All practice questions

Question 881Question →

During the calibration of a naval radar control system, engineers measure a continuous current of $1.5\text{ A}$ flowing through a series safety loop. The loop is energized by a $60\text{ V}$ direct current source and contains three elements: a $15\ \Omega$ current-limiting resistor ( $R_1$ ), a calibration rheostat ( $R_2$ ), and a primary sensor coil ( $R_3$ ). If the primary sensor coil dissipates exactly $36\text{ W}$ of power, what is the required resistance of the calibration rheostat ( $R_2$ )?

1\ \Omega

40\ \Omega

9\ \Omega

59\ \Omega

Show answer & explanation

Answer: $9\ \Omega$

Answer

The required resistance of the calibration rheostat is

9\ \Omega

.

In this series circuit, the total resistance must equal the source voltage divided by the total current (

60\text{ V} / 1.5\text{ A} = 40\ \Omega

). Since components are in series, they share the same

1.5\text{ A}

current. The sensor coil dissipates

36\text{ W}

, meaning its resistance is

P / I^2

(

36 / 2.25 = 16\ \Omega

). Subtracting the

15\ \Omega

resistor and the

16\ \Omega

coil from the

40\ \Omega

total leaves

9\ \Omega

for the calibration rheostat.

Step-by-Step Solution

1

Calculate the total equivalent resistance of the series circuit using Ohm's Law.

R_T = \frac{V_T}{I_T} = \frac{60\text{ V}}{1.5\text{ A}} = 40\ \Omega

Total resistance is needed to find the missing component value in a series string.

2

Determine the resistance of the primary sensor coil (

R_3

) using the power formula

P = I^2R

.

R_3 = \frac{P}{I^2} = \frac{36\text{ W}}{(1.5\text{ A})^2} = \frac{36}{2.25} = 16\ \Omega

Since the circuit is in series, the full

1.5\text{ A}

current flows through all components, allowing the coil's resistance to be calculated from its power dissipation.

3

Subtract the known series resistances from the total resistance to find the rheostat's value (

R_2

).

R_2 = R_T - R_1 - R_3 = 40\ \Omega - 15\ \Omega - 16\ \Omega = 9\ \Omega

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

R_T = R_1 + R_2 + R_3

).

Key Concept

Series Circuit Resistance and Power Dissipation

Question 882Question →

An artillery spotter is recording the impact distances (in meters) from a target center for five test rounds. The distances are $34$ , $41$ , $28$ , $52$ , and $40$ . A sixth round is fired. If the mean distance for all six rounds is exactly $41$ meters, what was the impact distance of the sixth round?

10

39

40

51

Show answer & explanation

Answer: $51$

Answer

The correct impact distance of the sixth round is

51

meters.

To find a missing value when the mean is known, you must work backward using the total sum. Since the mean of

6

rounds is

41

, their total sum must be

41 \times 6 = 246

. The sum of the first five rounds is

195

. Subtracting the current sum from the required total sum (

246 - 195

) reveals that the sixth round must be

51

.

Step-by-Step Solution

1

Calculate the required sum of the distances for all six rounds.

41 \times 6 = 246

The mean of a data set is its total sum divided by the number of items. To find the required total sum, multiply the target mean (

41

) by the total number of rounds (

6

).

2

Calculate the sum of the distances for the first five rounds.

34 + 41 + 28 + 52 + 40 = 195

This establishes the total distance that has already been accounted for by the initial test rounds.

3

Subtract the sum of the first five rounds from the required total sum to find the sixth round.

246 - 195 = 51

The difference between the total needed for all six rounds and the total already recorded gives the exact value required for the final round.

Key Concept

Solving for a missing value given a target mean

Question 883Question →

An AC step-down transformer has $600\text{ turns}$ on its primary winding and $50\text{ turns}$ on its secondary winding. The primary winding is connected to a $120\text{ V}$ AC source, and the secondary circuit delivers current to a load drawing $2.4\text{ A}$ . What is the current flowing in the primary winding?

