All practice questions

Question 801Question →

A direct current (DC) emergency lighting circuit operates from a $28\text{ V}$ power supply. An in-line ammeter measures a steady current flow of $350\text{ mA}$ through the circuit. What is the total electrical resistance of the lighting circuit?

0.08\text{ }\Omega

9.8\text{ }\Omega

12.5\text{ }\Omega

80\text{ }\Omega

Show answer & explanation

Answer: $80\text{ }\Omega$

Answer

The total electrical resistance of the lighting circuit is

80\text{ }\Omega

.

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

R = V / I

). Converting

350\text{ mA}

into amperes yields

0.35\text{ A}

. Dividing

28\text{ V}

by

0.35\text{ A}

gives the correct value of

80\text{ }\Omega

.

Step-by-Step Solution

1

Convert the current measurement from milliamperes to amperes.

I = 350\text{ mA} = \frac{350}{1000}\text{ A} = 0.35\text{ A}

Standard Ohm's law calculations require current to be in base SI units (amperes) to produce resistance in ohms.

2

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

).

R = \frac{V}{I}

Resistance equals the electromotive force (voltage) across the circuit divided by the resulting current flow.

3

Substitute the known voltage and converted current values into the rearranged formula.

R = \frac{28\text{ V}}{0.35\text{ A}} = \frac{2800}{35}\text{ }\Omega = 80\text{ }\Omega

Dividing

28\text{ V}

by

0.35\text{ A}

determines the load resistance in ohms.

Key Concept

Ohm's Law (

V = I \times R

) defines the mathematical relationship between voltage, current, and resistance in DC circuits, requiring proper metric unit conversion before calculation.

Estimated Time:1m 0s

Question 802Question →

A technician is configuring a PNP bipolar junction transistor to operate as a switch. To forward-bias the base-emitter junction and turn the transistor ON, what electrical condition must be established at the base terminal relative to the emitter?

The base voltage must be higher (more positive) than the emitter voltage.

The base voltage must be lower (more negative) than the emitter voltage.

The base must receive conventional current flowing into it from the main positive supply rail.

The base voltage must be held at the exact same potential as the collector terminal.

Show answer & explanation

Answer: The base voltage must be lower (more negative) than the emitter voltage.

Answer

The base voltage must be lower (more negative) than the emitter voltage.

For a PNP bipolar junction transistor to turn ON, its base-emitter P-N junction must be forward-biased. Because the emitter is P-type semiconductor material and the base is N-type material, the base terminal must be held at a lower (more negative) electrical potential than the emitter terminal (typically by at least

0.7\text{ V}

in standard silicon transistors).

Step-by-Step Solution

1

Identify the transistor polarity and internal semiconductor structure.

A PNP transistor consists of an N-type base sandwiched between a P-type emitter and a P-type collector.

Understanding layer material types determines the required voltage polarities across the junctions.

2

Determine the biasing requirement for the base-emitter junction to allow conduction.

To forward-bias a P-N junction (P-type emitter to N-type base), the P-type region must be more positive than the N-type region.

Current can only flow across a forward-biased junction when the appropriate potential difference overcomes the barrier voltage (typically

0.7\text{ V}

for silicon).

3

Formulate the necessary base terminal condition relative to the emitter.

The base voltage must be pulled lower (more negative) relative to the emitter voltage, allowing conventional current to flow out of the base.

This establishes forward bias at the base-emitter junction, switching the PNP transistor ON.

Key Concept

PNP Transistor Biasing and Conduction Principles

Estimated Time:1m 0s

Question 803Question →

A surveillance camera heating assembly is powered by a $120\text{ V}$ DC power supply connected in series with three heating elements: $R_1 = 18\ \Omega$ , $R_2 = 22\ \Omega$ , and $R_3 = 40\ \Omega$ . What is the voltage drop across the $22\ \Omega$ element ( $R_2$ )?

120\text{ V}

33\text{ V}

40\text{ V}

1.5\text{ V}

Show answer & explanation

Answer: $33\text{ V}$

Answer

33\text{ V}

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

18\ \Omega + 22\ \Omega + 40\ \Omega = 80\ \Omega

). Applying Ohm's Law with a

120\text{ V}

supply gives a uniform loop current of

1.5\text{ A}

(

120\text{ V} / 80\ \Omega

). Multiplying this current by the resistance of the specified component yields

1.5\text{ A} \times 22\ \Omega = 33\text{ V}

.

