All practice questions

Question 921Question →

A diesel generator preheat circuit is powered by a $24\text{ V}$ DC source connected in series with a ballast resistor $R_1 = 1.8\ \Omega$ and two identical glow plugs ( $R_2$ and $R_3$ ). If an ammeter in the circuit measures a steady current of $4.0\text{ A}$ , what is the resistance of a single glow plug in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 2.1

Answer

The resistance of a single glow plug is

2.1\ \Omega

.

Applying Ohm's Law gives a total series loop resistance of

R_{\text{total}} = 24\text{ V} / 4.0\text{ A} = 6.0\ \Omega

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

R_{\text{total}} = R_1 + R_2 + R_3

). Subtracting the ballast resistor (

1.8\ \Omega

) leaves

4.2\ \Omega

across the two identical glow plugs. Dividing

4.2\ \Omega

equally between them gives

2.1\ \Omega

per glow plug.

Step-by-Step Solution

1

Calculate the total circuit resistance using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{24\text{ V}}{4.0\text{ A}} = 6.0\ \Omega

In a closed DC loop, total resistance equals total source voltage divided by total current.

2

Subtract the known ballast resistance from the total resistance to determine the combined resistance of both glow plugs.

R_2 + R_3 = R_{\text{total}} - R_1 = 6.0\ \Omega - 1.8\ \Omega = 4.2\ \Omega

Resistances in a series circuit are directly additive (

R_{\text{total}} = R_1 + R_2 + R_3

).

3

Divide the combined glow plug resistance equally between the two identical units.

R_{\text{plug}} = \frac{4.2\ \Omega}{2} = 2.1\ \Omega

Because the two glow plugs are identical, each provides exactly half of the remaining series resistance.

Key Concept

Series Total Resistance and Ohm's Law

Estimated Time:1m 0s

Question 922Question →

An HVAC technician is testing a motorized zone damper actuator powered by a $24\text{ V}$ direct current (DC) power supply. An ammeter connected in series measures a current draw of $400\text{ mA}$ through the actuator coil. What is the electrical resistance of the actuator coil?

0.06\text{ }\Omega

9.6\text{ }\Omega

60\text{ }\Omega

9.6\text{ k}\Omega

Show answer & explanation

Answer: $60\text{ }\Omega$

Answer

The electrical resistance of the damper actuator coil is

60\text{ }\Omega

.

According to Ohm's Law, resistance is calculated by dividing the voltage across the component by the current flowing through it (

R = \frac{V}{I}

). Converting

400\text{ mA}

to amperes yields

0.4\text{ A}

. Dividing

24\text{ V}

by

0.4\text{ A}

gives

60\text{ }\Omega

.

Step-by-Step Solution

1

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

I = 400\text{ mA} = \frac{400}{1000}\text{ A} = 0.4\text{ A}

Standard electrical calculations using Ohm's Law require current to be expressed in amperes.

2

State and apply the Ohm's Law formula solved for resistance (

R

)

R = \frac{V}{I}

Resistance equals electromotive force (voltage) divided by current.

3

Substitute the known values and calculate the resistance

R = \frac{24\text{ V}}{0.4\text{ A}} = 60\text{ }\Omega

Dividing

24

by

0.4

gives the coil resistance in ohms.

Key Concept

Ohm's Law (

R = \frac{V}{I}

) and metric prefix conversion

Estimated Time:1m 0s

Question 923Question →

A step-down transformer has a $10:1$ turns ratio between its primary and secondary windings. If the primary winding is connected across a steady $12\text{ V}$ direct current (DC) power supply, what voltage is measured across the secondary winding under steady-state conditions?