28.8\text{ A}

20\text{ mA}

0.2\text{ A}

0\text{ A}

Show answer & explanation

Answer: $0.2\text{ A}$

Answer

0.2\text{ A}

In an ideal transformer, the ratio of primary current to secondary current is inversely proportional to the ratio of primary turns to secondary turns (

I_p / I_s = N_s / N_p

). Since the transformer has a

12:1

step-down turns ratio (

600/50 = 12

), the secondary voltage decreases by a factor of 12 while the primary current must be 12 times smaller than the secondary current:

I_p = 2.4\text{ A} / 12 = 0.2\text{ A}

.

Step-by-Step Solution

1

Determine the turns ratio between the primary and secondary windings

\frac{N_p}{N_s} = \frac{600}{50} = 12

The turns ratio determines both the voltage transformation ratio and the inverse current transformation ratio.

2

Apply the transformer current relationship to solve for primary current (

I_p

)

\frac{I_p}{I_s} = \frac{N_s}{N_p} \implies I_p = 2.4\text{ A} \times \frac{50}{600} = \frac{2.4\text{ A}}{12} = 0.2\text{ A}

In an ideal transformer, power in equals power out (

V_p I_p = V_s I_s

), which means current is inversely proportional to the number of turns.

Key Concept

Transformer Current and Turns Ratio Relationship

Estimated Time:1m 0s

Question 884Question →

Match each specialized switching or relay device to its corresponding operational characteristic and actuation method.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Reed switch matches ferromagnetic flexible metal reeds sealed in glass; Rotary switch matches rotating shaft and wiper contact selecting between three or more positions; Contactor matches heavy-duty electromechanical device for high-current power loads; Thermal overload relay matches monitoring sustained overcurrent using a bimetallic strip.

Each switching component is matched according to its fundamental physics of operation: the reed switch relies on magnetic field induction on ferrous contacts; the rotary switch provides multi-position manual circuit routing; the contactor uses an electromagnetic solenoid for high-amperage switching; and the thermal overload relay uses thermal deflection of a bimetallic strip for motor overload protection.

Step-by-Step Solution

1

Identify the actuation mechanism of a reed switch.

Recognize that reed switches operate magnetically via enclosed flexible ferrous blades.

Reed switches rely on external magnetic flux rather than manual toggling or internal thermal deflection.

2

Analyze the mechanical multi-position selection function of a rotary switch.

Associate the rotary switch with angular shaft rotation routing a signal to multiple switch poles/throws.

Rotary switches provide multi-throw circuit selection through rotational movement.

3

Distinguish between standard low-power control relays and contactors.

Connect contactors to high-current power switching controlled via electromagnetic coils.

Contactors feature heavy-duty contacts built specifically to handle high current without contact welding.

4

Determine the protective operation of a thermal overload relay.

Pair the thermal overload relay with bimetallic thermal expansion sensing excessive motor current.

Thermal overload relays protect motor windings by converting persistent current into heat that mechanically triggers auxiliary contacts.

Key Concept

Specialized switch and relay actuation mechanisms and circuit roles

Estimated Time:1m 0s

Question 885Question →

When troubleshooting electrical circuits with a digital multimeter set to measure resistance ( $\Omega$ ), the display reading reflects the physical state of the circuit path. Match each digital multimeter display reading with its corresponding circuit condition.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Zero ohms (

0.0\ \Omega

) matches a continuous, closed path; over-limit (

\text{OL}

) matches an open circuit or broken conductor; and a finite numerical value (

150\ \Omega

) matches a functional resistive load.

A reading of

0.0\ \Omega

indicates that the path is continuous with zero opposition to current. An over-limit display (

\text{OL}

) indicates infinite resistance, which occurs when a conductor is broken or a switch is open. A finite numerical value such as

150\ \Omega

represents an intact resistive component operating normally.

Step-by-Step Solution

1

Interpret the zero resistance reading (

0.0\ \Omega

)

Identified as full continuity

When meter leads are connected across a continuous conductor or closed switch, opposition to current is zero.

2

Interpret the infinite reading (

\text{OL}

)

Identified as an open circuit

An interrupted path or broken filament does not permit electron flow, resulting in infinite resistance beyond the meter range.