Step-by-Step Solution

1

Calculate the total series resistance (

R_T

) by summing individual resistances.

R_T = R_1 + R_2 + R_3 = 18\ \Omega + 22\ \Omega + 40\ \Omega = 80\ \Omega

In a series circuit, total resistance equals the sum of all individual resistances in the loop.

2

Determine the total circuit current (

I_T

) using Ohm's Law (

I = V / R

).

I_T = \frac{V_T}{R_T} = \frac{120\text{ V}}{80\ \Omega} = 1.5\text{ A}

Current remains constant through every component in a single continuous series loop.

3

Calculate the voltage drop (

V_2

) across the target resistor (

R_2 = 22\ \Omega

) using

V = I \times R

.

V_2 = I_T \times R_2 = 1.5\text{ A} \times 22\ \Omega = 33\text{ V}

The voltage drop across any individual series resistor is proportional to its resistance and the loop current.

Key Concept

Series Circuit Resistance and Voltage Division

Estimated Time:1m 0s

Question 804Question →

An electronics technician is organizing replacement components for various power supply and control circuits. Pair each protective device with its designated operational response or circuit function. What is the correct match for each component?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fast-acting fuse matches with melting an internal sacrificial element almost instantaneously during overcurrent; Slow-blow fuse matches with tolerating momentary startup inrush currents while melting under sustained overload; Metal-oxide varistor (MOV) matches with clamping transient high-voltage spikes by rapidly dropping resistance; Bimetallic thermal circuit breaker matches with using heat-induced mechanical deflection of bonded metals to trip a reusable switch.

Each protective component operates on a specific physical mechanism: fast-acting fuses melt rapidly to safeguard delicate electronics; slow-blow fuses incorporate intentional thermal delays to permit harmless inrush currents; MOVs clamp high-voltage spikes by conducting heavily above their rated clamping voltage; and thermal circuit breakers employ bimetallic strip bending to offer resettable overcurrent interruption.

Step-by-Step Solution

1

Identify the primary fault type (overcurrent vs. overvoltage) addressed by each protective component.

Fast-acting fuses, slow-blow fuses, and thermal circuit breakers protect against excessive current, whereas metal-oxide varistors protect against transient high-voltage surges.

Categorizing devices by their primary protection role prevents confusing overvoltage suppressors with overcurrent interrupters.

2

Distinguish between the response speeds and mechanisms of the overcurrent protective devices.

Fast-acting fuses blow instantly; slow-blow fuses accommodate brief inrush surges; thermal breakers provide resettable mechanical tripping via bimetallic heating.

Different circuit loads, such as inductive motors versus sensitive solid-state electronics, require tailored opening times and reset capabilities.

3

Correlate each device with its matching functional description.

All four protective devices are correctly paired with their physical operating characteristics.

Accurate pairing verifies mastery of basic electronic protection principles.

Key Concept

Operational characteristics, response times, and functions of circuit protective devices

Estimated Time:1m 0s

Question 805Question →

A remote weather station is powered by a battery bank containing two identical parallel branches. Each branch is composed of three $12\text{ V}$ batteries connected in series aiding. What total electromotive force (EMF) is delivered across the main output terminals of the bank?

18\text{ V}

72\text{ V}

36\text{ V}

12\text{ V}

Show answer & explanation

Answer: $36\text{ V}$

Answer

36\text{ V}

In a battery bank containing series strings wired in parallel, each series branch sums the voltages of its three

12\text{ V}

batteries to produce

36\text{ V}

(

12\text{ V} + 12\text{ V} + 12\text{ V} = 36\text{ V}

). Connecting two identical

36\text{ V}

strings in parallel keeps the potential difference across the output terminals at

36\text{ V}

while doubling current capacity.

Step-by-Step Solution

1

Calculate the total electromotive force of a single series-connected branch.

12\text{ V} + 12\text{ V} + 12\text{ V} = 36\text{ V}

When DC voltage sources are wired in series aiding, their individual voltages add directly.

2

Determine the combined electromotive force delivered by the parallel configuration.

36\text{ V}

Connecting identical voltage branches in parallel provides the same terminal voltage as a single branch while expanding overall current capacity.

Key Concept

Series-parallel voltage principles dictate that series-connected sources sum their electromotive forces, whereas connecting identical strings in parallel maintains that branch voltage across the load terminals.

