Circuits and Magnetism Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 6

Question 101Question →

An electric horn connected to a $12\text{ V}$ direct current (DC) automotive battery has an internal resistance of $3\ \Omega$ . What is the steady current flowing through the horn circuit? Complete the statement below.

Fill in the blanks below

According to Ohm's law, the current flowing through the horn is amperes.

Show answer & explanation

Answer

4 amperes (or 4)

According to Ohm's law, current is calculated using the formula

I = \frac{V}{R}

. Dividing the applied voltage (

12\text{ V}

) by the circuit resistance (

3\ \Omega

) gives a current of

4\text{ A}

.

Step-by-Step Solution

1

Identify the given electrical parameters from the problem statement

Voltage (

V

) =

12\text{ V}

, Resistance (

R

) =

3\ \Omega

Ohm's law relates voltage, current, and resistance in a closed DC circuit.

2

State Ohm's law rearranged to solve for current (

I

)

I = \frac{V}{R}

Current is directly proportional to voltage and inversely proportional to resistance.

3

Substitute the given numerical values into the formula and calculate

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

Dividing

12

volts by

3

ohms yields

4

amperes.

Key Concept

Ohm's Law states that current (

I

) equals voltage (

V

) divided by resistance (

R

).

Estimated Time:45s

Question 102Question →

A direct-current circuit powers three identical $60\,\Omega$ indicator lamps connected in parallel across a $12\text{ V}$ DC bus. During a routine operational check, an ammeter placed in series with the main supply line reads $0.4\text{ A}$ instead of the expected normal current of $0.6\text{ A}$ . Which condition best explains this current reading?

One indicator lamp has suffered an internal short circuit.

One indicator lamp has burned out and created an open circuit.

A fourth identical indicator lamp was accidentally wired in parallel.

The DC bus supply voltage has dropped to zero volts.

Show answer & explanation

Answer: One indicator lamp has burned out and created an open circuit.

Answer

One indicator lamp has burned out and created an open circuit.

In a parallel circuit, each branch operates independently across the full

12\text{ V}

supply. Under normal conditions, each

60\,\Omega

lamp draws

I = \frac{12\text{ V}}{60\,\Omega} = 0.2\text{ A}

, yielding a total current of

3 \times 0.2\text{ A} = 0.6\text{ A}

. When one lamp fails open (such as a broken filament), its branch current falls to

0\text{ A}

while the other two branches continue drawing

0.2\text{ A}

each. The sum of the remaining branch currents is

0.2\text{ A} + 0.2\text{ A} = 0.4\text{ A}

, which matches the measured ammeter reading.

Step-by-Step Solution

1

Calculate the normal operating current for a single parallel branch using Ohm's Law.

I_{\text{branch}} = \frac{V}{R} = \frac{12\text{ V}}{60\,\Omega} = 0.2\text{ A}

In a parallel circuit, each branch receives the full source voltage of

12\text{ V}

independently.

2

Determine the expected total current under normal operating conditions.

I_{\text{total}} = 3 \times 0.2\text{ A} = 0.6\text{ A}

Total current in a parallel circuit is the sum of all individual branch currents (

I_{\text{total}} = I_1 + I_2 + I_3

).

3

Analyze the measured current to identify the active number of branches and diagnose the fault.

\text{Active branches} = \frac{0.4\text{ A}}{0.2\text{ A/branch}} = 2\text{ branches}

A drop from

0.6\text{ A}

to

0.4\text{ A}

indicates that exactly one

0.2\text{ A}

branch is no longer drawing current, which occurs when a lamp burns out into an open circuit.

