Basic Electrical Concepts Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 3

Soru 41Soru →

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

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

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

).

Adım Adım Çözüm

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.

Anahtar Kavram

Alternating Current (AC) Directional Reversal and Frequency

Soru 42Soru →

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.

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Cevabı ve açıklamayı göster

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Ohm's Law calculations relating voltage (

V

), current (

I

), and resistance (

R

), integrated with metric prefix conversions.

Tahmini Süre:2m 0s

Soru 43Soru →

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

Current

Voltage

Resistance

Inductance

Cevabı ve açıklamayı göster

Cevap: Voltage

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:30s

Soru 44Soru →

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.

Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Electric current is the time rate of charge flow (

I = Q/t

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

Tahmini Süre:45s

Soru 45Soru →

A technician connects the primary winding of an iron-core transformer to a $12\text{ V}$ direct current (DC) battery and then to a $12\text{ V}$ alternating current (AC) power supply. What voltage behavior will be observed at the secondary output terminals for each power source?

A continuous output voltage is produced by both sources because transformer windings respond identically to steady and changing currents.

A continuous output voltage is produced only with the DC battery because a static magnetic field transfers energy without inductive losses.

A continuous output voltage is produced only with the AC power supply because transformers require a continuously changing magnetic field to induce a voltage.

A continuous output voltage is produced by the DC battery, whereas the AC power supply causes the primary winding to short-circuit due to alternating polarity.

Cevabı ve açıklamayı göster

Cevap: A continuous output voltage is produced only with the AC power supply because transformers require a continuously changing magnetic field to induce a voltage.

Cevap

A continuous output voltage is produced only with the AC power supply because transformers require a continuously changing magnetic field to induce a voltage.

The correct answer identifies that only the alternating current (AC) supply produces a sustained secondary output voltage. A transformer functions via mutual induction, which fundamentally requires a dynamic, changing magnetic field. Because AC continuously alternates in direction and amplitude, it creates a constantly fluctuating magnetic flux in the iron core that induces an electromotive force (voltage) across the secondary winding. Steady direct current (DC) establishes an unvarying magnetic field, which cannot produce a sustained output voltage.

Adım Adım Çözüm

1

Analyze the electromagnetic requirements of a transformer

Transformers operate on Faraday's law of induction, which states that a voltage is induced in a secondary winding only when there is a changing magnetic flux (

\Delta \Phi / \Delta t

).

Induction requires relative movement between magnetic lines of force and the secondary conductor.

2

Evaluate the behavior under a Direct Current (DC) input

A steady DC battery creates a constant, unvarying magnetic field in the core after initial energization, yielding zero sustained induced voltage across the secondary winding.

With steady DC, the rate of change of magnetic flux is zero after the initial switch closure.

3

Evaluate the behavior under an Alternating Current (AC) input

An AC supply continuously changes magnitude and reverses direction periodically, producing a dynamic magnetic field that continuously induces an alternating voltage in the secondary winding.

The continuous cyclical change in AC satisfies the condition for ongoing mutual inductance.

Anahtar Kavram

Transformers require a continuously changing current (AC) to produce the dynamic magnetic field necessary for mutual induction; steady DC produces a static magnetic field that cannot sustain transformer operation.

Tahmini Süre:45s

Soru 46Soru →

A benchtop power supply plugged into a $120\text{ V}$ household wall outlet provides a steady $+12\text{ V}$ direct current (DC) output to test an electronic device. Which internal circuit is responsible for converting the incoming alternating current (AC) into direct current (DC)?

Inverter

Rectifier

Step-up transformer

Alternator

Cevabı ve açıklamayı göster

Cevap: Rectifier

Cevap

A rectifier is the electronic circuit responsible for converting alternating current (AC) into direct current (DC).

A rectifier contains diodes that restrict electron flow to a single direction. When AC voltage alternates polarity, the rectifier routes the current unidirectionally, turning alternating current into direct current.

Adım Adım Çözüm

1

Identify the electrical input type and the required electrical output type.

The input from the wall outlet is alternating current (AC), and the required output for the device is direct current (DC).

Determining the directionality and waveform of the power source dictates the specific conversion device needed.

2

Match the conversion requirement to the corresponding electronic device function.