Question 806Question →

When a freewheeling (flyback) diode is installed across an inductive DC relay coil to suppress transient voltage spikes, its cathode must be connected to the positive supply rail so that the diode remains reverse-biased during normal energized operation and becomes forward-biased only when the collapsing magnetic field inverts the coil's terminal polarity.

True

False

Show answer & explanation

Answer: True

Answer

True: In a flyback protection configuration, connecting the diode's cathode to the positive supply rail keeps it reverse-biased during steady-state energization, while allowing inductive collapse back-EMF to forward-bias the diode and clamp harmful voltage transients.

The statement is true because connecting the cathode of a freewheeling diode to the positive DC rail holds the diode in reverse bias during normal circuit operation, preventing unwanted conduction. When the control switch opens, the inductor's collapsing magnetic field inverts the terminal voltage polarity (back-EMF), which makes the anode positive relative to the cathode, placing the diode in forward bias to clamp voltage spikes and dissipate stored inductive energy.

Step-by-Step Solution

1

Analyze the biasing condition of the diode during normal steady-state energization of the inductive relay.

With the cathode tied to the positive DC supply rail and the anode connected to the switched low side, the diode is reverse-biased and draws only negligible leakage current, allowing full operating voltage across the coil.

A semiconductor diode conducts only when its anode is positive relative to its cathode; applying a higher potential to the cathode establishes reverse bias.

2

Determine the electrical response of the relay coil when the control switch opens to de-energize the circuit.

In accordance with Lenz's Law, the collapsing magnetic field induces a high-voltage back-EMF across the coil that reverses the terminal polarity, driving the switched terminal significantly positive relative to the supply rail.

An inductor opposes instantaneous changes in current by reversing its induced terminal voltage.

3

Evaluate the diode's conduction state during the induced back-EMF transient.

The reversed polarity places a higher positive voltage at the diode's anode than at its cathode, forward-biasing the diode to create a circulating discharge path that clamps the voltage spike to approximately

0.7\text{ V}

above the supply rail.

Forward conduction through the diode safely dissipates the stored magnetic energy and protects switching components from overvoltage breakdown.

Key Concept

Flyback diode polarity and inductive transient voltage clamping in DC circuits

Estimated Time:1m 30s

Question 807Question →

Electromechanical relays and manual switches use specific contact arrangements to govern how electrical circuits open, close, and divert current under various operating states. Which of the following correctly pairs each switch configuration or relay contact state on the left with its functional definition on the right?

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Normally Open (NO) Relay Contact matches with 'Remains open to block current when de-energized, closing to complete the circuit only when the coil is energized'; Normally Closed (NC) Relay Contact matches with 'Maintains continuity in the unpowered resting state and opens to interrupt current when the relay coil is energized'; Single-Pole, Double-Throw (SPDT) Switch matches with 'Routes current from a single common input terminal to either of two separate output paths depending on the actuator position'; and Double-Pole, Single-Throw (DPST) Switch matches with 'Simultaneously opens or closes two separate, electrically isolated circuits using a single mechanical actuator'.

Each contact type is paired based on standard electrical conventions: Normally Open contacts require coil excitation to establish continuity; Normally Closed contacts conduct by default until coil energization breaks continuity; SPDT switches route one source line between two alternate destinations; and DPST switches control two isolated lines in tandem.

Step-by-Step Solution

1

Analyze the Normally Open (NO) relay contact behavior.

Identified as an open circuit in the resting state that closes upon coil energization.

'Normal' in relay terminology denotes the de-energized or unpowered coil state; thus, an NO contact remains disconnected until magnetic attraction pulls the contacts shut.

2

Analyze the Normally Closed (NC) relay contact behavior.

Identified as a closed circuit in the resting state that opens when the coil is energized.

An NC contact provides continuous circuit conduction until an energized coil generates a magnetic field that pulls the armature away to break continuity.

3

Evaluate the Single-Pole, Double-Throw (SPDT) switch configuration.

Identified as one input pole with two selectable throw positions.

'Single-Pole' means one circuit path is controlled; 'Double-Throw' means the common terminal can connect to either of two output load contacts.

4

Evaluate the Double-Pole, Single-Throw (DPST) switch configuration.

Identified as two electrically isolated circuit poles that simultaneously turn on or off.

