Circuit Faults and Troubleshooting Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

An electronics maintenance panel is powered by a $120\text{ V}$ DC bus. The circuit features a primary $40\ \Omega$ load resistor ( $R_1$ ) connected in series with a parallel junction containing two secondary resistors: $R_2$ ( $60\ \Omega$ ) and $R_3$ ( $30\ \Omega$ ). During a routine inspection, a digital multimeter indicates a $48\text{ V}$ drop across $R_1$ . Assuming the source voltage is nominal, which specific component failure is causing this reading?

The

30\ \Omega

secondary resistor (

R_3

) has failed open.

The

60\ \Omega

secondary resistor (

R_2

) has failed open.

The primary resistor (

R_1

) has degraded and reduced in value to

24\ \Omega

.

The parallel junction has short-circuited to ground.

Show answer & explanation

Answer: The $30\ \Omega$ secondary resistor ( $R_3$ ) has failed open.

Answer

The

30\ \Omega

secondary resistor (

R_3

) has failed open.

Based on the

48\text{ V}

drop across the

40\ \Omega

series resistor, the actual circuit current is

1.2\text{ A}

. To draw exactly

1.2\text{ A}

from a

120\text{ V}

source, the total circuit resistance must be

100\ \Omega

. Since the series resistor accounts for

40\ \Omega

, the parallel junction must be contributing exactly

60\ \Omega

. The only way the parallel junction (normally containing a

60\ \Omega

and a

30\ \Omega

resistor) equals

60\ \Omega

is if the

30\ \Omega

resistor has failed open, leaving only the

60\ \Omega

resistor in the path.

Step-by-Step Solution

1

Calculate the actual current flowing through the primary resistor (

R_1

) under fault conditions.

I_{total} = \frac{48\text{ V}}{40\ \Omega} = 1.2\text{ A}

By Ohm's Law, the measured voltage drop across a known resistance reveals the actual current flowing through that component.

2

Determine the new total circuit resistance (

R_{total}

) required to restrict the current to

1.2\text{ A}

.

R_{total} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

The total resistance determines how much current is drawn from the main voltage bus.

3

Calculate the equivalent resistance of the parallel network (

R_p

) under the current faulted condition.

R_p = R_{total} - R_1 = 100\ \Omega - 40\ \Omega = 60\ \Omega

Subtracting the series resistance from the total circuit resistance isolates the effective resistance of the parallel branch.

4

Identify the component failure that changes the parallel network's resistance from its normal

20\ \Omega

to exactly

60\ \Omega

.

The

30\ \Omega

resistor (

R_3

) must be an open circuit.

If

R_3

is open (infinite resistance), no current flows through it. The only path through the junction is via

R_2

(

60\ \Omega

), perfectly matching the required

60\ \Omega

equivalent resistance.

Key Concept

Troubleshooting Series-Parallel Circuits using Voltage Drops and Ohm's Law

Estimated Time:3m 0s

Question 2Question →

A $24\text{ V}$ DC power supply energizes a circuit consisting of a $100\,\Omega$ series resistor ( $R_1$ ) connected in series with a parallel network composed of two $200\,\Omega$ branch loads. Under normal operating conditions, the circuit draws $120\text{ mA}$ , dropping $12\text{ V}$ across $R_1$ and $12\text{ V}$ across the parallel network. During a troubleshooting check, a technician records a total circuit current of $80\text{ mA}$ , a voltage drop of $8\text{ V}$ across $R_1$ , and a voltage drop of $16\text{ V}$ across the parallel network. Which fault condition in the circuit explains these diagnostic readings?

A direct short circuit occurred across one of the two parallel branch loads.

A partial short circuit developed across the series resistor

R_1

.

An open circuit occurred in one of the two parallel branch loads.

The series resistor

R_1

has opened completely.

Show answer & explanation

Answer: An open circuit occurred in one of the two parallel branch loads.

Answer

An open circuit occurred in one of the two parallel branch loads.

When one parallel branch develops an open circuit, current can only flow through the remaining

200\,\Omega

branch. This increases the equivalent parallel resistance from

100\,\Omega

to

200\,\Omega

, bringing total circuit resistance to

300\,\Omega

. By Ohm's law, total current drops to

I = \frac{24\text{ V}}{300\,\Omega} = 80\text{ mA}

. Consequently, the voltage across

R_1

drops to

8\text{ V}

(

80\text{ mA} \times 100\,\Omega

) while the parallel network drops the remaining

16\text{ V}

(

80\text{ mA} \times 200\,\Omega

).