Key Concept

Parallel Circuit Fault Diagnosis and Open-Circuit Current Reduction

Alternative Method

Calculate equivalent resistance before and after: Normal equivalent resistance is

R_{\text{eq}} = \frac{60}{3} = 20\,\Omega

, drawing

I = \frac{12\text{ V}}{20\,\Omega} = 0.6\text{ A}

. With the measured current of

0.4\text{ A}

, the circuit's total resistance is

R_{\text{eq}} = \frac{12\text{ V}}{0.4\text{ A}} = 30\,\Omega

. Since two

60\,\Omega

resistors in parallel equal

30\,\Omega

(

\frac{60}{2} = 30\,\Omega

), exactly one branch is open.

Estimated Time:1m 0s

Question 103Question →

According to the fundamental laws of magnetism, like magnetic poles attract one another, while opposite magnetic poles repel one another.

True

False

Show answer & explanation

Answer: False

Answer

False. Like magnetic poles repel each other, while opposite magnetic poles attract each other.

The statement is false because the fundamental law of magnetism states that like poles repel one another and opposite poles attract one another.

Step-by-Step Solution

1

Recall the basic rule of magnetic pole interaction.

Like poles repel each other, and unlike poles attract each other.

Magnetic flux lines emerging from identical poles push against each other causing repulsion, while flux lines between opposite poles flow together causing attraction.

2

Evaluate the statement against the fundamental rule.

The statement asserts that like poles attract and opposite poles repel, which is the exact inverse of the true physical principle.

Because the statement reverses the definitions of attraction and repulsion for magnetic poles, it is false.

Key Concept

Fundamental Law of Magnetic Attraction and Repulsion

Estimated Time:30s

Question 104Question →

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 105Question →

An electronics technician performs diagnostic checks on various malfunctioning direct-current (DC) systems. Match each electrical troubleshooting symptom or meter reading on the left with its most probable circuit fault condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The symptom of 0.0 ohms across a 250-ohm coil matches short-circuited coil windings; full source voltage measured across closed switch terminals matches an open circuit from failed switch contacts; voltage dropping severely under load matches a high-resistance or corroded connection; and 0.0 ohms between the ungrounded positive lead and metallic chassis matches a direct ground fault.

Each diagnostic reading correlates directly to fundamental circuit failure modes: a shorted coil eliminates component resistance (

0\,\Omega

); an open switch drops the entire source voltage (

12\text{ V}

) because current cannot bridge the open gap; a high-resistance joint creates excessive voltage drop only when current flows (

V = I \times R

); and continuity between a 'hot' power lead and the chassis represents a short to ground.

Step-by-Step Solution

1

Analyze the relay coil resistance reading of

0.0\,\Omega

.

Identifies an internal short circuit.

Inductive components such as relay coils are designed with specific continuous wire resistance. A reading of

0.0\,\Omega

means the insulation between windings broke down, creating a direct short.

2

Analyze the closed switch voltage reading showing full source voltage.

Identifies an open switch fault.

According to Kirchhoff's Voltage Law, an open point in a series circuit drops the entire source voltage because it creates infinite resistance and blocks current, whereas a properly closed switch drops

0\text{ V}

.

3

Analyze the loaded vs. unloaded voltage drop behavior.

Identifies a high-resistance connection fault.

When no current flows (

I = 0

), no voltage is dropped across bad connections (

V = 0\text{ V}

). When load current flows, parasitic resistance drops voltage in proportion to current, starving the load.

4

Analyze the zero-ohm reading between the positive supply rail and metal chassis.

Identifies a direct chassis ground fault.

Conductors intended to carry current must remain insulated from the metal chassis. A

0.0\,\Omega

reading confirms an unintended low-resistance bridge to the chassis frame.

Key Concept

Circuit fault diagnosis using voltage drop and resistance characteristics for opens, shorts, grounds, and high-resistance faults.

Estimated Time:1m 30s

Question 106Question →

A railway signal monitoring circuit is powered by a $72\text{ V}$ DC power supply. The circuit consists of three components connected in series: an $85\ \Omega$ relay coil, a $35\ \Omega$ surge-limiting resistor, and an unknown protective resistor $R_x$ . If the entire circuit draws a steady current of $0.4\text{ A}$ , what is the resistance of the protective resistor $R_x$ in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 60

Answer

The resistance of the protective resistor

R_x

is

60\ \Omega

.

