All practice questions

Question 641Question →

In direct-current (DC) power distribution, parallel circuit configurations govern how voltage, current, and resistance interact under various operating states. Match each parallel circuit modification on the left with its corresponding electrical outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding an additional resistive branch matches with decreasing total equivalent resistance and increasing total source current; an open-circuit fault matches with decreasing total circuit current while remaining branches operate normally; a short-circuit fault matches with dropping total resistance near zero and triggering circuit protection; and increasing main supply voltage matches with increasing the voltage drop and current across all branches simultaneously.

In parallel networks, the total equivalent resistance decreases when branches are added and increases when branches are removed. Full line voltage is maintained across every branch independently, meaning faults in one branch (such as an open) do not disrupt the operation of other branches, whereas a direct short bridges the common voltage bus and triggers circuit protection.

Step-by-Step Solution

1

Analyze the impact of adding a parallel branch

Additional current paths reduce equivalent resistance and raise total current without altering existing branch conditions.

Parallel branches operate independently across a constant voltage source.

2

Analyze an open circuit in a parallel branch

Only the opened branch stops conducting current; other branches retain full voltage and normal current.

An open in one parallel leg does not break the continuity of the remaining parallel paths.

3

Analyze a short circuit across a parallel network

A zero-ohm bridge shorts out all parallel loads and draws extreme current.

Current follows the path of least resistance, bypassing the loads entirely.

4

Analyze a change in primary DC source voltage

Every branch experiences the higher voltage and draws proportionally more current via Ohm's law (

I = V/R

).

The full source voltage is applied directly across every individual branch in a parallel circuit.

Key Concept

Parallel Circuit Governing Rules and Fault Characteristics

Estimated Time:1m 15s

Question 642Question →

A technician needs to replace the coil spring on a MacPherson strut assembly that has been removed from a vehicle. What is the correct chronological order of steps required to safely disassemble the strut assembly?

Drag items to arrange them in the correct order

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Answer

The correct sequence is: first secure the strut assembly in a bench fixture, next compress the coil spring with an approved compressor tool, then remove the damper shaft center retaining nut, and finally remove the upper strut mount and slide the spring off the strut body.

The correct procedure starts by securing the strut in a bench fixture. Next, the coil spring must be compressed with a spring compressor to relieve the preload acting against the upper spring seat. With spring tension relieved, the technician safely removes the center shaft retaining nut, and then removes the upper mount to slide the coil spring off the damper.

Step-by-Step Solution

1

Secure the strut assembly in a holding fixture or vise.

The strut assembly is stabilized for mechanical service.

Prevents the component from shifting or slipping during high-torque operations.

2

Compress the coil spring with a spring compressor.

Spring preload on the upper strut mount is fully relieved.

Crucial safety step to prevent stored spring energy from violently releasing when fasteners are removed.

3

Remove the center shaft retaining nut.

The upper mounting hardware is mechanically released from the damper rod.

The retaining nut holds the upper plate and damper shaft together.

4

Remove the upper strut mount and remove the spring.

The strut assembly is fully disassembled into its primary components.

Allows replacement of the worn or broken coil spring.

Key Concept

MacPherson strut spring disassembly safety and procedure

Estimated Time:1m 0s

Question 643Question →

A technician is testing an unlabeled three-terminal semiconductor component using a digital multimeter in diode-check mode. The meter indicates a forward voltage drop of approximately $0.7\text{ V}$ when the positive (red) meter lead is placed on Terminal 1 and the negative (black) lead is placed on either Terminal 2 or Terminal 3. All reverse probe combinations, as well as direct measurements between Terminal 2 and Terminal 3, show an open-circuit (over-limit) reading. Which device type and lead configuration does this component represent?

PNP bipolar junction transistor with Terminal 1 as the base

NPN bipolar junction transistor with Terminal 1 as the base

NPN bipolar junction transistor with Terminal 1 as the collector

PNP bipolar junction transistor with Terminal 1 as the emitter

Show answer & explanation

Answer: NPN bipolar junction transistor with Terminal 1 as the base

Answer

An NPN bipolar junction transistor with Terminal 1 as the base.

