Basic Electrical Concepts

189 soru

Soru 21Soru →

Rechargeable electronic devices store energy in batteries that require direct current (DC) to charge. Which device is specifically designed to convert the alternating current (AC) supplied by a wall outlet into direct current (DC)?

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Cevap: A rectifier

Cevap

A rectifier is the device that converts alternating current (AC) into direct current (DC).
A rectifier is an electrical circuit consisting of diodes that permits current flow in only one direction, transforming bidirectional alternating current (AC) into unidirectional direct current (DC).

Adım Adım Çözüm

1
Identify the type of incoming electrical supply and the required output format.
The source is alternating current (AC) from a utility outlet, and the target load requires direct current (DC).
Battery charging requires unidirectional current flow (DC), whereas grid power supplies bidirectional oscillating current (AC).
2
Select the electrical conversion device that turns bidirectional AC into unidirectional DC.
A rectifier (typically using semiconductor diodes) accomplishes AC-to-DC conversion.
Diodes conduct current in only one direction, converting the alternating wave into a unidirectional direct current.

Anahtar Kavram

Direct current (DC) flows continuously in one direction, whereas alternating current (AC) periodically reverses direction. Conversion from AC to DC is accomplished using a rectifier, while conversion from DC to AC is performed by an inverter.
Tahmini Süre:40s
Soru 22Soru →

Electric utility companies transmit electrical power across regional grids using high-voltage alternating current (AC) rather than direct current (DC). What is the primary technical advantage of using AC for long-distance power distribution?

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Cevap: AC voltage can be easily stepped up to high voltages using transformers to minimize transmission power loss.

Cevap

AC voltage can be easily stepped up to high voltages using transformers to minimize transmission power loss.
The primary advantage of alternating current (AC) in power distribution is its compatibility with electromagnetic transformers. By stepping voltage up to very high levels for long-distance transmission, the line current is substantially reduced, which minimizes I2RI^2 R heat losses along transmission lines. Step-down transformers then reduce the voltage to safe, usable levels at homes and businesses.

Adım Adım Çözüm

1
Analyze the relationship between transmission voltage, current, and line power loss.
Power lost as heat in transmission conductors is given by Ploss=I2RP_{\text{loss}} = I^2 R. Stepping up transmission voltage reduces current II for a given power level, which drastically reduces power loss across the line resistance RR.
High voltage is necessary to minimize resistive losses over long transmission distances.
2
Evaluate why AC is uniquely suited for voltage stepping compared to steady DC.
Transformers rely on electromagnetic induction driven by a continuously changing magnetic field, which requires alternating current (AC). Steady DC cannot induce a continuous secondary voltage in standard transformers.
AC operates transformers with high efficiency, allowing simple conversion between generation, transmission, and household voltage levels.

Anahtar Kavram

Alternating current (AC) produces fluctuating magnetic fields that allow transformers to efficiently step voltage up for low-loss transmission and down for end-user distribution.
Tahmini Süre:45s
Soru 23Soru →

An off-grid weather monitoring station stores electrical energy in a 24 V24\text{ V} direct current (DC) battery bank but must operate a 120 V120\text{ V} alternating current (AC) transmitter. Which component is required to power the transmitter from the battery bank, and how does it manipulate the electrical current?

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Cevap: An inverter, which converts unidirectional direct current into alternating current that periodically reverses polarity

Cevap

An inverter, which converts unidirectional direct current into alternating current that periodically reverses polarity
Batteries store and supply direct current (DC), in which electrons flow in only one direction with constant polarity. An inverter uses electronic switching circuitry (such as transistors or MOSFETs) to transform this unidirectional current into alternating current (AC), where the voltage and electron flow periodically alternate and reverse direction to match the requirements of AC equipment.