3

Interpret the finite numeric reading (

150\ \Omega

)

Identified as a normal resistive component

Operational electrical components such as heating elements or resistors have a rated non-zero resistance value.

Key Concept

Interpreting Digital Multimeter Resistance and Continuity Display Readings

Estimated Time:45s

Question 886Question →

Match each electromagnetic device on the left with its correct operating principle or primary functional description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electromechanical Relay matches with using an energized coil to pull an armature to open or close circuit contacts; Linear Solenoid matches with converting an electromagnetic field into linear mechanical motion to actuate a plunger; Step-Up Transformer matches with using mutual induction to increase AC voltage while decreasing AC current; Inductor Coil matches with opposing changes in electrical current via self-induced counter-EMF.

Each electromagnetic device corresponds to a distinct electromagnetic mechanism: the relay magnetically toggles switch contacts; the solenoid converts magnetic fields into linear mechanical stroke; the step-up transformer uses mutual induction to increase AC voltage; and the inductor coil generates self-induced counter-EMF to oppose changes in current.

Step-by-Step Solution

1

Identify the operating mechanism of an electromechanical relay

Matched to using an energized coil to pull a ferromagnetic armature and toggle electrical contacts.

Relays function as electrically operated switches, using electromagnetism to control contacts in a secondary circuit without direct electrical connection.

2

Identify the operating mechanism of a linear solenoid

Matched to converting magnetic flux into linear mechanical motion of an internal plunger.

Solenoids are actuators that translate electromagnetic coil energization into physical linear pushing or pulling work.

3

Identify the operating mechanism of a step-up transformer

Matched to using mutual induction to increase AC voltage while proportionally decreasing current.

Transformers rely on dynamic magnetic flux linkage between primary and secondary coils to scale alternating voltage levels according to turns ratios.

4

Identify the operating mechanism of an inductor coil

Matched to opposing current variations through self-induced counter-electromotive force.

According to Lenz's law and Faraday's law of self-induction, an inductor generates a counter-voltage that directly resists changes in the current passing through it.

Key Concept

Operating principles and applications of core electromagnetic components: relays, solenoids, transformers, and inductors

Estimated Time:1m 0s

Question 887Question →

In order for a standard PN junction diode to conduct conventional electric current, its cathode must be connected to a higher positive electrical potential than its anode.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because forward biasing a PN junction diode requires the anode (P-type material) to be at a higher positive potential than the cathode (N-type material). Applying a higher positive potential to the cathode places the diode in reverse bias, which widens the depletion layer and prevents current flow.

Step-by-Step Solution

1

Identify the relationship between diode terminals and semiconductor materials.

The anode is connected to the P-type semiconductor region, and the cathode is connected to the N-type semiconductor region.

Understanding terminal orientation is essential to determining how external voltage affects the internal PN junction barrier.

2

Evaluate the biasing condition described in the statement.

Connecting the cathode to a more positive potential than the anode creates a reverse-bias condition, widening the depletion region and preventing conventional current conduction.

For conduction to occur, the diode must be forward-biased by applying a more positive potential to the anode relative to the cathode.

Key Concept

Diode Biasing and Polarity

Estimated Time:45s

Question 888Question →

An electronics technician is analyzing the operational principles of a DC series circuit. Match each series circuit characteristic on the left with its correct electrical rule or behavioral description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Total circuit resistance matches with the sum of all individual load values; Current flow matches with remaining constant and identical at every point; Individual component voltage drop matches with being directly proportional to resistance (

I \times R_n

); Open-circuit fault matches with reducing total loop current to zero and disabling all components.

In a series circuit, all components are connected sequentially in a single loop. Therefore, total resistance is the direct sum of all resistor values, the current is uniform throughout, the individual voltage drops are proportional to component resistance via Ohm's Law, and an open circuit completely eliminates current flow across all components.

Step-by-Step Solution

1

Identify the relationship for total resistance in a series string

R_T = R_1 + R_2 + \dots + R_n

Series components add their oppositions consecutively to the flow of current.