Applying Ohm's Law (

R_T = \frac{V_T}{I}

) gives a total series resistance of

180\ \Omega

. Since resistances in series add directly (

R_T = 85\ \Omega + 35\ \Omega + R_x = 120\ \Omega + R_x

), solving for the unknown yields

R_x = 180\ \Omega - 120\ \Omega = 60\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law.

R_T = \frac{V_{\text{source}}}{I_{\text{total}}} = \frac{72\text{ V}}{0.4\text{ A}} = 180\ \Omega

In a series circuit, total resistance equals total source voltage divided by total circuit current.

2

Determine the combined resistance of the known series components.

R_{\text{known}} = R_{\text{relay}} + R_{\text{surge}} = 85\ \Omega + 35\ \Omega = 120\ \Omega

Resistances in a series circuit are directly additive.

3

Subtract the known series resistance from the total circuit resistance to solve for

R_x

.

R_x = R_T - R_{\text{known}} = 180\ \Omega - 120\ \Omega = 60\ \Omega

The sum of all individual series resistors must equal the total equivalent resistance (

R_T = R_1 + R_2 + R_x

).

Key Concept

In a direct-current (DC) series circuit, current is identical through every component (

I_T = I_1 = I_2 = I_3

), and the total equivalent resistance is the arithmetic sum of all individual resistances (

R_T = R_1 + R_2 + \dots + R_n

).

Alternative Method

Find the voltage drops across the known components first:

V_{\text{relay}} = 0.4\text{ A} \times 85\ \Omega = 34\text{ V}

and

V_{\text{surge}} = 0.4\text{ A} \times 35\ \Omega = 14\text{ V}

. Subtract these from the total source voltage using Kirchhoff's Voltage Law:

V_{Rx} = 72\text{ V} - (34\text{ V} + 14\text{ V}) = 24\text{ V}

. Then find

R_x = \frac{V_{Rx}}{I} = \frac{24\text{ V}}{0.4\text{ A}} = 60\ \Omega

.

Estimated Time:1m 0s

Question 107Question →

An industrial control module features a $36\text{ V}$ DC source driving a resistor network. The configuration consists of a $150\ \Omega$ input resistor ( $R_1$ ) wired in series with a parallel pair of $300\ \Omega$ load resistors ( $R_2$ and $R_3$ ). A test point (TP1) is located at the junction separating $R_1$ from the parallel pair.

Under nominal conditions, the circuit draws $120\text{ mA}$ from the source, and a multimeter reads $18\text{ V}$ at TP1 with respect to ground.

A system failure occurs, and a technician records new diagnostic readings:
- The voltage at TP1 with respect to ground is now $36\text{ V}$ .
- The total current drawn from the $36\text{ V}$ source is now $240\text{ mA}$ .

Based on these diagnostic readings, which of the following component failures has occurred?

R_1

has suffered an internal short circuit.

TP1 has short-circuited directly to the chassis ground.

The common ground connection for

R_2

and

R_3

has become an open circuit.

R_1

has burned out, creating an open circuit.

Show answer & explanation

Answer: $R_1$ has suffered an internal short circuit.

Answer

The component failure is that

R_1

has suffered an internal short circuit.

For the voltage at TP1 to equal the source voltage (

36\text{ V}

), there must be no voltage drop across the series resistor

R_1

. This occurs either when there is no current flowing, or when

R_1

has zero resistance. Because the circuit is drawing

240\text{ mA}

(a higher-than-normal current), the circuit is not open. Therefore,

R_1

must have zero resistance due to an internal short circuit. When

R_1

is shorted, the total circuit resistance drops to just the

150\ \Omega

equivalent resistance of the parallel pair, which mathematically accounts for the exactly

240\text{ mA}

current draw (

36\text{ V} / 150\ \Omega

).