'Double-Pole' designates two separate conductive lines controlled by a unified toggle/actuator; 'Single-Throw' indicates simple open/closed switching for both.

Key Concept

Switch pole/throw nomenclature and electromechanical relay contact states

Estimated Time:1m 0s

Question 808Question →

A $36\text{ V}$ direct-current power supply is connected across a parallel circuit containing two branches. If the first branch contains an $18\ \Omega$ resistor and the second branch contains a $9\ \Omega$ resistor, what is the current flowing through the second branch in amperes?

Show answer & explanation

Answer: 4

Answer

4 A

In a parallel circuit, the voltage across all individual branches is identical to the supply voltage (

36\text{ V}

). Using Ohm's law for the second branch, the current is calculated as

I = \frac{V}{R} = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}

.

Step-by-Step Solution

1

Determine the voltage drop across the second branch.

The voltage across the second branch is

36\text{ V}

.

In a parallel circuit, each individual branch experiences the full source voltage (

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

).

2

Calculate the branch current using Ohm's law.

I_2 = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}

.

Current through an individual resistor branch is found by dividing the branch voltage by the branch resistance (

I = \frac{V}{R}

).

Key Concept

In a parallel circuit, the voltage across every parallel branch is equal to the source voltage, and the current through any branch depends solely on that branch's resistance via Ohm's law (

I = \frac{V}{R}

).

Estimated Time:45s

Question 809Question →

An electronics technician tests a DC audio indicator module connected to a $24\text{ V}$ power supply. An in-line meter measures a steady current flow of $80\text{ mA}$ through the component. According to Ohm's law, what is the electrical resistance of the module in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 300

Answer

The total resistance of the module is

300\text{ }\Omega

.

According to Ohm's law (

V = I \times R

), resistance equals voltage divided by current (

R = \frac{V}{I}

). Converting

80\text{ mA}

to standard amperes gives

0.08\text{ A}

. Dividing the

24\text{ V}

source by

0.08\text{ A}

yields a resistance of

300\text{ }\Omega

.

Step-by-Step Solution

1

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

80\text{ mA} = 80 \times 10^{-3}\text{ A} = 0.08\text{ A}

Ohm's law requires current to be expressed in standard SI units (amperes) when working with volts and ohms.

2

Rearrange Ohm's law (

V = I \times R

) to isolate resistance (

R

)

R = \frac{V}{I}

Dividing both sides of the standard Ohm's law equation by current isolates the unknown resistance.

3

Substitute the known voltage (

24\text{ V}

) and current (

0.08\text{ A}

) into the formula

R = \frac{24\text{ V}}{0.08\text{ A}} = 300\text{ }\Omega

Performing the division calculates the exact resistance value in ohms.

Key Concept

Ohm's law resistance calculation with metric prefix conversion (

R = \frac{V}{I}

)

Estimated Time:1m 0s

Question 810Question →

A technician servicing a radio receiver needs to verify a fixed carbon-film resistor with four color bands: Brown, Black, Yellow, and Silver. Calculate the nominal resistance and the minimum allowable resistance under its tolerance rating, and complete the statement below.

Fill in the blanks below

The nominal resistance of the resistor is

\text{k}\Omega

, and its minimum allowable resistance within tolerance is

\text{k}\Omega

.

Show answer & explanation

Answer

The nominal resistance is

100\text{ k}\Omega

and the minimum allowable resistance is

90\text{ k}\Omega

.

According to the electronic color code system, Brown represents 1 and Black represents 0, forming the base number 10. The third band, Yellow, represents a multiplier of

10^4

(

10{}000

), giving a nominal resistance of

10 \times 10{}000 = 100{}000\ \Omega = 100\text{ k}\Omega

. The fourth band, Silver, designates a

\pm 10\%

tolerance. Ten percent of

100\text{ k}\Omega

is

10\text{ k}\Omega

, meaning the minimum allowable resistance under tolerance is

100\text{ k}\Omega - 10\text{ k}\Omega = 90\text{ k}\Omega

(and the maximum is

110\text{ k}\Omega

).

Step-by-Step Solution

1

Decode the significant digits from the first two color bands.

Band 1 (Brown) =

1

, Band 2 (Black) =

0

, giving significant digits

10

.

In standard 4-band EIA color coding, the first two bands represent the first two significant figures of the resistance value.