2

Determine the path and behavior of electrical current in a series loop

I_T = I_1 = I_2 = \dots = I_n

A single conducting path allows no branching, ensuring the rate of charge flow is uniform everywhere.

3

Apply Ohm's Law to find the voltage drop across an individual resistor

V_n = I \times R_n

With current constant, the voltage drop across each component depends directly on its resistance value.

4

Analyze the impact of a discontinuity (open circuit)

I = 0\text{ A}

throughout the entire loop

Breaking any point in a single closed loop stops all current from circulating.

Key Concept

Core governing laws of DC series circuits: additive resistance, uniform loop current, proportional voltage division, and single-path continuity dependency.

Estimated Time:1m 0s

Question 889Question →

An electronics technician connects three inductors in parallel across an AC filter circuit. The inductors have inductance values of $60\text{ mH}$ , $30\text{ mH}$ , and $20\text{ mH}$ , with negligible mutual inductance between them. What is the total equivalent inductance of the circuit in millihenries (mH)?

Show answer & explanation

Answer: 10

Answer

The total equivalent inductance of the parallel circuit is

10\text{ mH}

.

Inductors in parallel without mutual coupling combine according to the reciprocal formula

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

. Computing

\frac{1}{60} + \frac{1}{30} + \frac{1}{20} = \frac{6}{60} = \frac{1}{10}

, and inverting yields

10\text{ mH}

.

Step-by-Step Solution

1

State the reciprocal formula for inductors connected in parallel

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

When inductors are in parallel and uncoupled, the total rate of current change is the sum of the individual branch rates, resulting in a reciprocal combination rule identical to parallel resistors.

2

Substitute the individual inductance values

\frac{1}{L_{\text{total}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}

Direct substitution of the known branch values:

L_1 = 60\text{ mH}

,

L_2 = 30\text{ mH}

, and

L_3 = 20\text{ mH}

.

3

Find the common denominator and sum the terms

\frac{1}{L_{\text{total}}} = \frac{1 + 2 + 3}{60} = \frac{6}{60} = \frac{1}{10}\text{ mH}^{-1}

Using the lowest common denominator (

60

) to perform fractional addition.

4

Invert the reciprocal to obtain equivalent inductance

L_{\text{total}} = 10\text{ mH}

Taking the reciprocal of

\frac{1}{10}\text{ mH}^{-1}

gives

10\text{ mH}

.

Key Concept

Parallel Inductance Calculation

Estimated Time:1m 0s

Question 890Question →

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 891Question →

An avionics technician tests a direct current (DC) relay control circuit connected to a $36\text{ V}$ power supply. An in-line ammeter indicates that the relay coil draws a steady current of $75\text{ mA}$ . What is the total electrical resistance of the relay coil?

0.48\text{ }\Omega

480\text{ }\Omega

2.7\text{ }\Omega

2,700\text{ }\Omega

Show answer & explanation

Answer: $480\text{ }\Omega$

Answer

The total electrical resistance of the relay coil is

480\text{ }\Omega

.

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

R = \frac{V}{I}

). First, converting the current from milliamperes to amperes gives

75\text{ mA} = 0.075\text{ A}

. Dividing the

36\text{ V}

supply by

0.075\text{ A}

yields a resistance of

480\text{ }\Omega

.

Step-by-Step Solution

1

Convert the measured current from milliamperes to base amperes.

I = 75\text{ mA} = 75 \times 10^{-3}\text{ A} = 0.075\text{ A}

Standard electrical calculations using Ohm's law require values to be expressed in base SI units (volts, amperes, and ohms).

2

Apply Ohm's law rearranged to solve for resistance (

R = \frac{V}{I}

).

R = \frac{36\text{ V}}{0.075\text{ A}} = 480\text{ }\Omega

Resistance equals electromotive force (voltage) divided by current flow.

Key Concept

Calculating resistance in a DC circuit using Ohm's Law with metric prefix conversion.

Estimated Time:1m 0s

Question 892Question →

A quality assurance technician is inspecting a four-band carbon-composition resistor with the color bands in order: Yellow, Violet, Red, and Silver. To meet quality standards, the resistor must operate within its specified manufacturer tolerance. What is the minimum acceptable resistance value, in ohms ( $\Omega$ ), for this component?