Step-by-Step Solution

1

Analyze the voltage reading at TP1 during the fault.

TP1 measures

36\text{ V}

, which equals the source voltage.

This indicates there is zero voltage drop across

R_1

. This can only happen if no current is flowing through

R_1

(an open circuit downstream) or if

R_1

has zero resistance (a short circuit).

2

Analyze the total current reading during the fault.

Total current is

240\text{ mA}

, which is twice the normal operating current.

Since current is flowing (and is higher than normal), the circuit cannot be open. Therefore, the zero voltage drop across

R_1

must be caused by a short circuit across

R_1

itself.

3

Verify the shorted

R_1

hypothesis mathematically using Ohm's Law.

If

R_1 = 0\ \Omega

, total resistance is

300\ \Omega / 2 = 150\ \Omega

. Total current =

36\text{ V} / 150\ \Omega = 0.24\text{ A}

(

240\text{ mA}

). Voltage at TP1 =

36\text{ V} - (0.24\text{ A} \times 0\ \Omega) = 36\text{ V}

.

Both mathematical results precisely match the technician's fault measurements, confirming

R_1

is shorted.

Key Concept

Series-Parallel Circuit Troubleshooting and Ohm's Law

Question 108Question →

Which of the following materials is classified as a ferromagnetic substance that is strongly attracted to a magnetic field?

Copper

Nickel

Aluminum

Mica

Show answer & explanation

Answer: Nickel

Answer

Nickel is a ferromagnetic material that is strongly attracted to a magnetic field.

Ferromagnetic materials such as iron, cobalt, and nickel have permanent atomic magnetic dipoles that align easily with an external magnetic field, creating strong magnetic attraction.

Step-by-Step Solution

1

Identify the characteristic properties of ferromagnetic materials.

Ferromagnetic substances contain atomic magnetic domains that readily align with an external field, resulting in strong magnetic attraction.

Ferromagnetism is the basic mechanism by which certain materials form permanent magnets or are strongly attracted to magnets.

2

Evaluate the listed materials to identify the ferromagnetic element.

The primary ferromagnetic elements at room temperature are iron, cobalt, and nickel.

Common conductors like copper and aluminum, as well as insulators like mica, do not exhibit ferromagnetism.

Key Concept

Ferromagnetic materials and magnetic permeability

Estimated Time:35s

Question 109Question →

A technician measures a steady current of $50\text{ mA}$ flowing through a $120\ \Omega$ fixed resistor in a signal conditioning module. What is the voltage drop across the resistor?

2.4\text{ V}

6\text{ V}

6{}000\text{ V}

0.6\text{ V}

Show answer & explanation

Answer: $6\text{ V}$

Answer

6\text{ V}

According to Ohm's law, the voltage drop across a component is determined by multiplying current by resistance (

V = I \times R

). Converting

50\text{ mA}

to

0.05\text{ A}

and multiplying by

120\ \Omega

gives

6\text{ V}

.

Step-by-Step Solution

1

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

50\text{ mA} = 0.05\text{ A}

Standard electrical formulas require current in amperes.

2

Apply Ohm's law formula for voltage (

V = I \times R

).

V = 0.05\text{ A} \times 120\ \Omega = 6\text{ V}

Voltage across a resistor equals the product of current and resistance.

Key Concept

Ohm's Law states that voltage is equal to current multiplied by resistance (

V = I \times R

).

Estimated Time:45s

Question 110Question →

A mobile radar subsystem is powered by a $24\text{ V}$ direct-current source. Its internal circuitry contains a parallel configuration with three distinct branches.

The total equivalent resistance of this parallel section is exactly $3\ \Omega$ .
A technician measures the current through the first branch to be $3\text{ A}$ .
The second branch has a known resistance of $24\ \Omega$ .

What is the resistance of the third branch in ohms?