In a bipolar junction transistor (BJT), the base-emitter and base-collector regions behave as two separate PN junction diodes sharing a common base material. For an NPN transistor, the base is composed of P-type semiconductor material, while the collector and emitter are N-type. Applying a positive test lead to the P-type base (Terminal 1) and a negative lead to either N-type terminal (Terminals 2 and 3) forward-biases each junction, yielding the standard silicon forward drop of approximately

0.7\text{ V}

.

Step-by-Step Solution

1

Analyze the multimeter diode test behavior

Conduction occurs only when Terminal 1 is connected to the positive meter lead and Terminals 2 or 3 are connected to the negative lead.

A digital multimeter diode test applies a small forward voltage from the red (+) probe to the black (-) probe to measure the junction threshold drop.

2

Identify the semiconductor layer structure

Terminal 1 acts as a common P-type anode, while Terminals 2 and 3 act as separate N-type cathodes.

Silicon PN junctions drop approximately

0.7\text{ V}

when forward-biased (anode positive relative to cathode).

3

Determine the transistor type and lead identity

The component is an NPN transistor where Terminal 1 is the Base (P-region), Terminal 2 is the Collector/Emitter (N-region), and Terminal 3 is the Emitter/Collector (N-region).

An NPN transistor electrically models as two back-to-back diodes with anodes joined at the base.

Key Concept

BJT internal PN junction polarity and diode-test verification

Estimated Time:1m 15s

Question 644Question →

An electronics technician is analyzing how electromotive force (EMF) and terminal voltage behave across various operational states of a DC power source. Match each electrical condition on the left with its correct analytical relationship or physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Open-circuit cell voltage matches the full internal electromotive force without internal drop; loaded cell terminal voltage matches the electromotive force minus internal resistance drop; external battery charging voltage matches the electromotive force plus internal resistance drop; and the fundamental definition of the volt matches one joule of energy per coulomb of charge.

Electromotive force (EMF,

\mathcal{E}

) is the total energy per unit charge supplied by a source. Under no-load conditions (

I = 0

), the terminal voltage equals EMF. When delivering current (

I > 0

) to a load, internal resistance produces a drop (

I r

), lowering terminal voltage to

\mathcal{E} - I r

. During charging, the external supply must exceed EMF by

\mathcal{E} + I r

to force current inward. Fundamentally, one volt is defined as one joule of energy per coulomb of charge (

1\text{ V} = 1\text{ J/C}

).

Step-by-Step Solution

1

Analyze open-circuit state for a DC voltage source

Terminal voltage equals internal electromotive force (

V = \mathcal{E}

)

When no current flows (

I = 0\text{ A}

), no voltage is dropped across internal resistance (

I r = 0\text{ V}

).

2

Evaluate terminal voltage during active discharge under load

Terminal voltage drops to

V = \mathcal{E} - I r

Current passing through the battery's internal resistance consumes a portion of the source EMF.

3

Determine required external voltage for charging a cell

Applied charging voltage must be

V = \mathcal{E} + I r

The charger must provide higher potential than the cell's EMF to drive current in reverse against internal resistance.

4

State the fundamental physical definition of potential difference and electromotive force

1\text{ V} = 1\text{ J/C}

Voltage is defined as electrical potential energy (work in joules) per unit charge (coulombs).

Key Concept

Electromotive Force (EMF) vs. Terminal Voltage and Internal Resistance Relations

Estimated Time:1m 30s

Question 645Question →

An alternating current (AC) sinusoidal voltage with a peak value of $10\text{ V}$ delivers the same electrical power to a resistive heating element as a steady $10\text{ V}$ direct current (DC) supply.

True

False

Show answer & explanation

Answer: False

Answer

False. An alternating current (AC) sine wave with a peak voltage of

10\text{ V}

delivers an effective (RMS) voltage of approximately

7.07\text{ V}

, which provides less electrical power to a resistor than a constant

10\text{ V}

direct current (DC) supply.

Direct current (DC) maintains a continuous, constant voltage and unidirectional current flow over time. In contrast, sinusoidal alternating current (AC) fluctuates, reaching its maximum peak voltage only momentarily and crossing zero volts twice per cycle. The effective value of an AC voltage—representing the DC voltage that would deliver the exact same average heating power across an identical resistor—is defined by its Root-Mean-Square (RMS) value (

V_{\text{RMS}} = 0.707 \times V_{\text{peak}}

). For a

10\text{ V}

peak AC supply, the effective voltage is only

7.07\text{ V}

, resulting in half the power output of a steady

10\text{ V}

DC supply.