Show answer & explanation

Answer: 4230

Answer

The minimum acceptable resistance value is

4230\text{ }\Omega

.

The first two bands (Yellow = 4, Violet = 7) establish the significant figures of 47. The third band (Red) indicates a multiplier of

10^2 = 100

, giving a nominal rating of

4700\text{ }\Omega

. The fourth band (Silver) specifies a

\pm 10\%

tolerance. Subtracting

10\%

of

4700\text{ }\Omega

(

470\text{ }\Omega

) from the nominal rating gives the minimum allowable threshold of

4230\text{ }\Omega

.

Step-by-Step Solution

1

Decode the first two color bands to obtain the significant numerical digits.

Yellow represents the digit

4

, and Violet represents the digit

7

, forming the base value of

47

.

Standard EIA four-band resistor color coding assigns the first two bands as significant figures of the resistance value.

2

Multiply the base significant figures by the power of ten indicated by the third band.

Red indicates a multiplier of

10^2

(

100

), so the nominal resistance is

47 \times 100 = 4700\text{ }\Omega

.

The third band functions as the decimal multiplier for four-band fixed resistors.

3

Identify the tolerance rating from the fourth band.

Silver represents a tolerance of

\pm 10\%

.

The fourth band specifies the allowable deviation range from the nominal resistance value.

4

Calculate the lower allowable resistance threshold.

Tolerance amount is

4700\text{ }\Omega \times 0.10 = 470\text{ }\Omega

. Minimum acceptable value is

4700\text{ }\Omega - 470\text{ }\Omega = 4230\text{ }\Omega

.

The minimum acceptable value represents the nominal value minus the total tolerance deviation.

Key Concept

Four-Band Resistor Color Code Decoding and Tolerance Range Calculation

Alternative Method

Multiply the nominal resistance directly by the minimum percentage factor:

4700\text{ }\Omega \times (1.00 - 0.10) = 4700\text{ }\Omega \times 0.90 = 4230\text{ }\Omega

.

Estimated Time:45s

Question 893Question →

A technician connects a 12-volt direct-current (DC) power supply to a two-way radio circuit. According to electron flow theory, how do charge carriers physically move through the external wiring?

Electrons move from the positive terminal, through the radio, toward the negative terminal.

Electrons move from the negative terminal, through the radio, toward the positive terminal.

Electrons oscillate back and forth in place without any net directional movement through the circuit.

Electromotive force creates static potential across the wire without causing physical particle drift.

Show answer & explanation

Answer: Electrons move from the negative terminal, through the radio, toward the positive terminal.

Answer

Electrons move from the negative terminal, through the radio, toward the positive terminal.

In solid metallic conductors, electric current is the physical drift of free electrons. Because electrons have a negative charge, they are repelled by the negative terminal of the DC power source and attracted toward the positive terminal through the external circuit.

Step-by-Step Solution

1

Identify the nature of the power source and charge carriers

The circuit is powered by a direct-current (DC) source where charge carriers in metallic conductors are negatively charged electrons.

Understanding the carrier polarity is essential for determining movement under an electric field.

2

Apply electrostatic attraction and repulsion principles

Electrons carry a negative charge and are therefore repelled by the negative battery terminal and attracted toward the positive terminal.

Like charges repel and opposite charges attract.

3

Determine the physical direction in the external path

In the external circuit outside the power supply, electron flow proceeds from the negative terminal through the load to the positive terminal.

Electron flow theory describes the actual physical movement of electrons.

Key Concept

Electron flow theory dictates that in a DC circuit, electrons drift through the external conductor from the negative terminal to the positive terminal.

Estimated Time:35s

Question 894Question →

A specialized weather station telemetry circuit is powered by a $96\text{ V}$ direct current (DC) power source connected in series with four resistive components: a surge-limiting resistor $R_1 = 18\ \Omega$ , a status indicator $R_2 = 27\ \Omega$ , a variable humidity sensor $R_3$ , and a reference resistor $R_4 = 15\ \Omega$ . A technician measures the voltage drop across the status indicator ( $R_2$ ) to be $21.6\text{ V}$ . What is the resistance of the humidity sensor $R_3$ in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 60

Answer

The resistance of the humidity sensor

R_3

is

60\ \Omega

.