Step-by-Step Solution

1

Analyze normal baseline circuit parameters.

Normal equivalent parallel resistance is

R_{\text{parallel}} = \frac{200\,\Omega \times 200\,\Omega}{200\,\Omega + 200\,\Omega} = 100\,\Omega

. Total normal resistance is

R_{\text{total}} = 100\,\Omega + 100\,\Omega = 200\,\Omega

. Normal current is

I = \frac{24\text{ V}}{200\,\Omega} = 120\text{ mA}

, producing equal

12\text{ V}

drops across

R_1

and the parallel bank.

Establishing normal values is required to determine whether total circuit resistance increased or decreased under fault conditions.

2

Calculate the faulted circuit's total resistance and component values using the measured readings.

With

I_{\text{measured}} = 80\text{ mA}

and

V_{\text{source}} = 24\text{ V}

, total circuit resistance is

R_{\text{total, fault}} = \frac{24\text{ V}}{0.08\text{ A}} = 300\,\Omega

. Since

R_1

drops

8\text{ V}

at

80\text{ mA}

(

R_1 = \frac{8\text{ V}}{0.08\text{ A}} = 100\,\Omega

, unchanged), the parallel bank now exhibits

R_{\text{parallel, fault}} = \frac{16\text{ V}}{0.08\text{ A}} = 200\,\Omega

.

Ohm's law confirms where the additional resistance is located.

3

Determine the physical fault that produces a

200\,\Omega

parallel bank resistance.

Because each individual branch load is rated at

200\,\Omega

, an equivalent resistance of

200\,\Omega

across the bank means only one

200\,\Omega

branch is conducting, indicating the other branch has an open circuit.

Removing one branch of a two-branch identical parallel network doubles the network's equivalent resistance to the value of the remaining single branch.

Key Concept

In a series-parallel circuit, an open fault in a parallel branch increases total circuit resistance, reduces total line current, and causes a higher proportion of the source voltage to drop across the parallel section.

Estimated Time:1m 30s

Question 3Question →

A technician is troubleshooting a simple series circuit powered by a 12-volt DC battery and containing a single incandescent light bulb. When the switch is closed, the bulb does not light up. Measuring with a digital voltmeter directly across the bulb's socket terminals shows a reading of 12 volts. What fault condition does this measurement indicate?

A direct short circuit across the light bulb socket

An open circuit in the light bulb filament

A blown fuse at the battery positive terminal

An incorrect multimeter setting configured for resistance measurement

Show answer & explanation

Answer: An open circuit in the light bulb filament

Answer

An open circuit in the light bulb filament

When a filament burns out, it creates an open circuit. Because the circuit is open, current cannot flow (

I = 0\text{ A}

), meaning no voltage is dropped across the closed switch or connecting wires. Consequently, connecting a voltmeter across the open filament places the meter across the full potential of the 12-volt battery.

Step-by-Step Solution

1

Analyze the circuit state and the voltmeter reading.

The voltmeter measures the full supply voltage of

12\text{ V}

directly across the bulb terminals, yet no light is produced.

In a series circuit with an open fault, current flow ceases entirely (

I = 0\text{ A}

). With no current flowing, there are no voltage drops across any normal conductors or closed switches.

2

Apply voltage laws to identify the location of the open fault.

The full source potential difference (

12\text{ V}

) appears across the open gap in the filament.

According to Kirchhoff's Voltage Law, the sum of voltage drops must equal the source voltage. Because the open filament has infinite resistance compared to the wiring, the entire

12\text{ V}

drops across the break.

Key Concept

An open component in a series circuit stops all current flow and drops the full source voltage across its terminals.

Estimated Time:45s

Question 4Question →

A technician is troubleshooting a series circuit powered by a $12\text{ V}$ DC source that normally operates with three identical $10\,\Omega$ resistive loads ( $R_1$ , $R_2$ , and $R_3$ ). Match each specific multimeter reading or diagnostic symptom on the left to its corresponding circuit fault condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The multimeter readings correctly pair with the specific circuit faults as follows: measuring full supply voltage across a single component with zero current corresponds to an open circuit in that component; measuring zero volts across a component while current and remaining voltage drops increase corresponds to a shorted component; reduced voltage across all loads with an unexpected voltage drop at a junction corresponds to a high-resistance connection; and measuring zero ohms to chassis before loads corresponds to a direct short-to-ground fault.