Show answer & explanation

Answer: 6

Answer

The resistance of the third branch is

6\ \Omega

.

The correct value is derived by finding the total circuit current (

8\text{ A}

) and subtracting the known branch currents (

3\text{ A}

and

1\text{ A}

) to find the current in the third branch (

4\text{ A}

). Dividing the common

24\text{ V}

source by this

4\text{ A}

current yields

6\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit current using the total voltage and equivalent resistance.

I_{total} = 24\text{ V} / 3\ \Omega = 8\text{ A}

Ohm's law applies to the entire circuit, linking total voltage, total current, and equivalent resistance.

2

Determine the current flowing through the second branch.

I_2 = 24\text{ V} / 24\ \Omega = 1\text{ A}

In a parallel circuit, the full source voltage drops across every branch. Ohm's law (

I = V/R

) gives the branch current.

3

Calculate the current flowing through the third branch.

I_3 = 8\text{ A} - (3\text{ A} + 1\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current is the sum of the individual branch currents (

I_{total} = I_1 + I_2 + I_3

).

4

Calculate the resistance of the third branch.

R_3 = 24\text{ V} / 4\text{ A} = 6\ \Omega

Apply Ohm's law (

R = V/I

) specifically to the third branch using the source voltage and the calculated branch current.

Key Concept

Applying Ohm's law to both individual branches and the entire circuit, combined with parallel circuit rules for voltage and current.

Alternative Method

Use the parallel resistance reciprocal formula:

1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3

. First, find

R_1 = V/I_1 = 24\text{ V} / 3\text{ A} = 8\ \Omega

. Then, substitute the known values:

1/3 = 1/8 + 1/24 + 1/R_3

. Finding a common denominator of 24 gives

8/24 = 3/24 + 1/24 + 1/R_3

, which simplifies to

8/24 = 4/24 + 1/R_3

. Therefore,

1/R_3 = 4/24 = 1/6

, meaning

R_3 = 6\ \Omega

.

Estimated Time:1m 30s

Question 111Question →

A direct-current power supply in an automotive workshop delivers $24\text{ V}$ across two parallel branches. The first branch contains an $8\ \Omega$ inspection lamp, and the second branch contains a $24\ \Omega$ ventilation motor. What is the total current drawn by the circuit from the power supply?

0.75\text{ A}

4\text{ A}

3\text{ A}

6\text{ A}

Show answer & explanation

Answer: $4\text{ A}$

Answer

4\text{ A}

In a parallel circuit, the voltage across each branch is identical to the supply voltage (

24\text{ V}

). The current in each branch is calculated using Ohm's Law: the

8\ \Omega

load draws

3\text{ A}

(

24\text{ V} / 8\ \Omega

), and the

24\ \Omega

load draws

1\text{ A}

(

24\text{ V} / 24\ \Omega

). Summing these branch currents gives a total supply current of

4\text{ A}

.

Step-by-Step Solution

1

Calculate the current flowing through the first branch (

8\ \Omega

resistor).

I_1 = \frac{V}{R_1} = \frac{24\text{ V}}{8\ \Omega} = 3\text{ A}

In a parallel circuit, each branch receives the full source voltage of

24\text{ V}

.

2

Calculate the current flowing through the second branch (

24\ \Omega

resistor).

I_2 = \frac{V}{R_2} = \frac{24\text{ V}}{24\ \Omega} = 1\text{ A}

Ohm's Law applies independently to the second branch using the same

24\text{ V}

potential.

3

Sum the individual branch currents to find the total circuit current.

I_{\text{total}} = I_1 + I_2 = 3\text{ A} + 1\text{ A} = 4\text{ A}

According to Kirchhoff's Current Law, the total current supplied to a parallel network equals the sum of all branch currents.

Key Concept

In a parallel circuit, total current is the sum of the individual branch currents, each determined by dividing the constant source voltage by the branch resistance.