Step-by-Step Solution

1

Identify the relationship between peak voltage and effective (RMS) voltage for alternating current.

V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} \approx 0.707 \times V_{\text{peak}} = 0.707 \times 10\text{ V} = 7.07\text{ V}

.

Direct current supplies a continuous, constant potential, whereas alternating current oscillates, meaning the effective power output of AC must be calculated using its RMS value.

2

Compare the power dissipated by the

10\text{ V}

DC source and the

10\text{ V}

peak AC source across a resistor

R

.

P_{\text{DC}} = \frac{(10\text{ V})^2}{R} = \frac{100}{R}

, whereas

P_{\text{AC}} = \frac{(V_{\text{RMS}})^2}{R} = \frac{(7.07\text{ V})^2}{R} = \frac{50}{R}

.

Power delivered to a resistor is proportional to the square of the applied voltage; the

10\text{ V}

peak AC source delivers only

50\%

of the power delivered by the

10\text{ V}

DC source.

3

Evaluate the truth value of the stem statement.

The statement is false.

Equal peak and DC voltage values do not result in equal power output due to the cyclic nature and lower effective RMS value of AC.

Key Concept

AC RMS Voltage vs. DC Equivalent Heating Power

Estimated Time:35s

Question 646Question →

An electronics technician measures a potential difference of $3.5\text{ kV}$ across the secondary winding of an ignition transformer. What is this electromotive force expressed in volts ( $\text{V}$ )?

Show answer & explanation

Answer: 3500

Answer

3,500 V

The prefix 'kilo-' (abbreviated as 'k') indicates a factor of

1{}000

. Therefore, an electromotive force of

3.5\text{ kV}

is equivalent to

3.5 \times 1{}000 = 3{}500\text{ V}

.

Step-by-Step Solution

1

Identify the standard metric prefix multiplier for kilo (k)

The prefix kilo represents a multiplier of

10^3

, or

1{}000

.

Electromotive force expressed with standard engineering prefixes must be converted using the appropriate power of ten.

2

Multiply the numerical value in kilovolts by 1,000

3.5 \times 1{}000 = 3{}500\text{ V}

To convert from a larger prefixed unit (kilovolts) to the base electrical unit (volts), multiply by

1{}000

.

Key Concept

Metric prefix conversion of electrical potential difference and electromotive force (EMF)

Estimated Time:30s

Question 647Question →

An avionics technician is testing a dual-polarity DC power supply that provides $+12\text{ V}$ on the positive rail and $-12\text{ V}$ on the negative rail, both measured relative to chassis ground. What is the total voltage (potential difference) across a component connected directly between the positive rail and the negative rail?

0\text{ V}

12\text{ V}

24\text{ V}

24\text{ mV}

Show answer & explanation

Answer: $24\text{ V}$

Answer

24\text{ V}

Voltage (electromotive force) is fundamentally a measure of electric potential difference between two specific points. By subtracting the negative rail potential (

-12\text{ V}

) from the positive rail potential (

+12\text{ V}

), the resulting potential difference is

(+12\text{ V}) - (-12\text{ V}) = 24\text{ V}

.

Step-by-Step Solution

1

Identify the reference potentials of both supply rails relative to common ground

V_{\text{positive}} = +12\text{ V}

and

V_{\text{negative}} = -12\text{ V}

Voltage at any point in a circuit is defined with respect to a designated reference (chassis ground).

2

Calculate the potential difference across the load connected between the two rails

V_{\text{load}} = V_{\text{positive}} - V_{\text{negative}} = 12\text{ V} - (-12\text{ V}) = 24\text{ V}

Electromotive force / voltage across two terminals is the algebraic difference between their electric potentials.

Key Concept

Voltage as Electric Potential Difference

Estimated Time:45s

Question 648Question →

In an alternating current (AC) circuit, electrical current periodically reverses its direction of flow at a regular rate measured in hertz ( $Hz$ ).

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because alternating current (AC) is defined by its periodic reversal of direction and polarity, with the number of complete cycles per second measured in hertz (

Hz

).

Step-by-Step Solution

1

Identify the core definition of alternating current (AC).

Alternating current is defined as current that periodically reverses its direction of flow over time.

