Circuits and Magnetism Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 8

Question 141Question →

A $72\text{ V}$ direct-current (DC) power supply energizes an industrial sensor interface consisting of two resistors connected in parallel. The first resistor ( $R_1$ ) has a resistance of $24\ \Omega$ . If the power supply delivers a total current of $5\text{ A}$ to the circuit, what is the resistance of the second resistor ( $R_2$ )?

9.6\ \Omega

36\ \Omega

14.4\ \Omega

28.8\ \Omega

Show answer & explanation

Answer: $36\ \Omega$

Answer

36\ \Omega

In a parallel circuit, the voltage across every parallel branch is identical to the source voltage (

72\text{ V}

). The current through the known resistor is

I_1 = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

. Since the total current entering the parallel junction is

5\text{ A}

, the remaining current flowing through the second branch must be

I_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}

. Using Ohm's law for the second branch yields

R_2 = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega

.

Step-by-Step Solution

1

Calculate the current flowing through the first branch resistor (

R_1

)

I_1 = \frac{V}{R_1} = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

In a parallel circuit, each branch experiences the full supply voltage of

72\text{ V}

.

2

Determine the current flowing through the second branch (

I_2

)

I_2 = I_{\text{total}} - I_1 = 5\text{ A} - 3\text{ A} = 2\text{ A}

By Kirchhoff's Current Law, total current in a parallel circuit equals the sum of the individual branch currents.

3

Calculate the resistance of the second resistor (

R_2

) using Ohm's law

R_2 = \frac{V}{I_2} = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega

Applying Ohm's law (

R = \frac{V}{I}

) with the branch voltage and branch current gives the branch resistance.

Key Concept

Parallel branch voltage equality, Kirchhoff's Current Law, and Ohm's law branch calculations

Estimated Time:1m 0s

Question 142Question →

An automotive diagnostic technician connects an auxiliary relay coil to a $12\text{ V}$ direct-current (DC) power supply. An ammeter placed in series measures a current of $150\text{ mA}$ . What is the electrical resistance of the relay coil?

1.8\ \Omega

12.5\ \Omega

80\ \Omega

800\ \Omega

Show answer & explanation

Answer: $80\ \Omega$

Answer

The electrical resistance of the relay coil is

80\ \Omega

.

According to Ohm's law, electrical resistance is calculated by dividing voltage by current (

R = \frac{V}{I}

). Converting

150\text{ mA}

to

0.15\text{ A}

and dividing

12\text{ V}

by

0.15\text{ A}

gives

80\ \Omega

.

Step-by-Step Solution

1

Convert the current from milliamperes (

\text{mA}

) to base units of amperes (

\text{A}

).

I = 150\text{ mA} = 150 \times 10^{-3}\text{ A} = 0.15\text{ A}

Ohm's law requires values in standard SI base units (volts, amperes, and ohms) before performing calculations.

2

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

) and substitute the given values.

R = \frac{V}{I} = \frac{12\text{ V}}{0.15\text{ A}} = 80\ \Omega

Resistance is equal to the potential difference divided by the current flowing through the component.

Key Concept

Ohm's Law states that resistance equals voltage divided by current (

R = \frac{V}{I}

), requiring current to be expressed in amperes.

Estimated Time:45s

Question 143Question →

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 144Question →

A shipboard auxiliary alarm horn is connected across a $72\text{ V}$ direct-current (DC) power rail. When activated, the horn draws a steady current of $120\text{ mA}$ . Based on Ohm's Law, what is the total electrical resistance of the horn circuit in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 600

Answer

The total electrical resistance of the horn circuit is

600\ \Omega

.

According to Ohm's Law, resistance equals voltage divided by current (

R = \frac{V}{I}

). Converting

120\text{ mA}

to standard amperes gives

0.12\text{ A}

. Dividing

72\text{ V}

by

0.12\text{ A}

yields an electrical resistance of

600\ \Omega

.