A rectifier circuit uses semiconductor diodes to force current to flow in only one direction, converting bidirectional AC into unidirectional DC.

Rectification is the fundamental electrical process for producing DC from an AC source.

Anahtar Kavram

Conversion of AC to DC via Rectification

Soru 47Soru →

A marine communication device is connected to a $12\text{ V}$ direct current (DC) power source. If the circuit has a total resistance of $4\text{ }\Omega$ , what is the current flowing through the circuit in amperes ( $\text{A}$ )?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The current flowing through the circuit is

3\text{ A}

.

According to Ohm's law (

I = \frac{V}{R}

), the current in a direct current circuit is calculated by dividing the total voltage by the total circuit resistance. With a voltage of

12\text{ V}

across a resistance of

4\text{ }\Omega

, the current is

\frac{12\text{ V}}{4\text{ }\Omega} = 3\text{ A}

.

Adım Adım Çözüm

1

Identify the given circuit parameters.

Voltage

V = 12\text{ V}

and resistance

R = 4\text{ }\Omega

.

Extracting the known quantities is necessary before applying the formula.

2

Select the appropriate form of Ohm's law to solve for current.

I = \frac{V}{R}

Ohm's law defines current as the ratio of electromotive force (voltage) to resistance.

3

Perform the division.

I = \frac{12}{4} = 3\text{ A}

Dividing

12\text{ V}

by

4\text{ }\Omega

computes the total current.

Anahtar Kavram

Ohm's law mathematical relationship where current is directly proportional to voltage and inversely proportional to resistance (

I = V / R

).

Tahmini Süre:45s

Soru 48Soru →

A series-connected thermopile module contains $14$ thermocouple junctions. If each junction generates an electromotive force (EMF) of $350\text{ mV}$ when heated, what is the total output potential difference of the module in volts ( $\text{V}$ )?

Cevabı ve açıklamayı göster

Cevap: 4.9

Cevap

The total output potential difference is

4.9\text{ V}

.

Because all

14

thermocouple junctions are connected in series aiding, their individual voltages sum directly:

14 \times 350\text{ mV} = 4{}900\text{ mV}

. Converting millivolts to volts by dividing by

1{}000

yields exactly

4.9\text{ V}

.

Adım Adım Çözüm

1

Calculate the total series electromotive force in millivolts

E_{\text{total}} = 14 \times 350\text{ mV} = 4{}900\text{ mV}

When voltage sources or EMF-generating elements are connected in a series-aiding string, their individual electromotive forces add directly.

2

Convert the total potential difference from millivolts to standard volts

V = \frac{4{}900\text{ mV}}{1{}000\text{ mV/V}} = 4.9\text{ V}

The metric prefix 'milli-' indicates one thousandth (

10^{-3}

) of a base unit, requiring division by

1{}000

to convert millivolts to volts.

Anahtar Kavram

Series electromotive forces (EMFs) add algebraically in a series-aiding configuration, and millivolts convert to standard volts by a factor of

10^{-3}

(

1{}000\text{ mV} = 1\text{ V}

).

Tahmini Süre:1m 0s

Soru 49Soru →

An electronics system contains a pulse-forming network where an electromotive force (EMF) source must do $0.24\text{ J}$ of work to move a charge of $2000\text{ \mu C}$ through the load. The power supply for this system is built by connecting identical $24\text{ V}$ battery modules in series. Assuming no internal resistance, how many $24\text{ V}$ battery modules must be connected in series to provide the required EMF?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5

The total EMF required is found by dividing the work done (

0.24\text{ J}

) by the charge in coulombs (

0.002\text{ C}

), which equals

120\text{ V}

. Since each battery module provides

24\text{ V}

and they are connected in series, the total EMF is divided by the module voltage.

120\text{ V}

divided by

24\text{ V}

per module equals exactly

5

modules.

Adım Adım Çözüm

1

Recall the physical definition of electromotive force.

E = \frac{W}{Q}

EMF is defined as the work done by a source per unit of charge it moves.

2

Convert the given charge into standard SI units (Coulombs).

2000\text{ \mu C} = 2000 \times 10^{-6}\text{ C} = 0.002\text{ C}

Standard units (Joules and Coulombs) are required to calculate voltage in Volts.