Each fault produces a distinct electrical signature in a series circuit: an open component exhibits infinite resistance and drops the full source voltage with zero current; a shorted component exhibits zero resistance and zero voltage drop while increasing circuit current; a high-resistance connection adds unwanted series resistance that drops voltage and reduces current to all loads; and a short-to-ground creates a direct zero-ohm path to ground before the load network.

Step-by-Step Solution

1

Analyze the voltage and current behavior for the open circuit symptom

Full source voltage (

12\text{ V}

) across

R_2

with

0\text{ A}

current indicates an open circuit at

R_2

.

An open circuit creates infinite resistance, stopping all current (

I = 0

) so no voltage drops occur across intact series components, placing total source EMF across the open terminals.

2

Analyze the voltage and current behavior for the shorted component symptom

Zero volts across

R_2

with increased current from

0.4\text{ A}

to

0.6\text{ A}

indicates a short circuit across

R_2

.

A shorted component has

0\,\Omega

, dropping

0\text{ V}

while decreasing total circuit resistance (

R_{total} = 20\,\Omega

) and increasing current (

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

).

3

Analyze the effect of an unwanted series resistance

Lower voltage drops across all loads combined with an unexpected drop across a terminal indicates a high-resistance connection.

Corrosion or a loose terminal adds resistance to the series string, lowering total current and taking a portion of the source voltage away from the loads.

4

Evaluate the de-energized ohmmeter reading to chassis

A reading of

0\,\Omega

from the power lead to chassis ground prior to the loads confirms a short-to-ground fault.

An ohmmeter reading continuity (

0\,\Omega

) to chassis ground before the loads shows an unintended low-resistance return path that bypasses the load network entirely.

Key Concept

Circuit Fault Diagnosis and Multimeter Reading Interpretation in DC Series Circuits

Estimated Time:1m 30s

Question 5Question →

A technician is troubleshooting a direct-current circuit that suddenly stopped operating. In what order should the technician perform the standard diagnostic steps to locate and resolve the fault safely?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct troubleshooting sequence begins with a visual inspection, followed by verifying supply voltage, then de-energizing the circuit, and finally measuring isolated component resistance.

Systematic troubleshooting follows a logical hierarchy: first perform a visual check for obvious physical damage; second, verify that input voltage is present; third, de-energize the circuit for safety; and fourth, perform resistance checks on isolated components to identify the specific faulty device.

Step-by-Step Solution

1

Inspect the physical condition of the circuit.

Identifies obvious signs of overheating, loose terminals, or mechanical damage without disturbing connections.

Visual checks are fast, safe, and can resolve obvious problems immediately.

2

Measure input voltage while the circuit is connected to power.

Confirms whether proper operational voltage reaches the circuit input terminals.

If supply voltage is missing, troubleshooting must shift to the upstream power source rather than the circuit itself.

3

Shut off and isolate the power supply.

Removes all live electrical energy from the circuit.

Safety protocol requires de-energizing the circuit before testing resistance or removing parts.

4

Use an ohmmeter to measure continuity and resistance across individual components.

Pinpoints open coils, shorted semiconductor junctions, or out-of-tolerance resistors.

Ohmmeters supply their own internal battery voltage and will be damaged if used on an active, energized circuit.

Key Concept

Standard systematic troubleshooting progresses from non-invasive visual inspection to live voltage verification, followed by complete power isolation before conducting ohmmeter resistance measurements.

Estimated Time:45s

Question 6Question →

An electrician tests a de-energized heating circuit using a digital multimeter set to measure resistance. When the test probes are placed across the disconnected heating element, the meter displays an infinite resistance reading (OL). What fault condition does this measurement indicate?

A short circuit across the heating element

An open circuit within the heating element

Excessive electrical current flowing through the circuit

A properly functioning element operating at its rated resistance

Show answer & explanation

Answer: An open circuit within the heating element

Answer

An open circuit within the heating element

An ohmmeter displaying infinite resistance (indicated as 'OL' for open loop or out of limits) across a component means there is a physical break in the internal conductor, which defines an open circuit.