Step-by-Step Solution

1

Convert the current measurement from milliamperes to base amperes.

I = 120\text{ mA} = 0.12\text{ A}

Ohm's Law requires standard SI units (volts, amperes, and ohms) for direct substitution.

2

Rearrange Ohm's Law (

V = I \times R

) to isolate resistance (

R

).

R = \frac{V}{I}

The unknown quantity in this circuit problem is resistance.

3

Divide the applied voltage by the circuit current.

R = \frac{72\text{ V}}{0.12\text{ A}} = 600\ \Omega

Evaluating the division yields the equivalent circuit resistance in ohms.

Key Concept

Ohm's Law defines the proportional relationship between voltage, current, and resistance (

R = \frac{V}{I}

).

Estimated Time:45s

Question 145Question →

A technician is testing a telemetry circuit board and measures a steady direct current of $40\text{ mA}$ flowing through a $750\ \Omega$ load resistor. What is the voltage drop across this resistor?

300\text{ V}

18.75\text{ V}

30\text{ V}

3.0\text{ V}

Show answer & explanation

Answer: $30\text{ V}$

Answer

30\text{ V}

According to Ohm's Law (

V = I \times R

), the voltage drop across a component is calculated by multiplying the current flowing through it by its resistance. Converting

40\text{ mA}

to standard amperes gives

0.040\text{ A}

. Multiplying

0.040\text{ A}

by

750\ \Omega

yields a potential difference of

30\text{ V}

.

Step-by-Step Solution

1

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

I = 40\text{ mA} = 40 \times 10^{-3}\text{ A} = 0.040\text{ A}

Standard electrical calculations using Ohm's Law require base units of amperes, ohms, and volts.

2

Apply Ohm's Law to determine the potential difference (

V = I \times R

)

V = 0.040\text{ A} \times 750\ \Omega = 30\text{ V}

Voltage drop across a resistive element is directly proportional to the product of the current and resistance.

Key Concept

Ohm's Law (

V = I \times R

) and metric unit conversion

Estimated Time:45s

Question 146Question →

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 147Question →

An aircraft instrument panel contains a direct-current (DC) indicator circuit powered by a $120\text{ V}$ source. What are the resulting circuit currents under each resistance condition? Fill in the blanks with the correct numerical values.

Fill in the blanks below

When connected to a

2.4\text{ k}\Omega

resistor, the current is

\text{mA}

. If the resistance is increased to

4.8\text{ k}\Omega

while maintaining constant voltage, the new current is

\text{mA}

.

Show answer & explanation

Answer

The current with a

2.4\text{ k}\Omega

resistor is

50\text{ mA}

, and the current with a

4.8\text{ k}\Omega

resistor is

25\text{ mA}

.

According to Ohm's Law (

I = \frac{V}{R}

), converting

2.4\text{ k}\Omega

to

2400\ \Omega

yields an initial current of

\frac{120\text{ V}}{2400\ \Omega} = 0.05\text{ A} = 50\text{ mA}

. When the resistance doubles to

4.8\text{ k}\Omega

(

4800\ \Omega

) with constant voltage, the current is halved to

25\text{ mA}

.

Step-by-Step Solution

1

Convert the first resistance value from kilohms to ohms and apply Ohm's Law to calculate the initial current

R_1 = 2.4\text{ k}\Omega = 2400\ \Omega

;

I_1 = \frac{V}{R_1} = \frac{120\text{ V}}{2400\ \Omega} = 0.05\text{ A} = 50\text{ mA}

Ohm's Law states that current equals voltage divided by resistance (

I = \frac{V}{R}

).

2

Convert the second resistance value or apply the principle of inverse proportionality to find the new current

R_2 = 4.8\text{ k}\Omega = 4800\ \Omega

;

I_2 = \frac{120\text{ V}}{4800\ \Omega} = 0.025\text{ A} = 25\text{ mA}

(or since resistance doubled at constant voltage, current halved:

\frac{50\text{ mA}}{2} = 25\text{ mA}

)

Current is inversely proportional to resistance when voltage remains constant.