2

Apply the multiplier from the third color band and convert to kilohms.

Band 3 (Yellow) = multiplier

10^4

(

10{}000

). Nominal resistance

= 10 \times 10{}000 = 100{}000\ \Omega = 100\text{ k}\Omega

.

The third band determines the power of ten by which the significant digits are multiplied.

3

Determine the tolerance percentage from the fourth color band.

Band 4 (Silver) =

\pm 10\%

tolerance.

Silver indicates a manufacturing variation of plus or minus ten percent from nominal value.

4

Calculate the minimum allowable resistance limit.

Tolerance amount

= 100\text{ k}\Omega \times 0.10 = 10\text{ k}\Omega

. Minimum resistance

= 100\text{ k}\Omega - 10\text{ k}\Omega = 90\text{ k}\Omega

.

Subtracting the tolerance deviation from the nominal value establishes the lower acceptable boundary.

Key Concept

Standard 4-band resistor color decoding and tolerance range calculation

Estimated Time:45s

Question 811Question →

When an auxiliary lighting circuit in an electrical system is active, a meter records a current of $1.5\text{ A}$ . What does this electrical current measurement physically represent?

The electromotive force or electrical pressure driving charges through the conductor

The rate at which electric charge flows past a given point in the circuit

The physical opposition that restricts the passage of electrons through the wire

The rapid back-and-forth oscillation of charge carriers between opposite terminals

Show answer & explanation

Answer: The rate at which electric charge flows past a given point in the circuit

Answer

The rate at which electric charge flows past a given point in the circuit

Electric current (

I

) is defined as the rate at which electric charge flows past a cross-sectional point in a conductor. One ampere equals one coulomb of charge passing a point per second (

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

).

Step-by-Step Solution

1

Identify the physical definition of electrical current (

I

)

Current represents the quantity of electrical charge (

Q

) passing through a cross-section of a conductor per unit time (

t

), expressed as

I = \frac{Q}{t}

.

Understanding the base definition of the ampere (

1\text{ A} = 1\text{ Coulomb/second}

) establishes what is being measured.

2

Distinguish current from related electrical properties like voltage and resistance

Voltage is the driving pressure (EMF), resistance is the opposition to flow, while current is the actual movement rate of the charges.

Eliminates common misconceptions regarding electrical circuit parameters.

Key Concept

Electric current is the rate of flow of electric charge over time, quantified in amperes.

Estimated Time:35s

Question 812Question →

An automotive spotlight operates on a $12\text{ V}$ direct current (DC) electrical system and draws a steady current of $4\text{ A}$ . How much electrical power is consumed by the spotlight?

3\text{ W}

48\text{ W}

16\text{ W}

192\text{ W}

Show answer & explanation

Answer: $48\text{ W}$

Answer

The power consumed by the spotlight is

48\text{ W}

.

According to Watt's Law, electrical power (

P

) is determined by multiplying voltage (

V

) by current (

I

), using the formula

P = V \times I

. Substituting

12\text{ V}

and

4\text{ A}

gives

P = 12\text{ V} \times 4\text{ A} = 48\text{ W}

.

Step-by-Step Solution

1

Identify the given electrical values from the problem statement.

Voltage (

V

) =

12\text{ V}

, Current (

I

) =

4\text{ A}

.

Establishing the known quantities determines which form of Watt's Law to apply.

2

Select the appropriate Watt's Law power formula relating power, voltage, and current.

P = V \times I

Electrical power in watts equals the product of potential difference in volts and current in amperes.

3

Substitute the values into the formula and calculate.

P = 12\text{ V} \times 4\text{ A} = 48\text{ W}

Carrying out the multiplication yields the total power dissipated.

Key Concept

Watt's Law relates electrical power (

P

) to voltage (

V

) and current (

I

) through the formula

P = V \times I

.

Estimated Time:45s

Question 813Question →

A technician connects a step-up transformer to a $30\text{ V}$ AC power supply. The transformer has $200\text{ turns}$ on its primary winding and $800\text{ turns}$ on its secondary winding. Assuming ideal operating conditions with no energy losses, what is the output voltage at the secondary winding?

7.5\text{ V}

12\text{ V}

120\text{ V}

0\text{ V}

Show answer & explanation

Answer: $120\text{ V}$

Answer

The secondary output voltage is

120\text{ V}

.