Alternative Method

Find the equivalent parallel resistance first:

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\ \Omega

. Then apply Ohm's Law to the whole circuit:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{24\text{ V}}{6\ \Omega} = 4\text{ A}

.

Estimated Time:1m 0s

Question 112Question →

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 113Question →

An electronics technician is troubleshooting an inoperative single-phase ventilation blower that fails to operate when energized. In what sequence should the technician perform the following diagnostic and repair actions to follow standard, safe troubleshooting procedures from initial assessment to final verification?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct logical sequence is: verify supply voltage at the input terminals, measure the voltage drop across the closed control switch, de-energize and lock out the power source, isolate and measure motor winding resistance with an ohmmeter, and finally replace the defective motor and test operation.

Standard electrical troubleshooting begins with verifying the source power, followed by live circuit voltage checks across active control elements. Once the faulty branch or load is localized, the circuit must be safely de-energized before isolating components for resistance testing. The final step is component replacement followed by an operational verification test.

Step-by-Step Solution

1

Check the input power supply with a voltmeter.

Confirms whether electrical power is reaching the system.

Eliminates external power failure or tripped upstream breakers before diagnosing internal faults.

2

Check the closed control switch for voltage drop under live conditions.

Determines if the switch contacts are conducting correctly (

0\text{ V}

drop across good closed contacts; full source voltage across open contacts).

Allows quick live-circuit localization of series circuit faults.

3

Disconnect and lock out the circuit power.

De-energizes the system to establish an electrically safe work condition.

Essential for personnel safety and required before using an ohmmeter, which can be damaged by live voltage.

4

Isolate one winding terminal and measure resistance using an ohmmeter.

Provides an accurate internal resistance reading (infinite ohms indicating an open winding).

Isolating one lead prevents parallel circuit paths from yielding false resistance readings.

5

Replace the faulty motor and restore power to test the unit.

Restores system operation and validates that the root fault is fully resolved.

Every repair procedure concludes with an operational check under normal load conditions.

Key Concept

Logical Troubleshooting and Safety Sequence

Estimated Time:1m 30s

Question 114Question →

A field communication station is powered by a $72\text{ V}$ direct-current supply connected across three parallel load branches. The resistance of the first branch is $18\ \Omega$ , the resistance of the second branch is $36\ \Omega$ , and the resistance of the third branch is $72\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), supplied to the circuit?

Show answer & explanation

Answer: 7

Answer

7 A

Each of the three branches is connected directly across the

72\text{ V}

source. Using Ohm's Law (

I = V/R

), the branch currents are

4\text{ A}

,

2\text{ A}

, and

1\text{ A}

. Adding these currents together yields the total current of

7\text{ A}

.

Step-by-Step Solution

1

Calculate the current in the first parallel branch using

I_1 = \frac{V}{R_1}

.

I_1 = \frac{72\text{ V}}{18\ \Omega} = 4\text{ A}

In a parallel DC circuit, the full supply voltage of

72\text{ V}

appears across every parallel branch.

2

Calculate the current in the second parallel branch using

I_2 = \frac{V}{R_2}

.

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

Applying Ohm's Law to the second branch with its specific resistance.

3

Calculate the current in the third parallel branch using

I_3 = \frac{V}{R_3}

.

I_3 = \frac{72\text{ V}}{72\ \Omega} = 1\text{ A}

Applying Ohm's Law to the third branch with its specific resistance.

4

Add all three individual branch currents together to determine the total current.

I_{\text{total}} = 4\text{ A} + 2\text{ A} + 1\text{ A} = 7\text{ A}

By Kirchhoff's Current Law, the total current leaving the power supply equals the sum of the currents through all parallel branches.

Key Concept

In a parallel circuit, the voltage across each parallel branch is identical to the source voltage, and the total circuit current is the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

).