Adım Adım Çözüm

1
Identify the power source and the load power requirements
The energy source is a 24 V24\text{ V} battery bank providing steady direct current (DC), while the load requires 120 V120\text{ V} alternating current (AC).
Determining the input waveform (unidirectional DC) and output waveform (alternating bidirectional AC) establishes the necessary power conversion process.
2
Evaluate the function of conversion and transformation devices
Transformers require alternating current to induce voltage and cannot operate on steady DC; rectifiers convert AC to DC; inverters convert DC to AC.
An inverter electronically switches and modulates DC to generate an alternating voltage and current waveform that periodically reverses direction.
3
Select the component that performs DC-to-AC conversion
An inverter changes the unidirectional DC into the required AC waveform.
The transmitter requires an AC waveform with reversing polarity, which only an inverter can synthesize from a DC storage bank.

Anahtar Kavram

Direct Current (DC) flows continuously in a single direction, whereas Alternating Current (AC) periodically reverses direction and polarity. Converting DC to AC requires an inverter, whereas converting AC to DC requires a rectifier.
Tahmini Süre:45s
Soru 24Soru →

A technician needs to measure the electromotive force (EMF) across an operating direct-current (DC) circuit component. Which of the following describes the correct meter configuration and the property being measured?

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Cevap: Connect the meter in parallel with the component to measure the potential difference in volts

Cevap

Connect the meter in parallel with the component to measure the potential difference in volts
Electromotive force (EMF) is the measure of electrical pressure or potential difference between two points, quantified in units of volts. Because voltage is a relative measurement across two nodes, the meter must be connected in parallel across the component under test.

Adım Adım Çözüm

1
Identify the electrical property defined by electromotive force (EMF)
EMF represents electrical pressure or potential difference, which is quantified in volts (VV).
Electromotive force is the energy per unit charge that drives electrons between two points in a circuit.
2
Determine the proper meter placement for measuring potential difference
The voltmeter must be connected in parallel (across) the component.
Potential difference is measured between two distinct points; connecting in parallel allows the high-impedance meter to sample the voltage drop without disrupting circuit current.

Anahtar Kavram

Electromotive force (EMF) is the potential difference or electrical pressure across two points, measured in volts using a voltmeter connected in parallel.
Tahmini Süre:45s
Soru 25Soru →

Match each voltage and electromotive force concept on the left with its correct definition or description on the right.

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Öğeler

Electromotive Force (EMF)
Potential Difference (Voltage Drop)
Volt
Voltmeter

Eşleşmeler

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Cevap

Electromotive Force (EMF) pairs with the total electrical pressure supplied by an energy source; Potential Difference pairs with the difference in electrical potential energy across a circuit component; Volt pairs with the standard unit defined as one joule per coulomb; and Voltmeter pairs with the measuring instrument connected in parallel across a load.
Electromotive Force (EMF) provides the driving electrical pressure from a source; Potential Difference (Voltage Drop) is the change in electrical energy across a component; the Volt is the standard unit representing one joule per coulomb; and a Voltmeter is the high-resistance meter connected in parallel to measure electrical potential.

Adım Adım Çözüm

1
Identify the source of electrical pressure in a circuit
Electromotive force (EMF) represents the energy provided by a source (such as a battery or generator) to establish electron flow.
EMF originates at the power source and represents the maximum potential available to do work.
2
Distinguish component potential difference from source EMF
Potential difference (voltage drop) describes the energy consumed per unit charge between two points across a load.
As electrons flow through resistance, potential energy decreases across each element.
3
Match the fundamental unit and measurement tool for electrical potential
The volt is the SI unit (1 V=1 J/C1\text{ V} = 1\text{ J/C}), and the voltmeter is the high-impedance tool connected in parallel to measure voltage.
Correct identification of units and test instruments is essential for electrical analysis and troubleshooting.

Anahtar Kavram

Voltage, Electromotive Force, and Measurement Fundamentals
Tahmini Süre:45s
Soru 26Soru →

Match each electrical component or power source to its corresponding operational characteristic regarding alternating current (AC) and direct current (DC).

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Öğeler

Automotive battery
Residential wall outlet
Rectifier
Inverter

Eşleşmeler

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Cevap

Automotive battery pairs with steady direct current flowing in one constant direction; Residential wall outlet pairs with supplying alternating current that periodically reverses direction; Rectifier pairs with converting AC into DC; Inverter pairs with converting DC into AC.
An automotive battery supplies direct current (DC) flowing in a single fixed direction, while a residential wall outlet provides alternating current (AC) that constantly reverses direction. For power conversion, a rectifier changes AC into DC, and an inverter changes DC into AC.