This bidirectional oscillation distinguishes AC from unidirectional direct current (DC).

2

Determine the unit of measurement for the rate of these reversals.

The number of complete forward-and-backward cycles per second is frequency, measured in hertz (

Hz

).

One hertz equals one complete cycle per second.

3

Evaluate the truth value of the stem statement.

The statement is true because it correctly describes both the periodic reversal and the unit of frequency for AC.

All facts in the statement align with standard electrical definitions.

Key Concept

Alternating Current (AC) Directional Reversal and Frequency

Question 649Question →

On a standard electronic schematic diagram, the arrow on the emitter terminal of an NPN bipolar junction transistor points outward, away from the base.

True

False

Show answer & explanation

Answer: True

Answer

True: The emitter arrow on an NPN transistor schematic symbol points outward, away from the base.

In standard schematic diagrams, all bipolar junction transistors place a directional arrow on the emitter lead. For an NPN transistor, this arrow points outward, away from the base terminal, representing conventional current leaving the device. A widely used technician mnemonic is 'NPN = Not Pointing iN'.

Step-by-Step Solution

1

Identify the component type and schematic lead conventions.

The component is an NPN bipolar junction transistor (BJT) consisting of Base, Collector, and Emitter leads.

Determining the transistor family and terminal configuration is required to verify the symbol markings.

2

Determine the arrow placement and orientation for the NPN configuration.

The arrow is always positioned on the emitter lead and points outward (away from the base line).

Schematic symbols use the emitter arrow to indicate the direction of conventional current flow, which flows out of the emitter in an NPN transistor (commonly recalled with the mnemonic 'Not Pointing iN').

3

Evaluate the accuracy of the provided statement.

The statement matches standard schematic conventions exactly.

An outward-pointing emitter arrow uniquely defines the NPN transistor symbol.

Key Concept

BJT Schematic Symbols and Emitter Arrow Orientation

Estimated Time:40s

Question 650Question →

An uncharged capacitor is connected in series with a resistor and a DC voltage source through an open switch. When the switch is suddenly closed, how do the voltage drop across the capacitor and the circuit current behave from the initial instant ( $t = 0$ ) to steady-state condition ( $t \ge 5\tau$ )?

Initial capacitor voltage equals the source voltage with zero circuit current; steady-state capacitor voltage is zero with maximum circuit current.

Initial and steady-state capacitor voltage both remain at zero while continuous DC current flows indefinitely through the circuit.

Initial capacitor voltage is zero with maximum circuit current; steady-state capacitor voltage equals the source voltage with zero circuit current.

Initial capacitor voltage equals the source voltage with maximum current; steady-state voltage and current both drop to zero.

Show answer & explanation

Answer: Initial capacitor voltage is zero with maximum circuit current; steady-state capacitor voltage equals the source voltage with zero circuit current.

Answer

At the initial instant (

t = 0

), the uncharged capacitor has zero voltage drop and allows maximum circuit current, while at steady state (

t \ge 5\tau

), the capacitor voltage reaches the full source voltage and circuit current drops to zero.

Because a capacitor stores energy in an electrostatic field between its plates, its voltage cannot change instantaneously. When connected to a DC source at

t = 0

, an uncharged capacitor exhibits zero potential difference (

0\text{ V}

) across its plates and behaves momentarily like a short circuit, allowing peak current (

I = V_{\text{source}} / R

) to flow through the series circuit. As charge accumulates over five time constants (

5\tau

), the potential difference across the capacitor rises to match the source voltage (

V_C = V_{\text{source}}

), completely balancing the applied EMF. At this steady state, the capacitor acts as an open circuit, reducing direct current flow to zero.

Step-by-Step Solution

1

Analyze the electrical state of an uncharged capacitor at the exact instant the DC switch is closed (

t = 0

).

Because charge cannot accumulate instantaneously (

Q = C \times V

), the initial capacitor voltage is

V_C(0) = 0\text{ V}

. The capacitor acts momentarily as a short circuit, allowing maximum current determined entirely by the series resistor:

I_{\text{max}} = \frac{V_{\text{source}}}{R}

.

Capacitors oppose instantaneous changes in voltage.

2

Analyze the transient charging phase over time (

0 < t < 5\tau

).