3

Calculate the total required EMF.

E = \frac{0.24\text{ J}}{0.002\text{ C}} = 120\text{ V}

Substituting the work and charge values yields the total potential difference required by the network.

4

Calculate the number of battery modules needed.

120\text{ V} / 24\text{ V} = 5

In a series connection, the total voltage is the sum of the individual voltages, requiring division.

Anahtar Kavram

Electromotive Force (EMF) as Work per Unit Charge and Series Circuit Addition

Soru 50Soru →

In direct-current circuit analysis, various terms describe electrical potential and force. Match each voltage-related concept on the left with its accurate operational definition on the right.

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Cevabı ve açıklamayı göster

Cevap

Electromotive Force (EMF) pairs with source energy per unit charge; Voltage Drop pairs with reduction in potential across a load; Reference Ground pairs with the common zero-volt baseline; and Open-Circuit Voltage pairs with the terminal potential measured under zero-current conditions.

Each electrical term uniquely characterizes a specific aspect of potential difference: Electromotive Force (EMF) is the driving energy provided by a source; Voltage Drop is the decrease in potential across an operating load; Reference Ground provides the

0\text{ V}

datum point; and Open-Circuit Voltage is the unloaded potential difference across source terminals.

Adım Adım Çözüm

1

Identify the concept of source electrical pressure

Electromotive Force (EMF) provides the driving energy per coulomb from a non-electrical source.

EMF represents the source's generated potential prior to internal losses.

2

Identify the behavior of potential across circuit loads

Voltage Drop reflects the potential decrease across circuit components.

Energy is dissipated or converted by resistive elements according to current flow.

3

Identify baseline circuit references and terminal conditions

Reference ground is the

0\text{ V}

node, while open-circuit voltage describes unloaded terminal potential.

Establishing a reference node allows meaningful relative voltage measurements throughout a schematic.

Anahtar Kavram

Voltage and Electromotive Force (EMF) Definitions and Circuit Roles

Soru 51Soru →

In electrical theory and technician practice, understanding how charge carriers behave, how current is quantified, and how current is measured in a circuit is essential. Match each electrical current concept on the left with its correct fundamental characteristic or definition on the right.

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Cevabı ve açıklamayı göster

Cevap

Electron flow direction matches the physical drift of negatively charged particles from negative to positive; Conventional current direction matches the model flowing from positive to negative; One ampere matches the transfer of one coulomb per second (

1\text{ C/s}

); Ammeter circuit placement matches a series configuration inserted directly into the conductive path.

Each concept pairs with its precise physical and operational principle: electron flow tracks actual electron drift (negative to positive), conventional current tracks theoretical positive charge flow (positive to negative), the ampere is defined as coulombs per second, and current measurement requires a series ammeter connection.

Adım Adım Çözüm

1

Differentiate physical charge motion from conventional flow

Identify that electrons carry negative charge and travel from the negative potential toward positive potential, whereas conventional current models flow from positive to negative.

This establishes the distinction between true physical electron transport and circuit engineering convention.

2

Apply the SI definition of current intensity

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

(

I = Q/t

).

Current is defined strictly as the time rate of charge flow through a conductor cross-section.

3

Determine proper instrumentation connection for measuring current

An ammeter must be placed in series within the circuit branch.

Because current is the flow of charge through a path, the meter must be part of that continuous path to measure the entire stream of electrons.

Anahtar Kavram

Electrical Current Definition, Carrier Flow Directions, and Measurement

Tahmini Süre:45s

Soru 52Soru →

An avionics technician tests a DC monitoring circuit connected to a constant $450\text{ V}$ power supply. The circuit currently draws $25\text{ mA}$ of current. To meet new operating specifications, the current must be reduced to $9\text{ mA}$ . What is the required new resistance of the circuit?

50\,\Omega

4.05\text{ k}\Omega

50\text{ k}\Omega

20\text{ k}\Omega

Cevabı ve açıklamayı göster

Cevap: $50\text{ k}\Omega$

Cevap

The required new resistance of the circuit is

50\text{ k}\Omega

.