For an ideal transformer, secondary voltage is calculated using the formula

V_s = V_p \times \left(\frac{N_s}{N_p}\right)

. Given a primary voltage of

30\text{ V}

and a turns ratio of

\frac{800}{200} = 4

, the secondary voltage is

30\text{ V} \times 4 = 120\text{ V}

.

Step-by-Step Solution

1

Identify the known values and the transformer voltage-turns relationship formula.

V_p = 30\text{ V}

,

N_p = 200\text{ turns}

,

N_s = 800\text{ turns}

, using

\frac{V_s}{V_p} = \frac{N_s}{N_p}

.

In an ideal transformer, the ratio of the secondary voltage to the primary voltage equals the ratio of secondary turns to primary turns.

2

Calculate the turns ratio (

N_s / N_p

).

\frac{800}{200} = 4

.

Finding the ratio simplifies multiplying the primary voltage to determine secondary voltage.

3

Multiply the primary input voltage by the turns ratio to find the secondary voltage (

V_s

).

V_s = 30\text{ V} \times 4 = 120\text{ V}

.

Since the secondary winding has four times as many turns as the primary, the secondary voltage must be four times greater than the primary voltage.

Key Concept

Transformer Turns and Voltage Ratio

Estimated Time:45s

Question 814Question →

An electrical technician is evaluating four substances for an electronics assembly based on how easily charge carriers move through their atomic lattices. What is the correct order of these materials when arranged from highest electrical conductivity (greatest electron flow) to lowest electrical conductivity (highest resistance to current flow) at room temperature ( $25^\circ\text{C}$ )?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Silver, followed by Intrinsic Germanium, Intrinsic Silicon, and finally Mica.

Electrical conductivity is determined by the density of free charge carriers available in a material's conduction band. Silver is an excellent metallic conductor with free electrons that readily drift under an applied electric field. Germanium and silicon are Group IV semiconductors; because germanium has a narrower forbidden bandgap (

0.67\text{ eV}

) than silicon (

1.12\text{ eV}

), more electron-hole pairs are thermally excited across the gap at room temperature, making germanium more conductive than silicon. Mica is an inorganic insulator with tightly held valence electrons and a very wide forbidden energy gap, giving it the lowest electrical conductivity in the sequence.

Step-by-Step Solution

1

Identify the metallic conductor

Silver is identified as the conductor with the highest concentration of free valence electrons.

Metals have overlapping valence and conduction bands, providing high electrical conductivity.

2

Compare the semiconductor bandgaps

Intrinsic Germanium (

E_g \approx 0.67\text{ eV}

) has higher conductivity than Intrinsic Silicon (

E_g \approx 1.12\text{ eV}

).

A smaller bandgap allows more valence electrons to bridge into the conduction band at room temperature.

3

Identify the electrical insulator

Mica is placed last with the lowest electrical conductivity.

Insulators have full valence bands separated from conduction bands by large forbidden energy gaps.

Key Concept

Ranking material electrical conductivity based on atomic structure, valence electron availability, and energy bandgap size across conductors, semiconductors, and insulators.

Estimated Time:1m 30s

Question 815Question →

A standard glass cartridge fuse is rated for a continuous operating current of $3\text{ A}$ . What occurs within this protective device if an electrical short circuit causes a sustained current of $10\text{ A}$ to flow through it?

The internal element temporarily flexes open and automatically snaps closed once the overload clears.

The internal metal element overheats and melts, creating an open circuit that permanently stops current flow.

The internal element increases its electrical resistance to convert the extra current into additional supply voltage.

The device alters its polarity to force excess electron flow back toward the power source.

Show answer & explanation

Answer: The internal metal element overheats and melts, creating an open circuit that permanently stops current flow.

Answer

The internal metal element overheats and melts, creating an open circuit that permanently stops current flow.

A fuse is a one-time sacrificial protective device wired in series with a load. When the circuit current (

10\text{ A}

) exceeds the rated threshold (

3\text{ A}

), resistive heating (

I^2R

) rapidly raises the temperature of the internal metal link past its melting point. The link separates, creating an open circuit (infinite resistance) that permanently halts current flow to prevent wiring damage or fire.

Step-by-Step Solution

1

Identify the type and rating of the protective device.

The device is a standard cartridge fuse rated for

3\text{ A}

.

Determining the device type establishes whether it is a one-time sacrificial device or a resettable mechanism.