1.2\text{ V}

12\text{ V}

0\text{ V}

120\text{ V}

Show answer & explanation

Answer: $0\text{ V}$

Answer

0\text{ V}

Transformers operate through mutual electromagnetic induction, requiring an alternating current (AC) to generate an expanding and collapsing magnetic field. A steady direct current (DC) produces a static, unvarying magnetic field once the circuit reaches steady state. Because Faraday's Law requires a changing magnetic field (

\frac{\Delta\Phi}{\Delta t} \neq 0

) to induce an electromotive force, the secondary winding produces exactly

0\text{ V}

.

Step-by-Step Solution

1

Identify the type of input voltage applied to the transformer primary winding.

The input source is a steady

12\text{ V}

direct current (DC) supply.

Transformers rely on mutual electromagnetic induction, which fundamentally requires a time-varying magnetic field.

2

Analyze the magnetic flux created in the core during steady-state DC operation.

A steady DC current produces a stationary, unchanging magnetic flux (

\frac{\Delta\Phi}{\Delta t} = 0

).

According to Faraday's Law of Induction, voltage is only induced in a conductor when magnetic flux lines expand or collapse across the winding turns.

3

Determine the resulting secondary output voltage.

The induced secondary voltage is

0\text{ V}

.

With zero rate of change of magnetic flux in steady state, no electromotive force (EMF) is induced in the secondary winding.

Key Concept

Transformers require alternating current (AC) to induce a secondary voltage via changing magnetic flux; steady direct current (DC) produces no secondary voltage.

Estimated Time:1m 0s

Question 924Question →

When diagnosing electrical circuits with a digital multimeter (DMM) set to measure resistance ( $\Omega$ ) or continuity, different display readings indicate specific circuit and component conditions. Match each multimeter reading to its corresponding electrical condition:

Click a left item, then click its matching right item

Show answer & explanation

Answer

The closed switch reading of

0.0\ \Omega

with a beep matches a continuous unbroken path; the

\text{OL}

reading on a

24\ \Omega

element matches an internal open circuit; the

4.7\ \text{k}\Omega

reading matches normal component resistance; and fluctuating readings with a voltage warning match measuring resistance on an energized circuit.

Each pair directly aligns the meter display with fundamental circuit states:

0.0\ \Omega

with a beep indicates unbroken continuity;

\text{OL}

on a load component signifies an open-circuit failure; a reading matching component ratings indicates proper health; and erratic values with a voltage warning indicate power was improperly left connected.

Step-by-Step Solution

1

Evaluate the zero-ohm reading with an audible tone

Identifies full continuity (closed path)

Multimeters use continuity mode to verify a continuous electrical path; zero ohms represents negligible resistance across closed switch contacts.

2

Evaluate the infinite resistance (OL) reading on a load component

Identifies an open circuit condition

A healthy

24\ \Omega

heating element should read approximately

24\ \Omega

; an OL reading indicates the internal filament or element wire is completely severed.

3

Evaluate the

4.7\ \text{k}\Omega

reading against the resistor color bands

Identifies normal nominal operation

Yellow (

4

), violet (

7

), and red multiplier (

10^2 = 100

) equals

4700\ \Omega

or

4.7\ \text{k}\Omega

, matching the measured reading.

4

Evaluate erratic values accompanied by a meter voltage warning

Identifies a live-circuit testing safety violation

Resistance must always be measured on fully de-energized circuits. Live current backfeeding into the meter distorts the internal test current and triggers warnings.

Key Concept

Interpreting digital multimeter resistance and continuity readings under open, closed, nominal, and live-circuit conditions

Estimated Time:1m 15s

Question 925Question →

A technician is testing a direct current (DC) cooling fan motor installed in an equipment rack. When powered by a $48\text{ V}$ DC supply, the motor draws a steady current of $250\text{ mA}$ . What is the electrical resistance of the cooling fan motor in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 192

Answer

The electrical resistance of the cooling fan motor is

192\text{ }\Omega

.

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

R = \frac{V}{I}

). Converting

250\text{ mA}

to

0.25\text{ A}

and dividing

48\text{ V}

by

0.25\text{ A}

gives exactly

192\text{ }\Omega

.