According to Ohm's law, resistance equals voltage divided by current (

R = \frac{V}{I}

). Converting the target current of

9\text{ mA}

to standard amperes yields

0.009\text{ A}

. Dividing the constant source voltage of

450\text{ V}

by

0.009\text{ A}

results in

50,000\,\Omega

, which simplifies to

50\text{ k}\Omega

.

Adım Adım Çözüm

1

Identify the target circuit parameters and convert metric units to base SI units

Voltage

V = 450\text{ V}

, Target Current

I = 9\text{ mA} = 9 \times 10^{-3}\text{ A} = 0.009\text{ A}

Ohm's law requires current to be in amperes when voltage is in volts to produce resistance in ohms.

2

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

)

R = \frac{V}{I}

Resistance is equal to the potential difference divided by the resulting current flow.

3

Calculate the target resistance and convert to kilohms

R = \frac{450\text{ V}}{0.009\text{ A}} = 50,000\,\Omega = 50\text{ k}\Omega

Dividing

450

by

0.009

gives

50,000\,\Omega

, which is expressed as

50\text{ k}\Omega

using standard metric prefixes.

Anahtar Kavram

Ohm's Law for Resistance (

R = \frac{V}{I}

) with Metric Prefix Conversion

Alternatif Yöntem

Using direct prefix arithmetic: since volts divided by milliamperes directly yields kilohms (

\frac{\text{V}}{\text{mA}} = \text{k}\Omega

), calculate

\frac{450\text{ V}}{9\text{ mA}} = 50\text{ k}\Omega

directly without manual decimal expansion.

Tahmini Süre:1m 15s

Soru 53Soru →

A technician is troubleshooting a $24 \text{ V}$ direct current (DC) control circuit. Currently, the circuit draws $8 \text{ mA}$ of current. In order to reduce the current draw to $2 \text{ mA}$ without changing the voltage source, how much additional resistance must be added in series to the existing circuit?

9 \text{ k}\Omega

12 \text{ k}\Omega

4 \text{ k}\Omega

9 \text{ }\Omega

Cevabı ve açıklamayı göster

Cevap: $9 \text{ k}\Omega$

Cevap

The correct amount of additional resistance needed is

9 \text{ k}\Omega

.

To find the additional resistance required, first determine the current resistance (

R = \frac{24 \text{ V}}{0.008 \text{ A}} = 3 \text{ k}\Omega

). Next, calculate the total target resistance needed to limit the current to

2 \text{ mA}

(

R = \frac{24 \text{ V}}{0.002 \text{ A}} = 12 \text{ k}\Omega

). Finally, subtract the existing resistance from the target total to find the additional series resistance required (

12 \text{ k}\Omega - 3 \text{ k}\Omega = 9 \text{ k}\Omega

).

Adım Adım Çözüm

1

Calculate the initial resistance of the circuit.

R_1 = \frac{24 \text{ V}}{0.008 \text{ A}} = 3,000 \text{ }\Omega

or

3 \text{ k}\Omega

Ohm's Law (

R = \frac{V}{I}

) is used to find the existing resistance before modifications.

2

Calculate the target total resistance required to limit the current to

2 \text{ mA}

.

R_2 = \frac{24 \text{ V}}{0.002 \text{ A}} = 12,000 \text{ }\Omega

or

12 \text{ k}\Omega

Ohm's Law is applied again to determine the total resistance needed for the new current requirement.

3

Determine the additional resistance needed by subtracting the initial resistance from the target total resistance.

12 \text{ k}\Omega - 3 \text{ k}\Omega = 9 \text{ k}\Omega

Resistors in series add together (

R_{total} = R_1 + R_{added}

). To find the added portion, subtract the existing resistance from the total.

Anahtar Kavram

Ohm's Law Calculations and Concepts

Soru 54Soru →

Match each electrical power source or practical application on the left with its corresponding current characteristic and operational principle on the right.

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Cevabı ve açıklamayı göster

Cevap

The correct matches pair the high-voltage transmission grid with AC transformer voltage stepping, the automotive lead-acid battery with unidirectional DC from chemical reactions, the residential wall receptacle with AC reversing at

60\text{ Hz}

, and the electroplating tank with unidirectional DC for steady metal ion deposition.