As electrons accumulate on the negative plate and depart the positive plate, an opposing electric field develops across the dielectric. The voltage

V_C(t)

increases exponentially according to

V_C(t) = V_{\text{source}}(1 - e^{-t/RC})

, while current decreases exponentially:

I(t) = I_{\text{max}} e^{-t/RC}

.

The accumulating charge creates a counter-voltage that reduces the net potential driving current through the series resistor.

3

Determine the steady-state circuit values after five time constants (

t \ge 5\tau

).

The capacitor reaches approximately

99.3\%

of full charge. The voltage across the capacitor equals the source voltage (

V_C = V_{\text{source}}

), and the net voltage across the resistor becomes

0\text{ V}

, causing circuit current to cease entirely (

I = 0\text{ A}

). The fully charged capacitor acts as an open circuit to DC.

Once the opposing electrostatic potential equals the applied DC source potential, no further charge transfer can occur.

Key Concept

DC Transient and Steady-State Behavior of Capacitors in RC Circuits

Estimated Time:1m 30s

Question 651Question →

In standard electronic schematic symbols for bipolar junction transistors (BJTs), the directional arrow indicating current flow is positioned on the emitter terminal.

True

False

Show answer & explanation

Answer: True

Answer

True. The directional arrow in a BJT schematic symbol is always located on the emitter terminal.

The statement is true because electronic drafting standards universally position the current direction arrow on the emitter terminal of a BJT. This arrow serves two primary purposes: it distinguishes the emitter lead from the collector lead, and its direction (pointing out for NPN, pointing in for PNP) identifies the polarity type of the transistor.

Step-by-Step Solution

1

Identify the standard terminals of a bipolar junction transistor (BJT).

A BJT has three terminals: base, collector, and emitter.

Recognizing the physical and schematic lead names is necessary to evaluate the statement.

2

Recall the schematic drawing conventions used to represent BJTs.

The base is drawn as a vertical bar, the collector is an un-arrowed diagonal branch, and the emitter is a diagonal branch with an arrow.

Standard electronics symbology uses the arrow to indicate the direction of conventional current flow through the device.

3

Verify the terminal location of the identifying arrow.

The arrow is placed exclusively on the emitter terminal for both NPN and PNP configurations.

The arrow differentiates the emitter from the collector and establishes the transistor type (pointing outward for NPN, inward for PNP).

Key Concept

BJT Schematic Symbol Conventions

Estimated Time:45s

Question 652Question →

A pulse-generator circuit contains two $50\text{ }\mu\text{F}$ capacitors connected in parallel. This parallel pair is then connected in series with a third capacitor rated at $25\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the circuit in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent capacitance of the circuit is

20\text{ }\mu\text{F}

.

The correct answer is

20\text{ }\mu\text{F}

. When two

50\text{ }\mu\text{F}

capacitors are wired in parallel, their total capacitance is

50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

. When this

100\text{ }\mu\text{F}

equivalent capacitance is connected in series with the

25\text{ }\mu\text{F}

capacitor, the product-over-sum formula yields

\frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two parallel capacitors (

C_1

and

C_2

).

C_p = 50\text{ }\mu\text{F} + 50\text{ }\mu\text{F} = 100\text{ }\mu\text{F}

Capacitors in parallel add their plate surface areas directly, so their individual capacitance values are summed (

C_p = C_1 + C_2

).

2

Calculate the total series capacitance combining the

100\text{ }\mu\text{F}

parallel equivalent with the

25\text{ }\mu\text{F}

third capacitor (

C_3

).

C_{\text{total}} = \frac{100 \times 25}{100 + 25} = \frac{2500}{125} = 20\text{ }\mu\text{F}

Capacitors in series combine reciprocally according to the product-over-sum rule for two components:

C_{\text{total}} = \frac{C_p \cdot C_3}{C_p + C_3}

.

Key Concept

Capacitors combine opposite to resistors: parallel capacitors add directly (

C_p = C_1 + C_2

), whereas series capacitors combine using reciprocal formulas (

C_s = \frac{C_1 \cdot C_2}{C_1 + C_2}

).

Alternative Method

Use the standard reciprocal formula for the series stage:

\frac{1}{C_{\text{total}}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{100} + \frac{1}{25} = \frac{1}{100} + \frac{4}{100} = \frac{5}{100} = \frac{1}{20}

, hence

C_{\text{total}} = 20\text{ }\mu\text{F}

.

