ }\Omega

heating element dissipates

360\text{ W}

.

Each electrical load's power dissipation is determined by applying the appropriate variation of Watt's Law based on the two given quantities:

P = I^2 R

for the solenoid yielding

24\text{ W}

,

P = V \times I

for the transceiver yielding

70\text{ W}

,

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

for the instrument backlight yielding

6\text{ W}

, and

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

for the de-icing element yielding

360\text{ W}

.

Step-by-Step Solution

1

Calculate power for the

150\text{ }\Omega

relay solenoid with

0.4\text{ A}

current.

P = I^2 R = (0.4)^2 \times 150 = 0.16 \times 150 = 24\text{ W}

When current and resistance are known, use Joule's law / Watt's law variation

P = I^2 R

.

2

Calculate power for the

28\text{ V}

transceiver drawing

2.5\text{ A}

.

P = V \times I = 28 \times 2.5 = 70\text{ W}

When voltage and current are known, use the fundamental power formula

P = V \times I

.

3

Calculate power for the

12\text{ V}

backlight with

24\text{ }\Omega

resistance.

P = \frac{V^2}{R} = \frac{12^2}{24} = \frac{144}{24} = 6\text{ W}

When voltage and resistance are known, use the power formula

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

.

4

Calculate power for the

120\text{ V}

heating element with

40\text{ }\Omega

resistance.

P = \frac{V^2}{R} = \frac{120^2}{40} = \frac{14400}{40} = 360\text{ W}

When voltage and resistance are known, use

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

.

Key Concept

Electrical power is calculated using the three primary variations of Watt's Law:

P = V \times I

,

P = I^2 R

, and

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

.

Estimated Time:1m 30s

Question 14Question →

A direct-current starter relay coil in a heavy transport vehicle has an internal resistance of $15\text{ }\Omega$ and draws a steady operating current of $6\text{ A}$ . What is the total electrical power dissipated by the relay coil?

90\text{ W}

540\text{ W}

180\text{ W}

2.5\text{ W}

Show answer & explanation

Answer: $540\text{ W}$

Answer

540\text{ W}

Electrical power dissipated by a resistor with a known current is found using the formula

P = I^2 \times R

. Squaring the current of

6\text{ A}

gives

36\text{ A}^2

, and multiplying by the coil resistance of

15\text{ }\Omega

results in

540\text{ W}

.

Step-by-Step Solution

1

Identify the given circuit parameters and required quantity.

Current

I = 6\text{ A}

, Resistance

R = 15\text{ }\Omega

, Unknown = Power (

P

).

Selecting the correct formula depends on the given parameters.

2

Select the appropriate power formula relating power, current, and resistance.

P = I^2 \times R

When current and resistance are known, Joule's/Watt's Law formula

P = I^2 R

directly determines electrical power.

3

Square the current value and multiply by resistance.

P = (6\text{ A})^2 \times 15\text{ }\Omega = 36 \times 15 = 540\text{ W}

Completing the arithmetic yields the power in watts.

Key Concept

Electrical power dissipation in a resistive component is directly proportional to resistance and the square of the current (

P = I^2 R

).

Estimated Time:1m 0s

Question 15Question →

A direct-current sensor heating element has an internal resistance of $12\text{ }\Omega$ and carries a steady current of $2.5\text{ A}$ . What is the electrical power dissipated by the heating element, in watts?

Show answer & explanation

Answer: 75

Answer

The electrical power dissipated by the heating element is

75\text{ W}

.

Electrical power dissipated by a resistive element is calculated using Watt's Law as

P = I^2 R

. Squaring the current of

2.5\text{ A}

yields

6.25\text{ A}^2

, and multiplying by the resistance of

12\text{ }\Omega

results in

75\text{ W}

.

Step-by-Step Solution

1

Identify the known circuit parameters

I = 2.5\text{ A}

and

R = 12\text{ }\Omega

These are the two direct measurements provided for the resistive load.

2

Select the appropriate power formula

P = I^2 R

Watt's Law specifies that power dissipation through a resistance carrying a known current equals the square of the current multiplied by the resistance.

3

Calculate the power dissipation

P = (2.5\text{ A})^2 \times 12\text{ }\Omega = 6.25 \times 12 = 75\text{ W}

Squaring

2.5

gives

6.25

, and multiplying

6.25

by

12

produces

75

watts.

Key Concept

Calculating electrical power using current and resistance (

P = I^2 R

)

Estimated Time:1m 0s

Question 16Question →

An electrical heating strip installed in an outdoor surveillance camera enclosure is connected across a $24\text{ V}$ direct-current (DC) power source and draws an operating current of $750\text{ mA}$ . What is the total electrical power consumed by the heating strip?

32\text{ W}

18{}000\text{ W}

18\text{ W}

13.5\text{ W}

Show answer & explanation

Answer: $18\text{ W}$

Answer

18\text{ W}

Electrical power is calculated using the formula

P = V \times I

. Converting

750\text{ mA}

to amperes yields

0.75\text{ A}

. Multiplying the supply voltage (

24\text{ V}

) by the current (

0.75\text{ A}

) gives

18\text{ W}

.

Step-by-Step Solution

1

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

I = \frac{750\text{ mA}}{1000} = 0.75\text{ A}

Standard electrical power equations require current to be expressed in amperes.

2

Apply Watt's Law for electrical power

P = V \times I

Electrical power equals the product of voltage and current.

3

Substitute the voltage and current values into the equation to calculate power

P = 24\text{ V} \times 0.75\text{ A} = 18\text{ W}

Multiplying the applied potential by the resulting current yields power in watts.

Key Concept

Calculation of electrical power in a direct-current circuit using Watt's Law (

P = V \times I

) with metric unit conversion.