Step-by-Step Solution

1

Analyze the multimeter mode and circuit condition

The meter is configured as an ohmmeter measuring a de-energized, isolated load.

Resistance must be tested with power removed to prevent meter damage and obtain true component resistance.

2

Interpret the display reading of infinite resistance (OL)

Infinite resistance signifies that there is no continuous electrical path between the meter leads.

An intact heating element has a continuous resistive wire that provides a measurable finite resistance.

3

Identify the corresponding circuit fault

A break or gap in the internal filament represents an open circuit.

Open circuits present infinite resistance and completely prevent current flow under normal operating voltage.

Key Concept

Identifying open circuits using ohmmeter resistance measurements

Estimated Time:45s

Question 7Question →

Match each standard electrical circuit fault with its correct definition and behavior.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Open circuit pairs with a complete break in the path that stops current flow; Short circuit pairs with an unintended zero-resistance bypass around a load that causes excessive current; Short to ground pairs with an unintended low-resistance connection to the chassis or frame; High-resistance connection pairs with an unwanted increase in opposition creating an unintended voltage drop.

Each electrical fault produces distinct physical and electrical symptoms: open circuits prevent current flow via continuity loss; short circuits draw maximum current by bypassing load resistance; ground faults divert current directly into the chassis/ground; and high-resistance joints cause partial voltage loss without completely breaking continuity.

Step-by-Step Solution

1

Analyze the definition of an open circuit.

Identify that an open creates infinite resistance and interrupts all current, matching the break in the conductive path.

An open breaks continuity.

2

Analyze the definition of a short circuit.

Identify that a short bypasses the load with near-zero resistance, matching excessive current flow and tripped protective devices.

Ohm's law dictates that near-zero resistance at supply voltage results in high current.

3

Analyze the definition of a short to ground.

Identify that contacting the equipment frame or chassis defines a ground fault.

Chassis/ground serves as the common return or protective reference.

4

Analyze the definition of a high-resistance connection.

Identify that dirty, loose, or corroded terminals act like unwanted series resistors that drop voltage.

Contact resistance reduces available voltage across the operating load.

Key Concept

Basic Circuit Fault Characteristics

Alternative Method

Use the fundamental relationship

I = \frac{V}{R}

: when

R = \infty

(open),

I = 0

; when

R \approx 0

(short),

I

is extremely large.

Estimated Time:45s

Question 8Question →

A $24\text{ V}$ DC bridge circuit consists of two parallel branches connected between a constant $+24\text{ V}$ DC supply rail and ground:
- Left branch: Resistor $R_1 = 1.0\,\text{k}\Omega$ connected in series with resistor $R_2 = 1.0\,\text{k}\Omega$ to ground, with Test Point A ( $V_A$ ) located between them.
- Right branch: Resistor $R_3 = 1.0\,\text{k}\Omega$ connected in series with resistor $R_4 = 1.0\,\text{k}\Omega$ to ground, with Test Point B ( $V_B$ ) located between them.

Under normal operation, the total current drawn from the supply is $24\text{ mA}$ , and $V_A = V_B = 12.0\text{ V}$ . During diagnostic testing of a faulty unit, a technician measures the following values:
- $V_A = 12.0\text{ V}$ (measured with respect to ground)
- $V_B = 24.0\text{ V}$ (measured with respect to ground)
- Total current drawn from the $24\text{ V}$ source = $36\text{ mA}$

Which of the following circuit faults accounts for these diagnostic readings?

An open circuit fault in resistor

R_4

A short circuit fault across resistor

R_4

A short circuit fault across resistor

R_3

An open circuit fault in resistor

R_3

Show answer & explanation

Answer: A short circuit fault across resistor $R_3$

Answer

A short circuit fault across resistor

R_3

A short circuit across resistor

R_3

bypasses the upper resistor, placing Test Point B directly at the

+24\text{ V}

supply rail. This leaves only resistor

R_4

(

1.0\,\text{k}\Omega

) in the right branch, increasing the right branch current to

24\text{ V} / 1.0\,\text{k}\Omega = 24\text{ mA}

. Combining this with the normal

12\text{ mA}

from the left branch produces the exact measured total current of

36\text{ mA}

.