2

Compare the actual fault current to the device's rated current.

The fault current of

10\text{ A}

significantly exceeds the

3\text{ A}

rating.

Current passing through the fuse element generates heat proportional to

I^2R

(

P = I^2R

).

3

Determine the physical response of the fuse element under this overcurrent condition.

The excessive current heats the thin metal wire or strip beyond its melting point, burning it open.

Melting breaks electrical continuity, producing an open circuit with infinite resistance that stops all downstream current.

Key Concept

Operating principle of a standard overcurrent fuse

Estimated Time:45s

Question 816Question →

An industrial controller module requires a total electromotive force (EMF) of $0.12\text{ kV}$ for its backup power rail. If standard $1.5\text{ V}$ alkaline cells are connected in series-aiding to supply the rail, how many cells are required?

Show answer & explanation

Answer: 80

Answer

80 cells

The total requirement is

0.12\text{ kV}

, which equals

120\text{ V}

. Since series-aiding cells add their electromotive forces directly, dividing the total required voltage of

120\text{ V}

by

1.5\text{ V}

per cell gives exactly

80

cells.

Step-by-Step Solution

1

Convert the required electromotive force from kilovolts (kV) to standard volts (V)

0.12\text{ kV} \times 1000 = 120\text{ V}

The prefix kilo- denotes a multiplier of

1000

, so the potential difference must be converted to base units to match individual cell ratings.

2

Calculate the number of series cells needed by dividing total EMF by the EMF of a single cell

n = \frac{120\text{ V}}{1.5\text{ V/cell}} = 80\text{ cells}

When electrical cells are connected in series-aiding configuration, their individual voltages add directly (

V_{\text{total}} = n \times V_{\text{cell}}

).

Key Concept

Series electromotive force summation and metric prefix voltage conversion

Estimated Time:1m 0s

Question 817Question →

An aircraft cockpit lighting circuit is connected to a $24\text{ V}$ direct current (DC) power source. The circuit consists of three loads connected in series: a $15\ \Omega$ panel lamp, a $25\ \Omega$ dimming rheostat, and a $20\ \Omega$ ballast resistor. What is the voltage drop across the $25\ \Omega$ dimming rheostat in volts?

Show answer & explanation

Answer: 10

Answer

The voltage drop across the

25\ \Omega

dimming rheostat is

10\text{ V}

.

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

15\ \Omega + 25\ \Omega + 20\ \Omega = 60\ \Omega

. Applying Ohm's Law to the entire circuit gives a current of

24\text{ V} / 60\ \Omega = 0.4\text{ A}

. Because current is constant throughout a series circuit, the voltage drop across the

25\ \Omega

rheostat is

0.4\text{ A} \times 25\ \Omega = 10\text{ V}

.

Step-by-Step Solution

1

Calculate the total series resistance (

R_T

) of the circuit.

R_T = 15\ \Omega + 25\ \Omega + 20\ \Omega = 60\ \Omega

In a series circuit, total resistance equals the algebraic sum of all individual series resistances.

2

Calculate the common series current (

I

) flowing through the circuit.

I = \frac{V_T}{R_T} = \frac{24\text{ V}}{60\ \Omega} = 0.4\text{ A}

Ohm's Law (

I = \frac{V}{R}

) relates the total supply voltage to the total circuit resistance.

3

Calculate the voltage drop (

V_2

) across the

25\ \Omega

resistor.

V_2 = I \times R_2 = 0.4\text{ A} \times 25\ \Omega = 10\text{ V}

Current is constant throughout all components in a series circuit, so the individual voltage drop is the product of loop current and component resistance.

Key Concept

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

R_T = R_1 + R_2 + \dots + R_n

), current is identical through all components (

I_T = I_1 = I_2 = \dots = I_n

), and individual voltage drops are directly proportional to component resistance (

V_n = I \times R_n

).

Estimated Time:1m 0s

Question 818Question →

Two resistors with values of $12\ \Omega$ and $6\ \Omega$ are wired in parallel across a circuit board. What is the total equivalent resistance of this combination?

18\ \Omega

4\ \Omega

9\ \Omega

72\ \Omega

Show answer & explanation

Answer: $4\ \Omega$

Answer

4\ \Omega

For two resistors in parallel, the equivalent resistance is calculated by dividing their product by their sum:

\frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega

. In parallel circuits, the equivalent resistance is always lower than any single branch resistance.