Step-by-Step Solution

1

Convert the current from milliamperes to amperes

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

Ohm's law calculations require current to be expressed in the base SI unit of amperes.

2

Calculate resistance using Ohm's law formula (

R = \frac{V}{I}

)

R = \frac{48\text{ V}}{0.25\text{ A}} = 192\text{ }\Omega

Dividing the supply voltage by the total circuit current yields the resistance in ohms.

Key Concept

Calculating resistance using Ohm's law (

R = \frac{V}{I}

) combined with metric prefix conversion from milliamperes to amperes.

Estimated Time:1m 0s

Question 926Question →

According to the principles of electromagnetic induction, what occurs in a closed wire loop when the surrounding magnetic field changes?

The electrical resistance of the wire drops completely to zero.

A voltage is generated only while the magnetic field remains completely stationary.

An electric current is induced to flow through the wire.

The wire is permanently converted into an electrical insulator.

Show answer & explanation

Answer: An electric current is induced to flow through the wire.

Answer

An electric current is induced to flow through the wire.

Electromagnetic induction occurs when a changing magnetic field cuts across a conductor, creating an electromotive force (voltage). In a closed loop, this induced voltage causes an electric current to flow.

Step-by-Step Solution

1

Identify the relationship between changing magnetic fields and conductors

A changing magnetic field cutting across a conductor generates an electromotive force (EMF).

Faraday's Law of Induction establishes that voltage is generated whenever magnetic flux lines cross a conductive path.

2

Apply the generated electromotive force to a closed circuit

The induced EMF causes electric current to circulate through the complete loop.

In any closed conductive loop, an induced potential difference creates electron flow.

Key Concept

Electromagnetic induction (generation of voltage and current via changing magnetic flux)

Estimated Time:40s

Question 927Question →

An apprentice technician connects digital multimeter test leads across an unpowered, unbroken length of copper wire to verify circuit continuity. Which multimeter response confirms that the wire provides a continuous, unbroken electrical path?

An overload or open-loop reading (

\text{OL}

) with no tone

A continuous audible tone accompanied by a resistance reading near

0\ \Omega

A very high resistance reading in the megaohm (

\text{M}\Omega

) range

A fluctuating DC voltage reading measured in millivolts (

\text{mV}

)

Show answer & explanation

Answer: A continuous audible tone accompanied by a resistance reading near $0\ \Omega$

Answer

A continuous audible tone accompanied by a resistance reading near

0\ \Omega

When a digital multimeter is set to continuity mode, it sends a tiny test current through the leads. When placed across an unbroken conductor, the resistance is nearly

0\ \Omega

, which closes the test loop and activates the meter's audible beeper.

Step-by-Step Solution

1

Identify the purpose of a continuity check

A continuity test verifies whether an unbroken path exists for electrical current to flow.

Continuity requires a low-resistance physical connection between two test points.

2

Determine the expected resistance and meter behavior for a good conductor

An intact copper conductor exhibits minimal resistance (close to

0\ \Omega

), which causes the multimeter's internal continuity buzzer to sound continuously.

Digital multimeters sound a continuous audible beep when the measured resistance across the leads is below a designated threshold (typically under

30\ \Omega

to

50\ \Omega

).

Key Concept

Multimeter continuity mode detects complete electrical paths by measuring near-zero resistance and emitting an audible tone.

Estimated Time:45s

Question 928Question →

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 929Question →

A continuous direct-current (DC) power feed delivers a steady current of $5\text{ amperes}$ through a heavy-duty copper busbar. Which statement correctly describes the physical flow of charge representing this $5\text{ A}$ current?

An electromotive force of five volts is continuously exerted across the length of the busbar.

Five coulombs of electric charge pass through a cross-section of the busbar each second.

Five coulombs of positive charge carriers travel from the negative terminal toward the positive terminal every minute.

Five alternating oscillations of charge direction occur back and forth along the busbar each second.