Step-by-Step Solution

1

Analyze the operational status and current of the left branch.

Left branch current

I_{\text{left}} = 12\text{ mA}

.

Since

V_A = 12.0\text{ V}

across

R_2 = 1.0\,\text{k}\Omega

, the left branch is normal and draws

I_{\text{left}} = 24\text{ V} / (1.0\,\text{k}\Omega + 1.0\,\text{k}\Omega) = 12\text{ mA}

.

2

Determine the current flowing through the right branch under fault conditions.

Right branch current

I_{\text{right}} = 24\text{ mA}

.

Subtract the normal left branch current from the measured total current:

I_{\text{right}} = I_{\text{total}} - I_{\text{left}} = 36\text{ mA} - 12\text{ mA} = 24\text{ mA}

.

3

Calculate the effective resistance of the right branch and evaluate voltage drop across components.

R_{\text{right}} = 1.0\,\text{k}\Omega

, with

0\text{ V}

dropped across

R_3

.

R_{\text{right}} = V_{\text{source}} / I_{\text{right}} = 24\text{ V} / 24\text{ mA} = 1.0\,\text{k}\Omega

. Since

R_4

is

1.0\,\text{k}\Omega

and

V_B = 24.0\text{ V}

, the full

24\text{ V}

drops across

R_4

, meaning

R_3

has

0\text{ V}

drop and zero resistance (a direct short circuit).

Key Concept

Distinguishing between open and short circuit faults using branch voltage and total current analysis

Estimated Time:1m 30s

Question 9Question →

A technician performs electrical diagnostic checks on various direct-current (DC) circuit components. Match each multimeter measurement or symptom to the correct circuit condition or fault.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Match the open coil to infinite resistance, intact fuse to 0 V drop, shorted load to 0 V with high current surge / blown fuse, and high-resistance contact fault to elevated closed-switch resistance.

Each diagnostic measurement corresponds to a fundamental circuit state: infinite coil resistance reflects an open circuit; zero volts across an intact fuse demonstrates normal low-resistance conduction; zero volts with blown protection indicates a short circuit; and elevated resistance across closed switch contacts indicates a high-resistance contact fault.

Step-by-Step Solution

1

Analyze the isolated coil resistance test reading

An infinite resistance (O.L.) measurement signifies an internal discontinuity or open circuit.

Current cannot pass through a severed conductor, resulting in infinite measured resistance.

2

Evaluate the in-circuit voltage drop across the closed fuse

A 0 V drop across an intact conducting fuse indicates normal low-resistance operation.

According to Ohm's law (

V = I \times R

), a conductor with negligible resistance produces virtually no voltage drop.

3

Identify the fault causing 0 V across a load with blown overcurrent protection

A short circuit parallel to or across the load bypasses normal resistance and draws excessive current.

Zero branch resistance eliminates voltage drop across the parallel terminals while causing current to surge past the fuse rating.

4

Interpret the elevated resistance reading across closed switch contacts

A reading of

180\,\Omega

represents an abnormal high-resistance fault.

Clean, closed mechanical contacts should read less than

1\,\Omega

; oxidation and pitting insert unwanted resistance into the path.

Key Concept

Circuit Fault Diagnosis and Multimeter Interpretation

Estimated Time:1m 15s

Question 10Question →

A $12\text{ V}$ direct-current (DC) circuit consists of a power source, a single-pole toggle switch, and an indicator lamp wired in series. When the switch is flipped to the ON (closed) position, the lamp fails to illuminate. A technician connects a digital multimeter across the two terminals of the closed switch and reads $12\text{ V}$ . Which of the following faults is indicated by this measurement?

The indicator lamp has a short circuit across its filament.

The circuit is operating normally and drawing maximum rated current.

The switch contacts have failed open internally.

The indicator lamp filament is burned open.

Show answer & explanation

Answer: The switch contacts have failed open internally.

Answer

The switch contacts have failed open internally.

In a series circuit, a closed switch should exhibit continuity (

0\ \Omega

) and produce a

0\text{ V}

drop. When a digital multimeter measures the entire source voltage (

12\text{ V}

) across a switch in the ON position, it proves that the switch contacts have failed to close electrically, acting as an open circuit that drops the entire available potential.

Step-by-Step Solution

1

Analyze the expected behavior of a properly closed switch in a series circuit.