In a series circuit, current is uniform throughout the entire loop. Using the measured voltage drop across

R_2

(

21.6\text{ V}

) and its known resistance (

27\ \Omega

), the loop current is

I = \frac{V_2}{R_2} = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}

. The total equivalent resistance connected across the

96\text{ V}

source is

R_T = \frac{V_{\text{source}}}{I} = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega

. Because series resistances are directly additive (

R_T = R_1 + R_2 + R_3 + R_4

), the unknown resistance is determined by subtracting the known values from the total:

R_3 = 120\ \Omega - (18\ \Omega + 27\ \Omega + 15\ \Omega) = 120\ \Omega - 60\ \Omega = 60\ \Omega

.

Step-by-Step Solution

1

Calculate loop current from the known component voltage drop and resistance.

I = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}

In a series circuit, current is identical through all components.

2

Calculate total circuit resistance using Ohm's Law and the source voltage.

R_T = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega

Total resistance equals source voltage divided by total circuit current.

3

Calculate the sum of all known resistances.

R_{\text{known}} = 18\ \Omega + 27\ \Omega + 15\ \Omega = 60\ \Omega

Resistances in a series circuit are directly additive.

4

Subtract the sum of known resistors from total circuit resistance to isolate

R_3

.

R_3 = 120\ \Omega - 60\ \Omega = 60\ \Omega

R_T = R_1 + R_2 + R_3 + R_4

, so

R_3 = R_T - (R_1 + R_2 + R_4)

.

Key Concept

Series Circuit Loop Current and Total Resistance Analysis

Alternative Method

Calculate individual voltage drops using loop current

I = 0.8\text{ A}

:

V_1 = 0.8\text{ A} \times 18\ \Omega = 14.4\text{ V}

and

V_4 = 0.8\text{ A} \times 15\ \Omega = 12.0\text{ V}

. By Kirchhoff's Voltage Law, the voltage across

R_3

is

V_3 = 96\text{ V} - (14.4\text{ V} + 21.6\text{ V} + 12.0\text{ V}) = 48.0\text{ V}

. Applying Ohm's Law directly to

R_3

yields

R_3 = \frac{V_3}{I} = \frac{48.0\text{ V}}{0.8\text{ A}} = 60\ \Omega

.

Estimated Time:1m 30s

Question 895Question →

An electromagnet is constructed by winding insulated copper wire around a soft iron core. Which of the following modifications will increase the strength of the magnetic field generated by this electromagnet?

Reversing the direction of current flow through the coil

Increasing the electric current flowing through the coil

Replacing the iron core with a non-magnetic plastic core

Maintaining a high voltage across an open circuit in the coil

Show answer & explanation

Answer: Increasing the electric current flowing through the coil

Answer

Increasing the electric current flowing through the coil

The magnetic field strength of an electromagnet depends directly on the magnitude of the electric current flowing through the windings, the number of wire turns, and the permeability of the core material. Increasing the current directly increases the magnetic flux density.

Step-by-Step Solution

1

Identify the factors determining the magnetic field strength of an electromagnet

The field strength of a solenoid or electromagnet depends directly on the electric current, the number of coil turns, and the magnetic permeability of the core.

Electromagnetic flux density is directly proportional to ampere-turns (

I \times N

) and core permeability.

2

Evaluate the proposed modification against the operational formula

Raising the current (

I

) increases the total magnetomotive force (

MMF = N \cdot I

), thereby producing a stronger magnetic field.

Greater electron movement generates more concentric magnetic lines of force around each loop of wire.

Key Concept

The strength of an electromagnet's magnetic field is directly proportional to the magnitude of current flowing through its windings.

Estimated Time:45s

Question 896Question →

An electronics technician is analyzing a circuit where a heavy-duty relay coil is controlled by a mechanical switch connected to a 24V DC power supply. Every time the switch is opened to turn off the relay, a visible spark arcs across the switch contacts, eventually causing them to fail. What electromagnetic principle primarily causes this arcing, and how does the coil's magnetic field behave to produce it?