Estimated Time:1m 0s

Question 17Question →

An electronics technician is evaluating four separate direct-current (DC) loads installed in an automated weather observatory. Match each DC load description with its calculated electrical power dissipation.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Telemetry data logger (

24\text{ V}

,

0.25\text{ A}

) matches

6\text{ W}

; Antenna de-icing heating element (

4\text{ A}

,

5\text{ }\Omega

) matches

80\text{ W}

; Warning beacon (

36\text{ V}

,

18\text{ }\Omega

) matches

72\text{ W}

; Ventilation blower motor (

48\text{ V}

,

2.5\text{ A}

) matches

120\text{ W}

.

Each electrical load is correctly paired by selecting and computing the appropriate power equation based on the given electrical parameters:

P = V \times I

for voltage and current,

P = I^2 \times R

for current and resistance, and

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

for voltage and resistance.

Step-by-Step Solution

1

Calculate the power dissipation for the telemetry data logger using given voltage and current.

P = V \times I = 24\text{ V} \times 0.25\text{ A} = 6\text{ W}

When voltage (

V

) and current (

I

) are given, power is the product of voltage and current.

2

Calculate the power dissipation for the antenna de-icing heating element using current and resistance.

P = I^2 \times R = (4\text{ A})^2 \times 5\text{ }\Omega = 16 \times 5 = 80\text{ W}

When current (

I

) and resistance (

R

) are given, power equals the square of the current multiplied by resistance.

3

Calculate the power dissipation for the warning beacon using voltage and resistance.

P = \frac{V^2}{R} = \frac{(36\text{ V})^2}{18\text{ }\Omega} = \frac{1296}{18} = 72\text{ W}

When voltage (

V

) and resistance (

R

) are given, power equals the square of the voltage divided by resistance.

4

Calculate the power dissipation for the ventilation blower motor using voltage and current.

P = V \times I = 48\text{ V} \times 2.5\text{ A} = 120\text{ W}

Multiplying the operating voltage by the operating current yields the total power consumption in watts.

Key Concept

Electrical power in direct-current (DC) circuits represents the rate at which electrical energy is converted into other energy forms (such as heat, light, or mechanical motion). The three standard formulas derived from Ohm's Law and Watt's Law are

P = V \times I

,

P = I^2 \times R

, and

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

.

Estimated Time:1m 30s

Question 18Question →

A high-intensity auxiliary floodlight installed on a naval patrol craft has an internal operating resistance of $6\text{ }\Omega$ and connects directly across a $24\text{ V}$ direct-current (DC) electrical bus. How much electrical power does the floodlight consume during operation?

4\text{ W}

96\text{ W}

144\text{ W}

0.25\text{ W}

Show answer & explanation

Answer: $96\text{ W}$

Answer

96\text{ W}

Electrical power is calculated using the formula

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

. Substituting

24\text{ V}

and

6\text{ }\Omega

gives

P = \frac{24^2}{6} = \frac{576}{6} = 96\text{ W}

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage

V = 24\text{ V}

and Resistance

R = 6\text{ }\Omega

These two values are provided directly in the problem statement.

2

Select the power formula relating voltage and resistance

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

Combining Watt's Law (

P = V \times I

) with Ohm's Law (

I = \frac{V}{R}

) allows direct calculation of power without first solving for current.

3

Substitute the values and calculate power

P = \frac{(24\text{ V})^2}{6\text{ }\Omega} = \frac{576}{6} = 96\text{ W}

Squaring

24

yields

576

, and dividing by

6

gives

96\text{ W}

.

Key Concept

Calculating electrical power from voltage and resistance using

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

Alternative Method

Calculate circuit current first using Ohm's Law:

I = \frac{V}{R} = \frac{24\text{ V}}{6\text{ }\Omega} = 4\text{ A}

. Then apply

P = V \times I = 24\text{ V} \times 4\text{ A} = 96\text{ W}

.

Estimated Time:45s

Question 19Question →

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 ( $\text{W}$ )?

Show answer & explanation

Answer: 360

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

1

State the given circuit parameters

V = 120\text{ V}

,

R = 40\text{ }\Omega

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

2

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.

3

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

Question 20Question →

A portable communications transceiver connected to a $12\text{ V}$ direct-current (DC) power source consumes $36\text{ W}$ of electrical power during operation. What is the effective resistance of the transceiver circuit?

3\text{ }\Omega

4\text{ }\Omega

0.33\text{ }\Omega

432\text{ }\Omega

Show answer & explanation

Answer: $4\text{ }\Omega$

Answer

The effective resistance of the transceiver circuit is

4\text{ }\Omega

.

Electrical power is related to voltage and resistance by the formula

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

. Rearranging this equation to solve for resistance gives

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

. Substituting the given values:

R = \frac{(12)^2}{36} = \frac{144}{36} = 4\text{ }\Omega

. Alternatively, calculating the current first gives

I = \frac{P}{V} = \frac{36}{12} = 3\text{ A}

, and applying Ohm's Law gives

R = \frac{V}{I} = \frac{12}{3} = 4\text{ }\Omega

.

Step-by-Step Solution

1

Identify the given values and appropriate power formula relating power (

P

), voltage (

V

), and resistance (

R

).

P = 36\text{ W}

,

V = 12\text{ V}

, using

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

.

This formula directly links the known electrical values to the unknown circuit resistance.

2

Rearrange the formula to solve for resistance (

R

).

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

Isolating the unknown variable allows direct calculation.

3

Substitute the given numerical values and calculate the final resistance.

R = \frac{(12\text{ V})^2}{36\text{ W}} = \frac{144}{36} = 4\text{ }\Omega

Squaring the voltage and dividing by total power yields the resistance in ohms.

Key Concept

Relationship between electrical power, voltage, and resistance in DC circuits

Estimated Time:1m 0s

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