An ideal closed switch provides continuity with approximately

0\ \Omega

resistance and a voltage drop of nearly

0\text{ V}

.

According to Ohm's Law (

V = I \times R

), current flowing through near-zero resistance produces negligible voltage drop.

2

Evaluate the meter reading across the switch terminals.

The meter measures

12\text{ V}

, which equals the full source voltage.

In a series circuit, an open point has infinite resistance and drops the entire supply voltage because no current can flow through the rest of the circuit.

3

Determine the exact fault location based on the voltage drop.

Because the full

12\text{ V}

appears across the switch itself, the break in continuity is inside the switch.

If the lamp were open instead, the voltmeter across the intact closed switch would read

0\text{ V}

.

Key Concept

Voltage Drop Across Open vs. Closed Series Components

Estimated Time:1m 0s

Question 11Question →

While troubleshooting a direct-current ( $12\text{ V}$ ) automotive auxiliary circuit, a technician observes that the branch fuse blows immediately every time the power switch is closed. Which of the following electrical faults is the most probable cause of this issue?

An open circuit in the wiring leading to the load

A short circuit directly to ground before the load

High contact resistance due to a corroded connector terminal

Reversed polarity connection at the power source

Show answer & explanation

Answer: A short circuit directly to ground before the load

Answer

A short circuit directly to ground before the load is the most probable cause because it creates a near-zero resistance path that draws excessive current, exceeding the fuse rating.

A short circuit directly to ground creates a zero- or near-zero-resistance bypass around the intended load. According to Ohm's law (

I = \frac{V}{R}

), as resistance approaches zero, the circuit current rapidly exceeds the protective fuse's current rating, instantly melting the fuse link when the switch is closed.

Step-by-Step Solution

1

Analyze the observed symptom and the operational role of a fuse.

A fuse is an overcurrent protection device designed to melt and open when current exceeds its specified ampere rating.

Understanding why fuses blow establishes that the root cause must be excessive current draw.

2

Apply Ohm's law (

I = \frac{V}{R}

) to evaluate how each potential fault affects circuit current.

To produce massive current (

I

) from a constant

12\text{ V}

source, the total resistance (

R

) must drop close to

0\,\Omega

.

Ohm's law dictates an inverse relationship between resistance and current.

3

Identify the circuit condition that causes near-zero resistance.

A short circuit bypasses the normal load resistance directly to ground, causing an instantaneous current spike that blows the fuse.

Open circuits and high-resistance connections reduce or eliminate current flow rather than increasing it.

Key Concept

A short circuit provides an unintended low-resistance bypass path that causes current to surge above safe levels, blowing protective fuses.

Estimated Time:45s

Question 12Question →

A multi-component series circuit has failed and is drawing zero current due to an open fault. Arrange the following troubleshooting steps in the correct chronological sequence to safely and logically locate the open component using a digital multimeter (DMM) while the circuit is energized.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is to first verify the source voltage, anchor the negative reference lead, systematically probe downstream with the positive lead, observe the full source voltage across intact paths, and pinpoint the fault where the voltage abruptly drops to zero.

The voltage drop method is the most effective and professional way to locate an open fault in an energized circuit. In an open circuit, current cannot flow. According to Ohm's Law (

V = I \times R

), if current (

I

) is zero, the voltage drop across any intact resistor is zero. Therefore, measuring anywhere between the positive terminal and the open fault will yield the full source voltage. Measuring anywhere after the open fault will yield zero volts because that section is physically isolated from the positive terminal but electrically connected to the negative reference. The steps outlined provide the logical, chronological, and safe procedure to trace this voltage path and isolate the break.

Step-by-Step Solution

1

Understand the electrical behavior of an open circuit.

Current is 0 Amps, meaning intact resistors will have a voltage drop of 0 Volts (

V = 0 \times R = 0

).

This principle dictates that the full source voltage will be measurable all the way up to the break.

2

Verify the baseline power supply.

The source voltage is confirmed to be present at the main terminals.

Troubleshooting downstream is useless if the main power source is dead or the battery is disconnected.

3

Establish a consistent measurement reference.

The black lead is fixed to ground, ensuring all readings are relative to 0 Volts.

Moving both leads simultaneously can cause confusion; anchoring one lead provides a steady reference point.

4

Trace the voltage potential through the circuit.