Adım Adım Çözüm

1
Classify the output type of the power sources.
An automotive battery produces a constant direct current (DC) flowing continuously from negative to positive terminals. A residential wall outlet delivers power that oscillates in polarity and reverses direction, which is alternating current (AC).
Electrochemical cells maintain constant terminal polarity, whereas utility power grids distribute AC power generated by rotating alternators.
2
Differentiate between the two electrical conversion devices.
A rectifier changes alternating current (AC) into direct current (DC), while an inverter changes direct current (DC) into alternating current (AC).
Rectifiers are used to power DC electronics from AC grid mains, whereas inverters allow standard AC appliances to operate from DC battery banks.

Anahtar Kavram

Distinction between AC and DC sources and the functional roles of rectifiers and inverters
Tahmini Süre:45s
Soru 27Soru →

In electrical and electronic systems, direct current (DC) and alternating current (AC) interact with specific devices for power generation, conversion, and voltage transformation. Match each electrical component or device on the left with its primary operational function or characteristic on the right.

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Öğeler

Electrochemical battery cell
Standard electromagnetic transformer
Silicon bridge rectifier
Solid-state electronic inverter

Eşleşmeler

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Cevap

Electrochemical battery cell matches with generates a continuous, unidirectional flow of electric charge; Standard electromagnetic transformer matches with steps voltage levels up or down using a changing magnetic field; Silicon bridge rectifier matches with converts alternating current into pulsating direct current; Solid-state electronic inverter matches with converts direct current into alternating current.
Each component is accurately paired with its fundamental relationship to AC or DC: batteries supply pure DC, transformers strictly require AC induction, rectifiers convert AC into DC, and inverters synthesize AC from DC.

Adım Adım Çözüm

1
Identify the source behavior of the electrochemical battery cell
Matched with producing continuous unidirectional current flow (DC).
Chemical reactions inside a battery maintain constant polarity at the terminals, forcing electrons to move in one direction.
2
Determine the operating principle of the electromagnetic transformer
Matched with stepping voltage up or down via a changing magnetic field.
Electromagnetic induction requires a constantly changing current, which is provided exclusively by alternating current (AC) rather than steady DC.
3
Distinguish between conversion devices (rectifier versus inverter)
Rectifier converts AC to DC; Inverter converts DC to AC.
Rectifiers use diodes to permit one-way conduction from an AC source, whereas inverters use electronic switching to create an alternating waveform from a DC supply.

Anahtar Kavram

Direct Current (DC) maintains unidirectional electron flow, while Alternating Current (AC) reverses direction periodically. Specific components are required to generate, transform, or convert between these two electrical modes.
Tahmini Süre:1m 0s
Soru 28Soru →

In electrical systems, direct current (DC) and alternating current (AC) exhibit distinct behaviors across chemical, capacitive, magnetic, and electromechanical applications. Match each electrical component or application on the left with its correct operational characteristic regarding AC or DC on the right.

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Öğeler

Secondary battery charging process
Iron-core step-down transformer
Capacitor connected across a steady DC source
Split-ring commutator in a rotating generator

Eşleşmeler

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Cevap

Secondary battery charging matches unidirectional chemical restoration; Iron-core transformer matches alternating flux mutual induction; Capacitor in steady DC matches open-circuit behavior at zero frequency; Split-ring commutator matches mechanical rectification of AC to DC.
Each component is correctly matched to its fundamental AC or DC operating mechanism: battery charging requires unidirectional DC electron flow for chemical storage, transformers depend on AC-induced time-varying magnetic flux, capacitors block steady DC at 0 Hz0\text{ Hz}, and commutators mechanically rectify internally induced AC into DC.