Show answer & explanation

Answer: Five coulombs of electric charge pass through a cross-section of the busbar each second.

Answer

Five coulombs of electric charge pass through a cross-section of the busbar each second.

Electric current (

I

) measures the rate at which charge passes through a conductor. One ampere (

1\text{ A}

) is defined as one coulomb (

1\text{ C}

) of electrical charge passing through a cross-sectional area in one second (

1\text{ s}

). Therefore, a steady current of

5\text{ A}

means that

5\text{ coulombs}

of charge pass through the cross-section of the busbar every second.

Step-by-Step Solution

1

Identify the fundamental definition of electric current and its standard unit.

Current (

I

) is defined as the rate of transfer of electric charge (

Q

) per unit of time (

t

), expressed by the formula

I = \frac{Q}{t}

.

The ampere is the base SI unit of electrical current, defined as one coulomb of charge passing a point per second (

1\text{ A} = 1\text{ C/s}

).

2

Apply the given current value of

5\text{ A}

to the rate formula.

I = 5\text{ A} = \frac{5\text{ coulombs}}{1\text{ second}}

.

A steady current of

5\text{ A}

directly translates to a charge flow rate of

5\text{ coulombs per second}

through the conductor.

Key Concept

Electric current is the time rate of charge flow through a conductor, where one ampere equals one coulomb per second (

I = Q/t

).

Estimated Time:40s

Question 930Question →

A $48\text{ V}$ direct-current telecommunications power bus supplies electricity to three parallel branches. The first branch draws a current of $2\text{ A}$ , the second branch contains a load with a resistance of $24\ \Omega$ , and the total current supplied by the power bus is $7\text{ A}$ . What is the electrical resistance of the third branch, in ohms?

Show answer & explanation

Answer: 16

Answer

The resistance of the third branch is

16\ \Omega

.

Because the circuit is parallel, the full supply voltage of

48\text{ V}

is present across every branch. The second branch with

24\ \Omega

draws

I_2 = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

. Subtracting the known currents of Branch 1 (

2\text{ A}

) and Branch 2 (

2\text{ A}

) from the total current of

7\text{ A}

leaves

3\text{ A}

flowing through Branch 3. Using Ohm's Law,

R_3 = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

.

Step-by-Step Solution

1

Calculate the current flowing through Branch 2

I_2 = \frac{V}{R_2} = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

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

V_1 = V_2 = V_3 = 48\text{ V}

).

2

Calculate the current flowing through Branch 3

I_3 = I_{\text{total}} - I_1 - I_2 = 7\text{ A} - 2\text{ A} - 2\text{ A} = 3\text{ A}

The total current entering a parallel network equals the sum of the individual branch currents (

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

).

3

Calculate the resistance of Branch 3

R_3 = \frac{V}{I_3} = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) using the common

48\text{ V}

supply voltage and the calculated

3\text{ A}

branch current yields the branch resistance.

Key Concept

In a parallel DC circuit, voltage remains constant across all branches, total current is the algebraic sum of individual branch currents, and branch resistance is determined by Ohm's Law (

R = \frac{V}{I}

).

Question 931Question →

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

Show answer & explanation

Answer: 72

Answer

The total electrical power consumed is

72\text{ W}

.

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

P = V \times I

. Substituting the given voltage of

24\text{ V}

and current of

3\text{ A}

yields

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

.

Step-by-Step Solution

1

Identify the given values from the circuit problem

Voltage

V = 24\text{ V}

and current

I = 3\text{ A}

Electrical power calculations require knowing at least two circuit parameters.

2

Select the appropriate electrical power formula

P = V \times I

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

3

Substitute the values and calculate the final result

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

Multiplying volts by amperes gives power directly in watts.

Key Concept

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

P = V \times I

).