Direct current (DC) flows in one unchanging direction with steady polarity, which is produced by chemical batteries and required for electroplating. Alternating current (AC) periodically alternates its direction of flow, enabling transformers to step voltage levels efficiently for transmission and supplying standard

60\text{ Hz}

household power.

Adım Adım Çözüm

1

Identify AC-dependent applications and their physical mechanisms

Recognize that power distribution grids and residential outlets utilize alternating current (AC). Grids depend on AC for transformer stepping, while outlets deliver cyclical

60\text{ Hz}

polarity reversals.

Transformers require time-varying current to induce secondary voltage, and utility generators produce sinusoidal alternating current.

2

Identify DC-dependent applications and their physical mechanisms

Recognize that lead-acid batteries and electroplating tanks operate on direct current (DC). Batteries generate unidirectional current chemically, while electroplating requires steady polarity to maintain ion migration.

Chemical cells produce static polarities, and reversing current in electroplating would dissolve rather than plate metal.

3

Match each application to its unique operational principle

Pair grid transmission to transformer stepping, battery to chemical unidirectional DC, wall outlet to

60\text{ Hz}

AC reversal, and electroplating to continuous ionic migration.

Each assignment links the fundamental behavior of alternating or direct current to its specific operational requirement.

Anahtar Kavram

Direct current (DC) maintains constant polarity and unidirectional electron flow, making it ideal for electrochemical storage (batteries) and electrodeposition. Alternating current (AC) reverses polarity periodically, enabling electromagnetic induction in transformers for long-distance power distribution and standard domestic utility delivery.

Tahmini Süre:1m 0s

Soru 55Soru →

An industrial control panel supplies a constant $24\text{ V}$ DC to a sensor circuit. During normal operation, the sensor draws a current of $16\text{ mA}$ . Following a component degradation, the measured current rises to $64\text{ mA}$ at the same operating voltage. By how many ohms ( $\Omega$ ) has the circuit's total resistance decreased?

Cevabı ve açıklamayı göster

Cevap: 1125

Cevap

The circuit's total resistance decreased by

1125\text{ }\Omega

.

According to Ohm's law, resistance equals voltage divided by current (

R = \frac{V}{I}

). Converting the currents to amperes gives

0.016\text{ A}

and

0.064\text{ A}

. The initial resistance is

R_1 = \frac{24\text{ V}}{0.016\text{ A}} = 1500\text{ }\Omega

, and the final resistance is

R_2 = \frac{24\text{ V}}{0.064\text{ A}} = 375\text{ }\Omega

. The decrease in resistance is

1500\text{ }\Omega - 375\text{ }\Omega = 1125\text{ }\Omega

.

Adım Adım Çözüm

1

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

I_1 = 0.016\text{ A}

and

I_2 = 0.064\text{ A}

Ohm's law requires current to be expressed in standard SI units (amperes) to yield resistance in ohms.

2

Calculate the initial resistance using Ohm's law (

R = \frac{V}{I}

)

R_1 = \frac{24\text{ V}}{0.016\text{ A}} = 1500\text{ }\Omega

Determines the baseline operating resistance of the sensor module.

3

Calculate the degraded state resistance using Ohm's law (

R = \frac{V}{I}

)

R_2 = \frac{24\text{ V}}{0.064\text{ A}} = 375\text{ }\Omega

Determines the lower operating resistance corresponding to the increased current draw.

4

Calculate the net difference between initial and degraded resistance

\Delta R = 1500\text{ }\Omega - 375\text{ }\Omega = 1125\text{ }\Omega

Finds the exact reduction in total circuit resistance.

Anahtar Kavram

Ohm's Law (

R = \frac{V}{I}

) applied to determine resistance changes in DC circuits with metric prefix conversion.

Tahmini Süre:1m 30s

Soru 56Soru →

An ammeter connected in series with a direct-current (DC) circuit registers a steady current of $250\text{ mA}$ . Which of the following statements correctly describes both the rate of electric charge transfer through the conductor and the physical direction of electron flow?

A charge of

0.25\text{ coulombs}

passes through the cross-section each second, with electrons drifting from the positive terminal toward the negative terminal.

The electromotive force drives

4\text{ coulombs}

of energy each second, with electron flow maintaining constant potential throughout the conductor.