Estimated Time:1m 0s

Question 653Question →

In basic electronics, the capacitance of a parallel-plate capacitor is determined by its physical construction and the dielectric material between its plates. Which of the following modifications will increase the capacitance of a parallel-plate capacitor?

Increasing the distance separating the two conductive plates

Increasing the surface area of the conductive plates

Winding the conductive material into tight coils to store energy in a magnetic field

Operating the component at higher AC frequencies to increase its Farad rating

Show answer & explanation

Answer: Increasing the surface area of the conductive plates

Answer

Increasing the surface area of the conductive plates increases the capacitance of a parallel-plate capacitor.

Capacitance (

C

) of a parallel-plate capacitor is given by

C = \frac{\varepsilon A}{d}

, where

\varepsilon

represents the permittivity of the dielectric material,

A

is the surface area of the plates, and

d

is the distance between the plates. Because capacitance is directly proportional to plate area, increasing the plate surface area allows more charge to be stored at any given voltage, directly increasing total capacitance.

Step-by-Step Solution

1

Recall the fundamental formula defining the capacitance of a parallel-plate capacitor.

The formula is

C = \frac{\varepsilon A}{d}

, where

C

is capacitance,

\varepsilon

is dielectric permittivity,

A

is the plate surface area, and

d

is the distance between plates.

Identifying the relationship between physical parameters and capacitance establishes which changes increase or decrease the value.

2

Analyze the proportional relationship between plate area and capacitance.

Capacitance is directly proportional to plate surface area (

C \propto A

) and inversely proportional to separation distance (

C \propto \frac{1}{d}

).

Because plate area

A

appears in the numerator of the formula, expanding

A

causes capacitance

C

to increase.

3

Identify the correct modification that increases capacitance.

Increasing the surface area of the conductive plates increases the total capacitance.

A larger plate area accommodates a greater quantity of charge for a given electric potential.

Key Concept

Physical factors affecting parallel-plate capacitance (

C = \frac{\varepsilon A}{d}

)

Estimated Time:1m 0s

Question 654Question →

Match each circuit operating scenario on the left with its corresponding calculated electrical quantity on the right by applying Ohm's law and metric unit conversions.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The relay coil matches

50\text{ mA}

, the cockpit indicator matches

80\ \Omega

, the diagnostic probe matches

100\text{ V}

, and the automotive control module matches

5\text{ mA}

.

Each scenario correctly applies the relevant permutation of Ohm's law (

V=IR

,

I=V/R

, or

R=V/I

) following proper metric prefix conversion to base electrical units.

Step-by-Step Solution

1

Calculate the current for the relay coil

I = \frac{24\text{ V}}{480\ \Omega} = 0.05\text{ A} = 50\text{ mA}

Apply

I = \frac{V}{R}

directly and convert amperes to milliamperes by multiplying by

1000

.

2

Calculate the resistance for the aircraft indicator

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

Convert

350\text{ mA}

to base units (

0.35\text{ A}

) and rearrange Ohm's law to

R = \frac{V}{I}

.

3

Calculate the voltage drop across the diagnostic probe

V = (40 \times 10^{-6}\text{ A}) \times (2.5 \times 10^{6}\ \Omega) = 100\text{ V}

Convert microamperes (

10^{-6}

) and megaohms (

10^{6}

) to standard base units and compute

V = I \times R

.

4

Calculate the current for the automotive control module

I = \frac{12\text{ V}}{2400\ \Omega} = 0.005\text{ A} = 5\text{ mA}

Convert

2.4\text{ k}\Omega

to

2400\ \Omega

, compute

I = \frac{V}{R}

, and express the result in milliamperes.

Key Concept

Ohm's Law calculations relating voltage (

V

), current (

I

), and resistance (

R

), integrated with metric prefix conversions.

Estimated Time:2m 0s

Question 655Question →

Review the four distinct practical situations below. Pair each descriptive problem with its mathematically correct calculated result.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are: Mean usage is

9

, median distance is

10

, mode size is

8

, and the required assignment score is

11

.

Each statistical measure is calculated according to its definition: the mean is the sum divided by the count, the median is the middle value of an ordered list (or average of two middle values), the mode is the most frequent value, and a missing value can be found algebraically using the mean formula.

Step-by-Step Solution

1

Determine the mean for the baker's sugar usage.