Key Concept

Ohm's Law (

I = \frac{V}{R}

) and inverse proportionality between current and resistance at constant voltage

Estimated Time:45s

Question 148Question →

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 149Question →

A radar transceiver module operates on a $45\text{ V}$ direct-current (DC) power line. When the module is active, a series ammeter registers a current of $15\text{ mA}$ . What is the total electrical resistance of the module?

3\ \Omega

3\text{ k}\Omega

675\ \Omega

333\ \Omega

Show answer & explanation

Answer: $3\text{ k}\Omega$

Answer

3\text{ k}\Omega

(or

3,000\ \Omega

)

According to Ohm's Law, resistance is the quotient of voltage divided by current (

R = \frac{V}{I}

). Converting

15\text{ mA}

to standard units gives

0.015\text{ A}

. Dividing the

45\text{ V}

potential difference by

0.015\text{ A}

results in

3,000\ \Omega

, which is equivalent to

3\text{ k}\Omega

.

Step-by-Step Solution

1

Convert the current measurement from milliamperes to base amperes.

I = 15\text{ mA} = 15 \times 10^{-3}\text{ A} = 0.015\text{ A}

Ohm's Law calculations require current to be expressed in standard SI units (amperes) to yield resistance in ohms.

2

Rearrange Ohm's Law (

V = I \times R

) to solve for resistance (

R

).

R = \frac{V}{I}

Isolating the unknown variable allows direct substitution of known circuit values.

3

Substitute the voltage and current values into the rearranged formula.

R = \frac{45\text{ V}}{0.015\text{ A}} = 3,000\ \Omega = 3\text{ k}\Omega

Dividing the potential difference by the current yields the total circuit resistance.

Key Concept

Ohm's Law defines the fundamental linear relationship between voltage, current, and resistance (

V = I \times R

).

Estimated Time:45s

Question 150Question →

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 151Question →

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 152Question →

An avionics technician is troubleshooting a direct-current (DC) audio preamplifier circuit. The circuit is powered by an $18\text{ V}$ DC source and draws a steady current of $2.5\text{ mA}$ . What is the total resistance of the circuit in kilohms ( $\text{k}\Omega$ )?

Show answer & explanation

Answer: 7.2

Answer

7.2 kΩ

According to Ohm's Law, resistance is determined by dividing voltage by current (

R = \frac{V}{I}

). Converting

2.5\text{ mA}

to

0.0025\text{ A}

and dividing

18\text{ V}

by

0.0025\text{ A}

yields

7200\ \Omega

, which equals

7.2\text{ k}\Omega

.

Step-by-Step Solution

1

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

I = 2.5\text{ mA} = 0.0025\text{ A}

Ohm's Law formulas require electrical quantities to be in standard base units (volts, amperes, and ohms) before computation.

2

Apply Ohm's Law rearranged to solve for resistance (

R = \frac{V}{I}

)

R = \frac{18\text{ V}}{0.0025\text{ A}} = 7200\ \Omega

Resistance equals the electromotive force divided by the resulting current flow.

3

Convert the calculated resistance from ohms (Ω) to kilohms (kΩ)

R = \frac{7200\ \Omega}{1000\ \Omega/\text{k}\Omega} = 7.2\text{ k}\Omega

The question specifically requests the final numerical value in kilohms.

Key Concept

Ohm's Law (

R = \frac{V}{I}

) with metric prefix conversion (

1\text{ mA} = 10^{-3}\text{ A}

and

1\text{ k}\Omega = 10^{3}\ \Omega

)

Estimated Time:45s

Question 153Question →

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 154Question →

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 155Question →

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 156Question →

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 157Question →

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

Question 158Question →

When a permanent bar magnet is broken into two separate pieces, each piece becomes an independent magnet containing both a north pole and a south pole.