A charge of

0.25\text{ coulombs}

passes through the cross-section each second, with electrons drifting from the negative terminal toward the positive terminal.

A charge of

250\text{ coulombs}

alternates direction each second, causing electrons to oscillate back and forth without net directional movement.

Cevabı ve açıklamayı göster

Cevap: A charge of $0.25\text{ coulombs}$ passes through the cross-section each second, with electrons drifting from the negative terminal toward the positive terminal.

Cevap

A charge of

0.25\text{ coulombs}

passes through the cross-section each second, with electrons drifting from the negative terminal toward the positive terminal.

One ampere equals one coulomb of electrical charge passing a point per second (

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

). Therefore, a direct current of

250\text{ mA}

(

0.25\text{ A}

) represents an electric charge transfer rate of

0.25\text{ coulombs per second}

. Furthermore, because electrons are negatively charged particles, electrostatic repulsion and attraction cause them to drift from the negative terminal of the DC source, through the external circuit, and into the positive terminal.

Adım Adım Çözüm

1

Convert the current measurement from milliamperes (

\text{mA}

) to standard SI units of amperes (

\text{A}

).

250\text{ mA} = 250 \times 10^{-3}\text{ A} = 0.25\text{ A}

The ampere is the standard unit of current where

1\text{ A} = 1000\text{ mA}

.

2

Relate the current in amperes to the rate of electric charge flow using the fundamental formula

I = \frac{Q}{t}

.

0.25\text{ A} = 0.25\text{ coulombs per second}

One ampere is defined strictly as the flow of one coulomb of electrical charge per second.

3

Determine the physical direction of electron flow in the external circuit.

Electrons flow from the negative terminal to the positive terminal.

Electrons carry a negative charge and are repelled by the negative terminal and attracted toward the positive terminal of the DC power source.

Anahtar Kavram

Electric current is the time rate of charge flow (

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

), and physical electron flow travels from negative to positive potential in a DC circuit.

Alternatif Yöntem

Calculate the total time required for a standard unit charge (

1\text{ C}

) to pass:

t = \frac{Q}{I} = \frac{1\text{ C}}{0.25\text{ A}} = 4\text{ seconds}

. This verifies that each second carries

\frac{1}{4}\text{ C} = 0.25\text{ C}

.

Tahmini Süre:45s

Soru 57Soru →

A handheld emergency flashlight is powered by four $1.5\text{ V}$ alkaline cells connected in series aiding. What is the total electromotive force (EMF) supplied to the flashlight bulb?

1.5\text{ V}

6.0\text{ V}

6.0\text{ A}

0.375\text{ V}

Cevabı ve açıklamayı göster

Cevap: $6.0\text{ V}$

Cevap

6.0\text{ V}

When DC voltage sources are connected in series aiding (positive terminal to negative terminal), their individual voltages add together directly. Four

1.5\text{ V}

cells in series provide a total electromotive force of

4 \times 1.5\text{ V} = 6.0\text{ V}

.

Adım Adım Çözüm

1

Identify the circuit configuration and the formula for total EMF in series.

E_{\text{total}} = E_1 + E_2 + E_3 + E_4

When electrical power sources (cells) are wired in series aiding, their individual voltages add directly together to produce the total potential difference.

2

Sum the voltages of the four identical

1.5\text{ V}

cells.

E_{\text{total}} = 1.5\text{ V} + 1.5\text{ V} + 1.5\text{ V} + 1.5\text{ V} = 6.0\text{ V}

Multiplying the cell voltage by the total number of cells (

4 \times 1.5\text{ V}

) yields the total electromotive force.

Anahtar Kavram

Series Cell Electromotive Force (EMF) Addition

Tahmini Süre:45s

Soru 58Soru →

A portable emergency radio beacon uses a power source consisting of three series-connected DC cells: a primary lithium cell with an electromotive force (EMF) of $3.6\text{ V}$ , a standard cell with an EMF of $1.5\text{ V}$ , and a photovoltaic booster cell with an EMF of $0.9\text{ V}$ . When all three cells are wired in a series-aiding configuration, what is the total electromotive force supplied to the beacon?