Step-by-Step Solution

1

Identify the given branch resistances and select the parallel equivalent resistance formula.

R_1 = 12\ \Omega

,

R_2 = 6\ \Omega

, and

R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}

The product-over-sum formula determines the total equivalent resistance of two parallel branches.

2

Calculate the product and sum of the two resistors, then divide.

R_{eq} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega

Completing the arithmetic yields the equivalent resistance, which must be less than the smallest individual branch.

Key Concept

Total equivalent resistance in a parallel circuit is always less than the lowest individual branch resistance and can be calculated using the reciprocal or product-over-sum formula.

Estimated Time:45s

Question 819Question →

An electronics technician is testing an optical pulse circuit and measures that a capacitor stores an electrostatic charge of $450\text{ }\mu\text{C}$ when connected across a potential difference of $15\text{ V}$ . What is the capacitance of this component?

6{}750\text{ }\mu\text{F}

30\text{ }\mu\text{H}

30\text{ }\mu\text{F}

0.03\text{ }\mu\text{F}

Show answer & explanation

Answer: $30\text{ }\mu\text{F}$

Answer

30\text{ }\mu\text{F}

Capacitance is defined by the fundamental relationship

C = \frac{Q}{V}

. Dividing the stored charge of

450\text{ }\mu\text{C}

by the applied potential difference of

15\text{ V}

yields

30\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Identify the relationship between charge, capacitance, and voltage

The governing equation is

Q = C \times V

, which rearranges to

C = \frac{Q}{V}

.

Capacitance is defined as the amount of electric charge stored per unit of electrical potential difference.

2

Substitute the given values into the formula and solve for capacitance

C = \frac{450\text{ }\mu\text{C}}{15\text{ V}} = 30\text{ }\mu\text{F}

Dividing the charge in microcoulombs by the potential difference in volts directly yields the capacitance in microfarads.

Key Concept

The capacitance (

C

) of a component is the ratio of electric charge (

Q

) stored on its plates to the applied voltage (

V

), measured in Farads (F).

Estimated Time:1m 0s

Question 820Question →

An aircraft avionics sensor circuit is powered by a $28\text{ V}$ direct current (DC) power source connected in series with three resistors: $R_1 = 30\ \Omega$ , $R_2 = 45\ \Omega$ , and an unknown resistor $R_3$ . A technician measures the voltage drop across $R_2$ and finds it to be $9\text{ V}$ . Based on these circuit measurements, what are the total loop current and the resistance value of $R_3$ ?

Fill in the blanks below

The total current flowing through the circuit is A, and the resistance value of

R_3

is

\Omega

.

Show answer & explanation

Answer

The circuit current is 0.2 A and the resistance of resistor R3 is 65 ohms.

Because the circuit components are connected in series, the current flowing through

R_2

(

I = 9\text{ V} / 45\ \Omega = 0.2\text{ A}

) is the exact same current passing through the entire loop. Applying Ohm's Law to the full circuit gives a total resistance of

R_T = 28\text{ V} / 0.2\text{ A} = 140\ \Omega

. Subtracting the known resistances (

30\ \Omega + 45\ \Omega = 75\ \Omega

) leaves

65\ \Omega

for

R_3

.

Step-by-Step Solution

1

Calculate the constant loop current using the known values for resistor

R_2

I = \frac{V_2}{R_2} = \frac{9\text{ V}}{45\ \Omega} = 0.2\text{ A}

In a series circuit, current is identical through every component.

2

Calculate the total equivalent resistance of the entire series circuit using Ohm's Law and the source voltage

R_T = \frac{V_{\text{source}}}{I} = \frac{28\text{ V}}{0.2\text{ A}} = 140\ \Omega

Total circuit resistance equals the total applied voltage divided by the common series current.

3

Determine the resistance of

R_3

by subtracting known series resistances from total resistance

R_3 = R_T - (R_1 + R_2) = 140\ \Omega - (30\ \Omega + 45\ \Omega) = 140\ \Omega - 75\ \Omega = 65\ \Omega

Total series resistance is the direct algebraic sum of all individual resistances in the loop.

Key Concept

In a single-loop series circuit, current is constant at all points (

I_T = I_1 = I_2 = I_3

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

R_T = R_1 + R_2 + R_3

).

Estimated Time:1m 15s

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