Self-induction; the sudden interruption of current causes the coil's magnetic field to collapse rapidly, inducing a high-voltage counter-EMF that bridges the switch gap.

Capacitive discharge; the relay coil stores electrostatic energy while powered and instantly releases this accumulated charge when the circuit is broken.

Continuous electromagnetic induction; the steady DC source creates an oscillating magnetic field in the coil that sustains the electron flow even after the contacts separate.

Short circuit generation; opening the switch removes the circuit's resistance, causing the coil to draw infinite current from the power supply which arcs across the gap.

Show answer & explanation

Answer: Self-induction; the sudden interruption of current causes the coil's magnetic field to collapse rapidly, inducing a high-voltage counter-EMF that bridges the switch gap.

Answer

Self-induction; the sudden interruption of current causes the coil's magnetic field to collapse rapidly, inducing a high-voltage counter-EMF that bridges the switch gap.

The phenomenon described is known as inductive kickback or back-EMF, which is a direct result of self-induction. An inductor (like a relay coil) opposes any change in current. When the switch is opened, the current attempts to drop to zero instantly. The magnetic field surrounding the coil collapses rapidly. According to Faraday's Law, this rapid change in magnetic flux induces a very large voltage spike across the coil. This spike is often much higher than the original 24V supply and is strong enough to ionize the air in the small gap of the opening switch, creating an electrical arc.

Step-by-Step Solution

1

Analyze the components and their state before the switch opens.

The relay coil acts as an inductor. When powered by 24V DC, a steady current flows, and energy is stored in a stable magnetic field around the coil.

Understanding the initial steady-state condition is required to determine what changes.

2

Determine the effect of opening the switch.

Opening the switch forces the current from the power supply to drop toward zero almost instantly.

Inductors oppose sudden changes in current flow.

3

Apply Faraday's Law of Induction to the changing field.

The rapid decrease in current causes the magnetic field to collapse at high speed. This high rate of change induces a very large voltage (counter-EMF) across the coil.

Induced voltage is directly proportional to the rate of change of the magnetic flux.

4

Relate the induced voltage to the physical arcing.

The induced high voltage (often hundreds of volts) easily exceeds the breakdown voltage of the tiny air gap between the separating switch contacts, causing the air to ionize and a spark to jump.

This explains the visible symptom (the spark) and the mechanism of failure.

Key Concept

Inductive kickback (flyback voltage) caused by self-induction

Estimated Time:1m 30s

Question 897Question →

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 898Question →

What critical safety precaution must always be taken before using a digital multimeter to measure the resistance of a component in an electrical circuit?

Keep the circuit energized so the ohmmeter can utilize the system's operating current

Ensure all electrical power to the circuit is completely disconnected and de-energized

Insert the black test lead into the high-amperage terminal port instead of the common port

Wrap the component terminals with electrical insulating tape before applying the test probes

Show answer & explanation

Answer: Ensure all electrical power to the circuit is completely disconnected and de-energized

Answer

Ensure all electrical power to the circuit is completely disconnected and de-energized

A multimeter's ohmmeter or continuity function operates by applying a small internal voltage from its own battery across the component leads and measuring the resulting current flow. If external power is present, the external voltage interferes with the measurement circuitry, causes invalid readings, and risks blowing internal protective fuses or damaging the instrument. Therefore, disconnecting and verifying zero power is the primary safety rule.

Step-by-Step Solution

1

Identify the operating principle of a multimeter set to measure resistance (

\Omega

)

The meter outputs its own low internal DC voltage from an internal battery to pass a known test current through the component and calculate resistance via Ohm's Law (

R = \frac{V}{I}

).

Understanding how the meter measures resistance clarifies why external circuit power disrupts the measurement.

2

Determine the necessary condition of the circuit under test

The circuit must have zero external voltage and zero external current flow.

Any live voltage present on the test leads will alter the calculation, blow the meter's internal fuse, or pose an electrical hazard.