Estimated Time:45s

Question 932Question →

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 933Question →

An emergency backup lighting circuit is powered by a $48\text{ V}$ DC supply connected in series with a $30\ \Omega$ current-limiting resistor and two identical LED lamp modules. If a steady current of $0.4\text{ A}$ flows through the circuit during operation, what is the voltage drop across each individual LED lamp module?

12\text{ V}

36\text{ V}

18\text{ V}

24\text{ V}

Show answer & explanation

Answer: $18\text{ V}$

Answer

18\text{ V}

In a series circuit, the total voltage of

48\text{ V}

is distributed across all components according to Ohm's Law. With

0.4\text{ A}

flowing through the

30\ \Omega

resistor, that resistor drops

12\text{ V}

. Subtracting

12\text{ V}

from the total

48\text{ V}

source leaves

36\text{ V}

dropped across the two identical LED modules. Because the modules are identical and in series, they share this remaining voltage equally, resulting in

18\text{ V}

across each module.

Step-by-Step Solution

1

Calculate the voltage drop across the

30\ \Omega

series resistor using Ohm's Law (

V = I \times R

).

V_{R} = 0.4\text{ A} \times 30\ \Omega = 12\text{ V}

Current is identical through every component in a series circuit.

2

Determine the remaining voltage available for the two LED modules by applying Kirchhoff's Voltage Law (

V_{\text{total}} - V_R

).

V_{\text{remaining}} = 48\text{ V} - 12\text{ V} = 36\text{ V}

The sum of all voltage drops in a series loop must equal the total supply voltage.

3

Divide the remaining voltage equally between the two identical LED modules.

V_{\text{module}} = \frac{36\text{ V}}{2} = 18\text{ V}

Identical series components carry the same current and therefore drop equal portions of the remaining voltage.

Key Concept

In a series circuit, current is constant throughout the single loop, and the total supply voltage equals the sum of the individual component voltage drops.

Estimated Time:1m 0s

Question 934Question →

In an automotive electrical system, an electromechanical relay is installed between a dashboard switch and a high-current cooling fan. What is the primary purpose of using the relay in this circuit configuration?

It acts as a sacrificial protective device that permanently melts open to interrupt excessive circuit current.

It allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

It steps up the source voltage to deliver a higher electromotive force across the fan motor terminals.

It converts direct current from the battery into alternating current before reaching the fan motor.

Show answer & explanation

Answer: It allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

Answer

The correct answer states that it allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

An electromechanical relay utilizes a low current passing through an internal coil to generate a magnetic field. This magnetic field pulls an armature to close heavy-duty contacts, safely powering high-current loads like cooling fans without subjecting delicate dashboard switches to high current or requiring thick, heavy wiring runs through the passenger compartment.

Step-by-Step Solution

1

Identify the distinct circuit sections established by the relay.

The dashboard switch operates within a low-current control circuit connected to the relay coil, while the cooling fan operates within a separate high-current load circuit connected through the relay contacts.

Relays provide electrical isolation and remote switching between control circuits and power circuits.

2

Evaluate the technical necessity of isolating high-current loads from cabin switches.

Routing substantial current directly through the dashboard switch would require thick gauge wiring into the vehicle interior, cause significant voltage drop over long distances, and degrade manual switch contacts due to arcing.

Electromechanical relays permit small control currents to safely actuate high-amperage contacts placed near the battery and load.

3

Match the identified operational principle to the corresponding functional description.

The relay serves as a remote electromechanical switch allowing low-current input to control high-current output.

Relays do not act as fuses, voltage transformers, or DC-to-AC inverters.

Key Concept

Electromechanical relay function in high-current DC switching

Estimated Time:1m 0s

Question 935Question →

Match each oscilloscope front-panel control or display feature on the left with its corresponding measurement function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Volts/Div matches vertical scale (amplitude/voltage measurement), Time/Div matches horizontal sweep rate (period/frequency measurement), Trigger matches waveform stabilization, and Graticule grid matches the calibrated measurement overlay.