The sum is

36

, divided by

4

equals

9

.

The arithmetic mean requires dividing the sum of all values by the total count of values.

2

Determine the median for the delivery driver's distances.

The sorted list is

5, 7, 9, 11, 12, 14

. The average of the middle two values is

(9 + 11) \div 2 = 10

.

For an even number of data points, the median is the average of the two middle numbers in an ordered list.

3

Determine the mode for the hat sizes.

The mode is

8

.

The mode is the value that appears most often in a dataset.

4

Calculate the missing assignment score.

The required score is

11

.

Using the algebraic formula for mean:

(6 + 10 + x) \div 3 = 9

, which simplifies to

16 + x = 27

.

Key Concept

Measures of central tendency (mean, median, mode) and using algebraic substitution to find a missing data point given a target mean.

Question 656Question →

Match each capacitor term or specification on the left with its correct electrical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Farad (F) matches with the standard unit used to quantify electrical capacitance; Dielectric matches with the insulating material separating the conductive plates; Electrolytic capacitor matches with a polarized capacitor type sensitive to connection polarity; Working voltage rating matches with the maximum continuous voltage that can safely be applied.

Each term accurately pairs with its foundational physical definition: the farad is the SI unit of capacitance, the dielectric is the insulating medium between plates, electrolytic capacitors are polarized devices requiring correct terminal orientation, and the working voltage rating defines the maximum safe operating voltage before dielectric puncture.

Step-by-Step Solution

1

Identify the standard unit of capacitance

Farad (F) is the metric unit measuring charge storage ability.

Capacitance (

C = Q/V

) is quantified in units of farads.

2

Identify the insulating component inside a capacitor

The dielectric is the insulating layer.

Dielectrics block steady direct current conduction across the plates while concentrating the electric field.

3

Identify the polarized capacitor category

Electrolytic capacitors have designated positive and negative terminals.

Reverse polarization disrupts the chemical oxide layer, potentially destroying the component.

4

Identify the safe maximum voltage parameter

Working voltage rating represents the dielectric breakdown threshold.

Operating above this threshold risks arcing and dielectric breakdown.

Key Concept

Fundamental capacitor terminology, units, internal construction, and operational ratings

Estimated Time:45s

Question 657Question →

Once a capacitor placed in a direct current (DC) circuit reaches a fully charged state, how does it affect the continuous flow of current in that branch?

It blocks direct current completely because the accumulated charge creates an opposing voltage equal to the supply voltage.

It allows direct current to conduct freely because the dielectric spacer functions as an electrical conductor.

It transforms the incoming direct current into an oscillating alternating current output.

It sustains ongoing current flow by continuously storing energy within a magnetic field.

Show answer & explanation

Answer: It blocks direct current completely because the accumulated charge creates an opposing voltage equal to the supply voltage.

Answer

A fully charged capacitor completely blocks the continuous flow of direct current by acting as an open circuit.

The correct choice explains that as a capacitor charges, its plate voltage rises until it matches the applied direct current source voltage. At this equilibrium point, no net voltage difference exists to move additional electrons through the circuit branch, completely stopping current flow and causing the capacitor to act as an open circuit to steady DC.

Step-by-Step Solution

1

Identify the charging behavior of a capacitor connected to a DC source.

Current flows initially as electrons accumulate on one plate and leave the other, creating a potential difference across the dielectric.

Capacitors respond to changes in voltage by charging until their terminal voltage matches the applied source voltage.

2

Determine the state of the circuit when the capacitor is fully charged.

The voltage across the capacitor equals the supply voltage (

V_C = V_S

), so the net voltage driving current becomes zero (

I = 0

).

With no potential difference remaining across the circuit resistance, current ceases entirely, meaning the fully charged capacitor acts as an open circuit to DC.

Key Concept

DC Blocking Behavior of Capacitors

Estimated Time:45s

Question 658Question →

In an electromechanical relay, energizing the coil produces a magnetic field that pulls the armature, which causes the normally closed (NC) contacts to establish circuit continuity and the normally open (NO) contacts to break circuit continuity.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because the descriptor 'normally' refers to the unenergized resting state of the relay. When current energizes the electromagnetic coil, the armature moves to close normally open (NO) contacts—completing that circuit—and open normally closed (NC) contacts—interrupting that circuit.