True

False

Show answer & explanation

Answer: True

Answer

True

In classical electromagnetism and magnetic materials, magnetic monopoles do not exist. Every segment of a magnetized substance retains its aligned atomic domains, ensuring that any severed piece forms a complete magnet with its own distinct north and south poles.

Step-by-Step Solution

1

Analyze the fundamental structure of magnetic materials.

Ferromagnetic materials contain microscopic magnetic domains where atomic magnetic moments are aligned, each acting as a dipole with paired north and south poles.

Understanding the atomic origin of magnetism explains why isolated magnetic poles cannot be produced by mechanical division.

2

Determine the effect of physically separating a magnet into two pieces.

Each broken piece retains aligned internal domains, causing opposite poles to manifest at the newly exposed fractured ends.

Magnetic flux lines must form continuous closed loops, requiring every physical magnet segment to have both a north and a south pole.

Key Concept

Magnetic Dipoles and Non-existence of Monopoles

Estimated Time:45s

Question 159Question →

In electronics and electromagnetic theory, magnetic materials and circuits possess distinct measurable properties. Match each magnetic term on the left with its correct technical definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Permeability matches the measure of a material's ability to allow magnetic flux to pass; Retentivity matches the capability to retain residual magnetism; Reluctance matches the opposition to magnetic lines of force; and Flux Density matches the concentration of field lines per unit cross-sectional area.

Each fundamental magnetic term pairs with its standard physical definition: Permeability measures the ease of supporting magnetic flux, Retentivity is the capacity to hold residual magnetism, Reluctance is the opposition to magnetic lines of force, and Flux Density is the concentration of field lines per unit area.

Step-by-Step Solution

1

Define Permeability

Corresponds to the measure of ease with which magnetic flux develops in a material.

Permeability quantifies flux conductivity through a material.

2

Define Retentivity

Corresponds to the retention of residual magnetism after removing the magnetizing source.

Retentive materials stay magnetized after external fields are turned off.

3

Define Reluctance

Corresponds to the opposition against the creation of magnetic flux.

Reluctance behaves like resistance in an electric circuit.

4

Define Flux Density

Corresponds to the number of magnetic lines of force per unit area.

Flux density describes field concentration across a designated plane.

Key Concept

Fundamental magnetic properties and technical terminology governing magnetic circuits.

Question 160Question →

When an unmagnetized ferromagnetic bar is exposed to an external magnetic field, which of the following processes causes the bar to become magnetized?

Free valence electrons are permanently stripped away, transforming the metal into an electrical insulator.

The internal magnetic dipoles repel adjacent atoms until all internal poles reverse and cancel out.

The randomly oriented magnetic domains rotate and align in the direction of the external field.

The applied magnetic energy is converted into static electric charge stored across internal dielectric layers.

Show answer & explanation

Answer: The randomly oriented magnetic domains rotate and align in the direction of the external field.

Answer

The randomly oriented magnetic domains rotate and align in the direction of the external field.

In an unmagnetized ferromagnetic substance, microscopic groups of atoms called magnetic domains have their individual magnetic moments oriented in random directions, canceling each other out. When placed in an external magnetic field, these domains rotate and align parallel to the external field lines, giving the material an overall net magnetic polarity.

Step-by-Step Solution

1

Recall the microscopic structure of unmagnetized ferromagnetic materials.

Unmagnetized ferromagnetic materials consist of microscopic regions called magnetic domains, where magnetic dipoles are grouped but oriented in random directions, resulting in zero net magnetic field.

Understanding domain theory is essential for explaining how materials become magnetized.

2

Analyze the effect of an external magnetic field on these domains.

The external magnetic force exerts a torque on the individual domains, causing them to rotate and align parallel to the external magnetic flux lines.

Alignment of internal domains creates additive magnetic fields that produce distinct north and south poles in the material.

Key Concept

Magnetic domain theory and alignment in ferromagnetic materials

Estimated Time:45s

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