Each oscilloscope feature is mapped directly to its core operational role: the Volts/Div knob scales the vertical axis for voltage amplitude, Time/Div scales the horizontal axis for time intervals, the Trigger synchronizes the horizontal sweep for a stable trace, and the Graticule acts as the calibrated grid overlay.

Step-by-Step Solution

1

Identify the vertical axis control and its function

Volts/Div controls vertical scaling to measure voltage values such as peak or peak-to-peak amplitude.

Vertical displacement on an oscilloscope screen is directly proportional to signal voltage.

2

Identify the horizontal axis control and its function

Time/Div controls horizontal sweep time to measure period and calculate frequency (

f = \frac{1}{T}

).

Horizontal displacement represents elapsed time across the screen.

3

Identify the synchronization mechanism

The Trigger control stabilizes the signal display.

Without proper triggering, the trace starts at random points of the cycle, causing the waveform to scroll continuously.

4

Identify the physical reference markings

The Graticule grid provides the reference divisions for counting units.

The graticule lines allow the operator to multiply division counts by the scale settings.

Key Concept

Basic oscilloscope controls and display anatomy

Estimated Time:45s

Question 936Question →

An industrial direct-current (DC) sensor loop is powered by a $50\text{ V}$ power source. The loop consists of two series-connected components: a fixed resistor $R_1 = 10\ \Omega$ and an unknown resistor $R_2$ . An ammeter placed in the circuit measures a continuous current of $2\text{ A}$ . What is the resistance of $R_2$ , and what is the voltage drop across $R_2$ ?

Fill in the blanks below

The resistance of

R_2

is

\Omega

, and the voltage drop across

R_2

is

\text{V}

.

Show answer & explanation

Answer

The resistance of

R_2

is

15\ \Omega

, and the voltage drop across

R_2

is

30\text{ V}

.

In a series circuit, the total resistance is calculated as

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

. Since

R_{\text{total}} = R_1 + R_2

, the unknown resistance is

R_2 = 25\ \Omega - 10\ \Omega = 15\ \Omega

. The voltage drop across

R_2

is found by multiplying the common series current by

R_2

, yielding

V_2 = 2\text{ A} \times 15\ \Omega = 30\text{ V}

(which also satisfies Kirchhoff's Voltage Law:

50\text{ V} - 20\text{ V} = 30\text{ V}

).

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_{\text{total}}

) of the series circuit using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

In a closed DC loop, total resistance equals total source voltage divided by total current.

2

Determine the value of the unknown resistor

R_2

using the series resistance rule.

R_2 = R_{\text{total}} - R_1 = 25\ \Omega - 10\ \Omega = 15\ \Omega

Total resistance in a series circuit is the sum of individual resistances (

R_{\text{total}} = R_1 + R_2

).

3

Calculate the individual voltage drop across

R_2

(

V_2

).

V_2 = I \times R_2 = 2\text{ A} \times 15\ \Omega = 30\text{ V}

The same current flows through all series components; voltage drop across any component equals current multiplied by its resistance (or

V_{\text{total}} - V_1 = 50\text{ V} - 20\text{ V} = 30\text{ V}

).

Key Concept

Series Circuit Resistance and Voltage Division

Estimated Time:1m 0s

Question 937Question →

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 938Question →

On the cylindrical casing of a standard axial-lead semiconductor diode, what feature is identified by the painted stripe or band near one of the ends?

The cathode terminal

The anode terminal

The nominal resistance rating in ohms

The maximum operating alternating current (AC) frequency

Show answer & explanation

Answer: The cathode terminal

Answer

The cathode terminal

On standard through-hole axial diodes, the painted stripe marks the cathode terminal. In schematic diagrams, this corresponds to the vertical bar toward which the triangle points.

Step-by-Step Solution

1

Identify the standard packaging markings used on axial-lead semiconductor diodes.

Standard cylindrical diodes (such as the 1N4001 series) use a single painted band at one end to distinguish polarity.