Step-by-Step Solution

1

Identify the baseline resting (de-energized) state of relay contacts

Normally open (NO) contacts are open (no continuity) and normally closed (NC) contacts are closed (continuity present) when the coil carries no current.

The term 'normally' specifies the default mechanical condition of the contacts prior to coil energization.

2

Analyze the mechanical effect of coil energization

Current flowing through the coil generates an electromagnetic field that pulls the movable armature toward the core.

Electromagnetic attraction provides the physical force required to switch contact positions.

3

Determine the resulting contact states when energized

Armature displacement forces NO contacts to close (establishing continuity) and NC contacts to open (breaking continuity).

Energization transitions all contact pairs to the state opposite their resting configuration.

4

Compare the derived operation with the given statement

The statement asserts that energization causes NC contacts to establish continuity and NO contacts to break continuity, which is the exact inverse of actual relay operation.

Because the statement inverts the functional behavior of NO and NC contacts during coil activation, it is false.

Key Concept

Electromechanical Relay Contact States (NO vs. NC) During Coil Energization

Estimated Time:1m 15s

Question 659Question →

In an electrical circuit, which quantity represents the electromotive force (EMF) or electrical pressure that pushes electrons through a conductor?

Current

Voltage

Resistance

Inductance

Show answer & explanation

Answer: Voltage

Answer

Voltage represents the electromotive force (EMF) or electrical pressure that causes electrons to flow through a circuit.

Voltage, often referred to as electromotive force (EMF), is the potential difference between two points in a circuit that supplies the electrical pressure required to push electrons through a conductor.

Step-by-Step Solution

1

Identify the fundamental electrical quantity associated with electrical pressure and electromotive force.

Voltage (measured in volts) is defined as the electromotive force (EMF) or potential difference between two points.

Electromotive force provides the energy per unit charge needed to move free electrons through a conductive path.

2

Distinguish voltage from rate of flow, opposition to flow, and reactive properties.

Current is the rate of flow, resistance opposes flow, and inductance opposes changes in current.

Only voltage acts as the driving potential or pressure in the electrical circuit.

Key Concept

Electromotive force (EMF) and voltage represent the potential difference or electrical pressure that causes electric charge to move.

Estimated Time:30s

Question 660Question →

A direct current (DC) circuit is monitored while a total charge of $180\text{ C}$ passes through a cross-section of a copper conductor over an elapsed time of $2\text{ minutes}$ . What is the magnitude of the resulting current, and in which direction do the free electrons drift through the conductor?

The current is

90\text{ A}

, and electrons drift toward the positive terminal.

The current is

1.5\text{ A}

, and electrons drift toward the negative terminal.

The current is

1.5\text{ A}

, and electrons drift toward the positive terminal.

The current is

1.5\text{ V}

, and electrons drift toward the positive terminal.

Show answer & explanation

Answer: The current is $1.5\text{ A}$ , and electrons drift toward the positive terminal.

Answer

The current is

1.5\text{ A}

, and electrons drift toward the positive terminal.

Current is defined as charge per unit time (

I = \frac{Q}{t}

). Converting

2\text{ minutes}

gives

120\text{ seconds}

, yielding

I = \frac{180\text{ C}}{120\text{ s}} = 1.5\text{ A}

. In metallic conductors, the mobile charge carriers are electrons; because electrons are negatively charged, electrostatic forces drive them away from the negative terminal and toward the positive terminal.

Step-by-Step Solution

1

Convert the elapsed time from minutes to the SI base unit of seconds.

t = 2\text{ min} \times 60\text{ s/min} = 120\text{ s}

An ampere is defined as one coulomb per second (

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

), which requires time to be expressed in seconds.

2

Calculate the electric current using the charge rate formula

I = \frac{Q}{t}

.

I = \frac{180\text{ C}}{120\text{ s}} = 1.5\text{ A}

Current measures the rate at which electric charge flows past a specific point in a circuit.

3

Determine the physical direction of charge carrier movement in the metallic conductor.

Electrons drift toward the positive terminal.

In solid metal conductors like copper, current consists of free electrons that carry negative charge and are attracted toward the positive potential.

Key Concept

Electric current is the time rate of charge flow (

I = Q/t

), with physical electron flow moving from negative to positive potential.

Estimated Time:45s

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