The meter reads full source voltage at each test point prior to the open fault.

The circuit wiring is intact up to the fault, extending the positive potential forward.

5

Identify the point of failure.

The fault is located exactly between the last point reading full voltage and the first point reading zero volts.

The open break physically separates the positive potential from the remainder of the circuit, causing the downstream voltage reading to drop to zero relative to ground.

Key Concept

Voltage Drop Troubleshooting Method

Estimated Time:1m 30s

Question 13Question →

An electronics technician is troubleshooting a $24\text{ V}$ direct-current (DC) control circuit that repeatedly trips its circuit breaker due to a suspected short circuit to ground. Arrange the following diagnostic procedures in the correct chronological order to safely locate and resolve the fault.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct troubleshooting procedure starts with de-energizing the circuit and verifying zero voltage, followed by visual inspection, disconnecting branch loads for isolation, measuring resistance to ground using an ohmmeter, and finally repairing the shorted component and verifying resistance before restoring power.

Effective and safe troubleshooting follows a structured logical progression: electrical safety isolation must occur first, followed by visual inspection, systematic branch isolation, ohmmeter resistance testing to locate the shorted path, and pre-power verification following repair.

Step-by-Step Solution

1

De-energize and verify zero voltage on the circuit.

Protects the technician from electrical shock and prevents damaging the ohmmeter.

Ohmmeters supply internal sensing voltage and will be damaged if connected to an energized circuit.

2

Visually inspect conductors, terminals, and components.

Identifies obvious burned insulation, pinched wires, or loose metallic debris.

Visual inspection is the fastest and least invasive diagnostic method before mechanical disassembly.

3

Disconnect branch loads to isolate parallel circuit sections.

Separates parallel current paths from one another.

In parallel networks, a single short circuit to ground will reflect across all interconnected branches unless they are isolated.

4

Measure resistance from each isolated conductor to ground using an ohmmeter.

Locates the exact branch exhibiting abnormal near-zero resistance (

0\ \Omega

).

A dead short creates an unintended low-resistance connection directly to chassis ground.

5

Repair the defective component and verify resistance prior to restoring power.

Ensures normal circuit resistance is restored and prevents immediate breaker tripping upon restart.

Verifying high resistance to ground on the repaired branch confirms the short has been eliminated before power is re-applied.

Key Concept

Systematic Isolation and Troubleshooting of a Short Circuit to Ground

Estimated Time:1m 15s

Question 14Question →

A series circuit consists of a $48\text{ V}$ direct-current (DC) power source and three identical $16\,\Omega$ resistors labeled $R_1$ , $R_2$ , and $R_3$ . Under normal operation, each resistor drops $16\text{ V}$ . During troubleshooting, a technician measures $24\text{ V}$ across $R_1$ , $0\text{ V}$ across $R_2$ , and $24\text{ V}$ across $R_3$ , while current continues to flow through the circuit. Which of the following conditions explains this circuit fault?

Resistor

R_2

has developed an open circuit.

Resistor

R_2

has developed a short circuit.

Resistor

R_2

has doubled its internal resistance value.

The power supply voltage has decreased by half.

Show answer & explanation

Answer: Resistor $R_2$ has developed a short circuit.

Answer

Resistor

R_2

has developed a short circuit.

When a component becomes shorted in a series circuit, its effective resistance drops to zero, resulting in a measured voltage drop of

0\text{ V}

across its terminals. The total circuit resistance decreases from

48\,\Omega

to

32\,\Omega

, increasing the circuit current to

1.5\text{ A}

. This higher current causes the remaining two

16\,\Omega

resistors to each drop

1.5\text{ A} \times 16\,\Omega = 24\text{ V}

, exactly matching the technician's meter readings.

Step-by-Step Solution

1

Analyze the voltage drop across the faulted component (

R_2

).

The voltage across

R_2

is measured at

0\text{ V}

.

A component with zero resistance (a short circuit) drops zero volts according to Ohm's Law (

V = I \times R = I \times 0 = 0\text{ V}

).

2

Evaluate the remaining circuit voltages across

R_1

and

R_3

.

The sum of voltages is

24\text{ V} + 24\text{ V} = 48\text{ V}

, matching the source voltage.

Kirchhoff's Voltage Law confirms the full source voltage is distributed across the remaining active components in the loop.