3.6\text{ V}

6.0\text{ V}

6.0\text{ A}

1.2\text{ V}

Cevabı ve açıklamayı göster

Cevap: $6.0\text{ V}$

Cevap

The total electromotive force supplied to the beacon is

6.0\text{ V}

.

When multiple direct-current sources are connected in a series-aiding configuration, their individual electromotive forces (EMFs) add algebraically. Summing the three cell ratings gives

3.6\text{ V} + 1.5\text{ V} + 0.9\text{ V} = 6.0\text{ V}

.

Adım Adım Çözüm

1

Identify the individual electromotive force (EMF) values of the series-connected sources.

E_1 = 3.6\text{ V}

,

E_2 = 1.5\text{ V}

, and

E_3 = 0.9\text{ V}

.

Extracting the component voltages is necessary to determine the overall potential difference across the power source.

2

Apply the rule for DC voltage sources connected in series-aiding polarity (

E_{\text{total}} = E_1 + E_2 + E_3

).

E_{\text{total}} = 3.6\text{ V} + 1.5\text{ V} + 0.9\text{ V} = 6.0\text{ V}

.

When cells are wired series-aiding (positive terminal of one cell connected to the negative terminal of the next), their individual electromotive forces combine additively.

Anahtar Kavram

Series-Aiding Electromotive Force (EMF) Summation

Tahmini Süre:1m 0s

Soru 59Soru →

A technician connects a $24\text{ V}$ direct current (DC) power source across an auxiliary heating element with a measured resistance of $8\text{ }\Omega$ . What is the amount of current flowing through the heating element?

192\text{ A}

0.33\text{ A}

3\text{ A}

24\text{ A}

Cevabı ve açıklamayı göster

Cevap: $3\text{ A}$

Cevap

3\text{ A}

According to Ohm's law, current (

I

) is calculated by dividing voltage (

V

) by resistance (

R

). For a

24\text{ V}

supply connected across an

8\text{ }\Omega

load, the calculation is

I = \frac{24\text{ V}}{8\text{ }\Omega} = 3\text{ A}

.

Adım Adım Çözüm

1

Identify the given values from the problem statement.

Voltage (

V

) =

24\text{ V}

, Resistance (

R

) =

8\text{ }\Omega

.

Ohm's law relates voltage, current, and resistance in a DC circuit.

2

Select the appropriate form of Ohm's law to solve for current (

I

).

I = \frac{V}{R}

Current is directly proportional to voltage and inversely proportional to resistance.

3

Substitute the known values into the equation and compute the result.

I = \frac{24\text{ V}}{8\text{ }\Omega} = 3\text{ A}

Dividing

24

by

8

gives the circuit current in amperes.

Anahtar Kavram

Ohm's Law relationship between voltage, current, and resistance (

I = \frac{V}{R}

)

Tahmini Süre:45s

Soru 60Soru →

Match each electrical current term with its correct definition or technical description.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Electron Flow matches the movement of negative charge carriers from negative to positive; Conventional Current matches the flow modeled from positive to negative; Ampere matches the unit defined as one coulomb per second; Ammeter matches the instrument placed in series to measure charge flow rate.

Each electrical term is paired with its precise technical role: electron flow represents real electron drift from negative to positive; conventional current is the established positive-to-negative analysis standard; the ampere represents a charge flow rate of one coulomb per second; and the ammeter is the tool configured in series to monitor that current.

Adım Adım Çözüm

1

Identify physical charge transport direction versus convention.

Electrons carry negative charge and move from negative to positive (Electron Flow), whereas engineering circuit analysis traditionally assumes positive-to-negative flow (Conventional Current).

Differentiating between physical particle drift and standardized schematic conventions is fundamental to electronics.

2

Identify the unit of rate of charge flow.

Current (

I

) is measured in amperes (

A

), where

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

.

Current is defined quantitatively as the rate of charge transfer over time (

I = Q/t

).

3

Identify the measurement device.

An ammeter measures current and is placed in series so the circuit current passes through it.

Measuring current requires intercepting the stream of charge carriers in a series configuration.

Anahtar Kavram

Fundamental definitions and flow conventions of electrical current

Tahmini Süre:45s

Basic Electrical Concepts

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