Key Concept

Multimeter Ohmmeter Safety and De-energization Requirement

Estimated Time:45s

Question 899Question →

An electrical technician is testing an electromagnetic induction system where a straight copper conductor moves through a uniform, stationary magnetic field. According to Faraday's law of induction, which of the following actions will increase the magnitude of the electromotive force (EMF) induced across the conductor?

Moving the conductor parallel to the magnetic flux lines rather than perpendicular to them

Increasing the velocity at which the conductor cuts across the magnetic flux lines

Applying a steady direct current (DC) voltage across the conductor to maintain a static electric field

Replacing the copper conductor with a dielectric insulator to store electrostatic charge

Show answer & explanation

Answer: Increasing the velocity at which the conductor cuts across the magnetic flux lines

Answer

Increasing the velocity at which the conductor cuts across the magnetic flux lines

According to Faraday's law of electromagnetic induction, the magnitude of induced voltage (

e

) across a conductor is determined by the formula

e = B \cdot l \cdot v \cdot \sin\theta

. Higher velocity (

v

) increases the rate of cutting magnetic flux lines (

\frac{\Delta \Phi}{\Delta t}

), thereby directly increasing the induced electromotive force.

Step-by-Step Solution

1

Identify the governing formula for induced electromotive force in a moving conductor.

e = B \cdot l \cdot v \cdot \sin\theta

, where

e

is induced EMF,

B

is magnetic flux density,

l

is conductor length,

v

is velocity, and

\theta

is the angle between velocity and the magnetic field.

Faraday's law dictates that induced voltage depends directly on the rate at which magnetic flux lines are cut.

2

Analyze how each parameter affects the magnitude of the induced EMF.

Increasing conductor velocity (

v

) directly increases

\frac{\Delta \Phi}{\Delta t}

(the rate of magnetic flux cutting), which proportionally increases the induced voltage (

e

).

Faster relative motion between the conductor and magnetic field lines generates greater electrical potential across the conductor.

Key Concept

Faraday's Law of Electromagnetic Induction

Estimated Time:1m 0s

Question 900Question →

A technician is testing a direct-current (DC) control circuit connected to a $20\text{ V}$ power supply. An ammeter in the loop indicates a steady current of $0.5\text{ A}$ . The circuit contains three resistors wired in series: two have fixed values of $12\ \Omega$ and $16\ \Omega$ , while the third is an unlabelled component. What is the resistance of the unlabelled resistor?

28\ \Omega

12\ \Omega

10\ \Omega

40\ \Omega

Show answer & explanation

Answer: $12\ \Omega$

Answer

12\ \Omega

The correct answer of

12\ \Omega

is found by first calculating the total circuit resistance using Ohm's Law (

R_T = \frac{20\text{ V}}{0.5\text{ A}} = 40\ \Omega

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

R_T = R_1 + R_2 + R_3

). Subtracting the two known values (

12\ \Omega + 16\ \Omega = 28\ \Omega

) from

40\ \Omega

leaves exactly

12\ \Omega

for the unlabelled resistor.

Step-by-Step Solution

1

Calculate the total resistance (

R_T

) of the series circuit using Ohm's Law.

R_T = \frac{V_{\text{total}}}{I} = \frac{20\text{ V}}{0.5\text{ A}} = 40\ \Omega

The total resistance of a DC circuit is equal to the total applied voltage divided by the total circuit current.

2

Sum the resistances of the two known series resistors.

R_{\text{known}} = 12\ \Omega + 16\ \Omega = 28\ \Omega

In a series circuit, individual resistances add directly to form the total equivalent resistance.

3

Subtract the sum of the known resistors from the total circuit resistance to find the unknown resistance (

R_3

).

R_3 = R_T - R_{\text{known}} = 40\ \Omega - 28\ \Omega = 12\ \Omega

Because

R_T = R_1 + R_2 + R_3

, the unlabelled resistor's value is the remaining resistance needed to reach

40\ \Omega

.

Key Concept

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

R_T = R_1 + R_2 + \dots + R_n

), and the total current is determined by Ohm's Law (

I = \frac{V_T}{R_T}

).

Estimated Time:1m 0s

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