Alternative Method

Find the total equivalent resistance (

R_{\text{eq}}

) first:

\frac{1}{R_{\text{eq}}} = \frac{1}{18} + \frac{1}{36} + \frac{1}{72} = \frac{4}{72} + \frac{2}{72} + \frac{1}{72} = \frac{7}{72}\ \Omega^{-1}

, so

R_{\text{eq}} = \frac{72}{7}\ \Omega

. Then find total current using

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{72}{72/7} = 7\text{ A}

.

Estimated Time:1m 0s

Question 115Question →

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 116Question →

A technician is troubleshooting a circuit where two identical heating elements, $R_1$ and $R_2$ , are connected in series across a $120\text{ V}$ AC supply. When the circuit is energized, neither element produces heat. Using a digital multimeter set to AC voltage, the technician measures $0\text{ V}$ across $R_1$ and $120\text{ V}$ across $R_2$ . Based on these measurements, which of the following is the most likely fault in the circuit?

R_1

has developed a short circuit.

R_2

has developed an open circuit.

R_2

has developed a short circuit.

R_1

has developed excessive contact resistance.

Show answer & explanation

Answer: $R_2$ has developed an open circuit.

Answer

The heating element

R_2

has an open circuit.

In a series circuit containing an open fault, the current drops to zero throughout the loop. According to Ohm's Law (

V = I \times R

), any intact component drops

0\text{ V}

because no current flows through it. Consequently, the entire source potential of

120\text{ V}

appears directly across the terminals of the open element (

R_2

).

Step-by-Step Solution

1

Analyze current flow in the series circuit based on operating symptoms.

Because neither series element produces heat, the total circuit current is zero (

I = 0\text{ A}

).

In a series circuit, any break in continuity stops current through all components.

2

Apply Ohm's Law to the intact component

R_1

.

V_{R1} = I \times R_1 = 0\text{ A} \times R_1 = 0\text{ V}

.

Without current flow, no potential difference develops across an intact resistor.

3

Evaluate the voltage drop across the faulted component

R_2

.

V_{R2} = 120\text{ V}

(entire supply voltage).

An open component represents infinite resistance, placing the multimeter in series with the line voltage across the open gap, thereby reading full source voltage.

Key Concept

In a series circuit, an open component prevents current flow through all other components and drops the entire applied source voltage across the open point.

Question 117Question →

A technician needs to protect a sensitive electronic component from stray external DC magnetic fields by placing it inside a soft iron enclosure. How does this magnetic shield prevent external magnetic flux from reaching the component?

The enclosure acts as a magnetic insulator, stopping and absorbing magnetic flux before it penetrates the surface.

The enclosure provides a low-reluctance path with high permeability, channeling the flux lines through its walls and around the interior.

The enclosure develops opposing magnetic poles that repel external magnetic field lines away from both sides of the casing.

The enclosure converts incoming magnetic lines of force into an electrostatic charge that is dissipated as heat.

Show answer & explanation

Answer: The enclosure provides a low-reluctance path with high permeability, channeling the flux lines through its walls and around the interior.

Answer

The enclosure provides a low-reluctance path with high permeability, channeling the flux lines through its walls and around the interior.

Magnetic shielding relies on high permeability and low reluctance. Because there is no known insulator that blocks magnetic flux lines, the soft iron shield provides an easy path of low reluctance that channels the magnetic field lines through its walls, directing them safely around the internal cavity where the sensitive component is placed.

Step-by-Step Solution

1

Identify the magnetic characteristics of the shielding material.

Soft iron is a ferromagnetic substance characterized by high magnetic permeability and low magnetic reluctance.

Permeability describes how readily a material conducts magnetic flux compared to surrounding air, while reluctance represents opposition to flux lines.

2

Determine the path magnetic flux lines follow when encountering the enclosure.

Magnetic flux lines naturally follow the path of least resistance (lowest reluctance), concentrating within the thick iron walls.

Flux lines take the easiest path available rather than crossing the higher-reluctance air gap in the center.

3

Conclude how the interior component is protected.