3

Calculate the total circuit resistance and current under shorted conditions.

R_{\text{total}} = 16\,\Omega + 0\,\Omega + 16\,\Omega = 32\,\Omega

, producing a current of

I = \frac{48\text{ V}}{32\,\Omega} = 1.5\text{ A}

.

A shorted component decreases total circuit resistance, which increases total current flow through the remaining healthy components.

Key Concept

Effect of a short circuit in a series circuit

Alternative Method

Use Kirchhoff's Voltage Law (

V_{\text{source}} = V_1 + V_2 + V_3

). Substituting the measured values gives

48\text{ V} = 24\text{ V} + 0\text{ V} + 24\text{ V}

. Because

V_2 = 0\text{ V}

while current is actively flowing,

R_2

must have zero resistance (

R_2 = \frac{0\text{ V}}{I} = 0\,\Omega

), which is the definition of a short circuit.

Estimated Time:1m 0s

Question 15Question →

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 16Question →

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 17Question →

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 18Question →

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 19Question →

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 20Question →

A $24\text{ V}$ direct-current (DC) power source is connected to a circuit where resistor $R_1$ ( $4\ \Omega$ ) is in series with a parallel pair consisting of $R_2$ ( $12\ \Omega$ ) and $R_3$ ( $6\ \Omega$ ). Under normal operation, the total current is $3.0\text{ A}$ and the voltage drop across the parallel node is $12\text{ V}$ . An electronics technician records abnormal multimeter readings indicating different single-component faults. Match each set of diagnostic symptoms to the corresponding circuit fault.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The symptoms map to the faults as follows: 2.4 A and 14.4 V indicates R2 is open; 1.5 A and 18.0 V indicates R3 is open; 6.0 A and 0 V indicates a short in the parallel section (R2 or R3); 0 A and 24 V across R1 indicates R1 is open.

Each diagnostic reading mathematically corresponds to a specific component failure. When a component fails open, its resistance becomes infinite, altering the equivalent circuit resistance and redistributing voltage drops. When a component short-circuits, it bypasses that section of the circuit with near-zero resistance, causing a current spike and eliminating the voltage drop across that section. Placing a voltmeter across an open component in a series path will read the full source voltage.

Step-by-Step Solution

1

Analyze the normal circuit parameters to establish a baseline for comparison.

Normal equivalent parallel resistance (

R_p

) is

4\ \Omega

. Total resistance (

R_t

) is

8\ \Omega

. Normal current is

3.0\text{ A}

.

Baseline parameters are necessary to evaluate how mathematical deviations correlate to specific component failures.

2

Evaluate the symptom with

I = 2.4\text{ A}

and

V_p = 14.4\text{ V}

.

Total resistance increased to

10\ \Omega

(

24\text{ V} / 2.4\text{ A}

). Since

R_1

is

4\ \Omega

, the parallel section must now be

6\ \Omega

. This matches the value of

R_3

alone.

If

R_2

fails open, its branch has infinite resistance, leaving only

R_3

(

6\ \Omega

) to carry current in the parallel node.

3

Evaluate the symptom with

I = 1.5\text{ A}

and

V_p = 18.0\text{ V}

.

Total resistance increased to

16\ \Omega

(

24\text{ V} / 1.5\text{ A}

). The parallel section must now be

12\ \Omega

, which matches the value of

R_2

alone.

If

R_3

fails open, its branch has infinite resistance, leaving only

R_2

to carry current in the parallel section.

4

Evaluate the symptom with

I = 6.0\text{ A}

and

V_p = 0\text{ V}

.

Total resistance dropped to

4\ \Omega

, exactly equal to

R_1

. The parallel section resistance is

0\ \Omega

.

A short circuit in either

R_2

or

R_3

bypasses the resistance of the parallel node entirely, dropping its voltage to zero and causing a current spike.

5

Evaluate the symptom with

I = 0\text{ A}

and

24\text{ V}

measured directly across

R_1

.

An open circuit exists at

R_1

.

When a component fails open, the entire source voltage drops across the break because there is no voltage drop across the rest of the intact circuit components (since current is zero).

Key Concept

Circuit Fault Diagnostics using Ohm's Law and Series-Parallel Analysis

Estimated Time:3m 0s

Circuit Faults and Troubleshooting

Answer

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