Because diodes are polarized components that conduct current primarily in one direction, physical orientation must be identifiable.

2

Correlate the physical stripe with the corresponding electronic terminal.

The stripe aligns with the cathode (negative lead during forward conduction), matching the vertical bar in the standard schematic symbol.

Correct alignment ensures the diode is wired in forward bias rather than accidental reverse bias.

Key Concept

Physical diode lead identification and polarity marking

Estimated Time:45s

Question 939Question →

A magnetic field is generated around a conductor only when alternating current (AC) flows through it, whereas direct current (DC) produces no magnetic field.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because any flow of electric current through a conductor creates a magnetic field. Direct current (DC) produces a steady, non-fluctuating magnetic field around the wire, while alternating current (AC) produces an expanding and collapsing magnetic field.

Step-by-Step Solution

1

Identify the fundamental relationship between electric current and magnetic fields.

Electric current is the movement of electric charge, and any moving charge produces a magnetic field around its path.

This is the core principle of electromagnetism discovered by Hans Christian Oersted.

2

Evaluate the behavior of direct current (DC) in a conductor.

A steady direct current produces a steady, stationary magnetic field forming concentric circular lines of flux around the conductor.

A constant current still consists of moving electrical charges, which continuously maintain a magnetic field.

3

Distinguish between magnetic field generation and electromagnetic induction.

While electromagnetic induction (generating a voltage in another conductor) requires a changing magnetic field (such as from AC), generating a magnetic field itself requires only the presence of electric current.

Confusing field generation with transformer induction leads to the false belief that DC cannot produce magnetism.

Key Concept

Magnetic Field Generation by Electric Current

Estimated Time:35s

Question 940Question →

While troubleshooting a de-energized printed circuit board, a technician measures the resistance across a $10\text{ k}\Omega$ resistor that remains fully soldered in place. The digital multimeter displays a steady reading of $2.4\text{ k}\Omega$ . What is the primary cause of this lower resistance reading?

The circuit must be energized so that the multimeter internal sensing circuit receives sufficient operating voltage.

The test current from the multimeter is flowing through alternate parallel paths connected across the circuit board.

The resistor has experienced an internal breakdown and formed a complete electrical short circuit.

The multimeter leads were placed across the component in reverse polarity without zeroing the test leads.

Show answer & explanation

Answer: The test current from the multimeter is flowing through alternate parallel paths connected across the circuit board.

Answer

The test current from the multimeter is flowing through alternate parallel paths connected across the circuit board.

A digital multimeter measures resistance by injecting a small internal test voltage and measuring the resulting current. When measuring a component that is still soldered into a circuit, that test current splits across the component and any other interconnected parallel branches. Because parallel resistance combinations always result in an equivalent resistance smaller than the smallest branch (

R_{total} < R_{branch}

), the meter registers a lower value (

2.4\text{ k}\Omega

) than the resistor's nominal rating (

10\text{ k}\Omega

). To obtain an accurate reading, at least one lead of the component must be disconnected or desoldered.

Step-by-Step Solution

1

Analyze the measurement environment and the component's circuit state.

The resistor is measured while fully connected (in-circuit) to other components on the board.

Components soldered onto a circuit board are rarely isolated and typically connect to adjacent traces, semiconductors, or resistors.

2

Apply parallel circuit principles to the ohmmeter test current.

The ohmmeter supplies a small internal test current across the probes, which divides among all available conductive paths between those two nodes.

According to the laws of parallel resistance, adding parallel paths always reduces the total equivalent resistance below the value of the lowest single branch (

R_{eq} < R_1

).

3

Determine the necessary troubleshooting procedure for an accurate reading.

Isolate at least one terminal/lead of the resistor from the circuit board before measuring.

Desoldering one leg breaks parallel loop paths, ensuring 100% of the meter's test current flows strictly through the component under test.

Key Concept

In-circuit resistance measurement, component isolation, and parallel path loading

Estimated Time:1m 0s

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