Because flux lines are guided through the perimeter walls and exit on the opposite side, the internal cavity remains essentially free of magnetic field lines.

Since true magnetic insulators do not exist, diversion through high-permeability conduits is the fundamental method of magnetic shielding.

Key Concept

Magnetic Shielding, Permeability, and Reluctance

Estimated Time:1m 0s

Question 118Question →

A security alarm control panel operates on a $12\text{ V}$ direct-current power supply connected across two parallel load branches. The first branch has a resistance of $20\ \Omega$ , and the second branch has a resistance of $30\ \Omega$ . What is the total current drawn by the alarm panel from the power supply?

0.24\text{ A}

0.60\text{ A}

1.0\text{ A}

144\text{ A}

Show answer & explanation

Answer: $1.0\text{ A}$

Answer

The total current drawn by the alarm panel is

1.0\text{ A}

.

The total current of

1.0\text{ A}

is correct because each parallel branch experiences the full source voltage of

12\text{ V}

. Dividing the voltage by each branch resistance yields branch currents of

0.60\text{ A}

and

0.40\text{ A}

. Adding these together gives the total circuit current of

1.0\text{ A}

.

Step-by-Step Solution

1

Calculate the current flowing through the first parallel branch (

I_1

)

I_1 = \frac{V}{R_1} = \frac{12\text{ V}}{20\ \Omega} = 0.60\text{ A}

In a parallel circuit, the full supply voltage of

12\text{ V}

appears across each individual branch.

2

Calculate the current flowing through the second parallel branch (

I_2

)

I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{30\ \Omega} = 0.40\text{ A}

Apply Ohm's Law to the second branch using the constant supply voltage.

3

Sum the branch currents to determine total circuit current (

I_{\text{total}}

)

I_{\text{total}} = I_1 + I_2 = 0.60\text{ A} + 0.40\text{ A} = 1.0\text{ A}

According to Kirchhoff's Current Law, total current entering a parallel network equals the sum of all individual branch currents.

Key Concept

In a parallel DC circuit, voltage remains equal across all branches, and total source current equals the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

).

Alternative Method

Find total equivalent resistance using the product-over-sum formula:

R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12\ \Omega

. Then calculate total current directly:

I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}

.

Estimated Time:1m 0s

Question 119Question →

In direct-current (DC) parallel circuit analysis, modifying circuit components or introducing faults alters electrical parameters according to fundamental circuit laws. Match each parallel circuit condition or parameter on the left with its correct electrical behavior or outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding a branch matches decreasing total equivalent resistance; an open branch matches dropping current to zero only in that branch; branch voltage drop matches remaining identical across all branches; unequal current division matches directing larger current through the smaller resistor.

In parallel circuits: (1) Adding branches decreases equivalent resistance and raises total current; (2) An open in one branch affects only that path; (3) Voltage is equal across all branches; and (4) Branch current is inversely proportional to resistance.

Step-by-Step Solution

1

Analyze equivalent resistance behavior when adding parallel paths.

R_{\text{eq}}

decreases as branches are added because

\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}

.

Each added branch provides an additional path for electron flow, reducing overall opposition to current.

2

Evaluate the effect of an open circuit in an isolated parallel branch.

Current ceases in the open branch, while voltage and current across other branches remain steady.

Independent branch connections across common rails preserve full line voltage across unaffected branches.

3

Determine voltage distribution across parallel loads.

Branch voltages are identical (

V_{\text{source}} = V_1 = V_2

).

All parallel elements share the same electrical junctions.

4

Determine current division rules for unequal resistances.

Current flows inversely to resistance (

I_1 = \frac{V}{R_1} > I_2 = \frac{V}{R_2}

when

R_1 < R_2

).

Lower resistance paths draw greater current from the constant branch voltage.

Key Concept

Parallel Circuit Voltage, Current, Resistance, and Fault Characteristics

Estimated Time:1m 15s

Question 120Question →

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.

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