Adım Adım Çözüm

1
Analyze the chemical requirement of battery charging
Battery charging requires continuous direct current (DC) so that ions travel consistently toward their designated electrodes without reversing every half-cycle.
AC would alternate between charging and discharging every cycle, resulting in zero net chemical energy storage.
2
Analyze transformer operational physics
Transformers require alternating current (AC) to produce a continuously changing magnetic field (ΔΦ/Δt\Delta\Phi/\Delta t) across the core to induce an electromotive force in the secondary coil.
Steady DC produces a static magnetic field that cannot induce continuous secondary voltage.
3
Analyze capacitive reactance under DC conditions
With capacitive reactance XC=12πfCX_C = \frac{1}{2\pi f C}, a steady DC source has a frequency of f=0 Hzf = 0\text{ Hz}, yielding infinite reactance (XC→∞X_C \rightarrow \infty) once charged.
Direct current cannot cross the insulating dielectric layer once charge equilibrium is established.
4
Analyze the electromechanical function of a split-ring commutator
Rotating loops in a magnetic field inherently produce AC; the split rings reverse contact connections at each half-rotation, converting internal AC into pulsating DC.
The commutator serves as a mechanical rectifier.

Anahtar Kavram

Fundamental distinctions in how alternating current (time-varying direction/magnitude) and direct current (unidirectional flow) govern induction, electrochemistry, capacitive opposition, and mechanical waveform rectification.
Tahmini Süre:1m 30s
Soru 29Soru →

In an electronic flash unit, a high-voltage trigger circuit transfers 0.72 J0.72\text{ J} of electrical energy by moving a charge of 1.6 mC1.6\text{ mC} across a gas-discharge tube. What is the electromotive force (voltage) developed across the tube during this discharge?

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Cevap: 450 V450\text{ V}

Cevap

450 V450\text{ V}
Electromotive force (EMF), or voltage, is defined as the amount of work or energy transferred per unit of electric charge, expressed by the formula V=WQV = \frac{W}{Q}, where 1 volt=1 joule per coulomb1\text{ volt} = 1\text{ joule per coulomb}. Converting the charge from millicoulombs to coulombs gives 1.6 mC=0.0016 C1.6\text{ mC} = 0.0016\text{ C}. Dividing the total transferred energy (0.72 J0.72\text{ J}) by the charge (0.0016 C0.0016\text{ C}) produces 0.720.0016=450 V\frac{0.72}{0.0016} = 450\text{ V}.

Adım Adım Çözüm

1
Identify the given physical quantities and their base units.
Energy W=0.72 JW = 0.72\text{ J} and charge Q=1.6 mCQ = 1.6\text{ mC}.
Standard electrical calculations require values expressed in base SI units.
2
Convert the electric charge from millicoulombs to coulombs.
Q=1.6×10−3 C=0.0016 CQ = 1.6 \times 10^{-3}\text{ C} = 0.0016\text{ C}.
The prefix milli- represents 10−310^{-3}, so 1.6 mC=0.0016 C1.6\text{ mC} = 0.0016\text{ C}.
3
Apply the definition formula of electromotive force and voltage (V=WQV = \frac{W}{Q}).
V=0.72 J0.0016 C=450 VV = \frac{0.72\text{ J}}{0.0016\text{ C}} = 450\text{ V}.
One volt is defined as one joule of energy transferred per coulomb of charge.

Anahtar Kavram

Electromotive force (EMF) and voltage represent the potential energy transferred per unit electric charge (1 V=1 J/C1\text{ V} = 1\text{ J/C}).
Tahmini Süre:1m 30s
Soru 30Soru →

Match each voltage-related term or instrument to its correct electrical definition or operational principle.

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Öğeler

Electromotive Force (EMF)
Voltage Drop
Volt (1 V1\text{ V})
Voltmeter

Eşleşmeler

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Cevap

Electromotive Force matches the source energy converting non-electrical to electrical energy; Voltage Drop matches the potential difference across a passive load; Volt matches one joule of work per coulomb of charge; Voltmeter matches the high-resistance instrument connected in parallel.
Electromotive Force (EMF) specifically refers to the maximum potential supplied by an active source converting chemical or mechanical energy to electrical energy. Voltage drop represents potential difference dissipated across a load. The volt is defined as one joule of energy per coulomb of charge (1 J/C1\text{ J/C}). A voltmeter is a high-resistance meter connected in parallel across components to measure potential difference.

Adım Adım Çözüm

1
Differentiate between source potential generation and passive circuit potential reduction.
Electromotive Force (EMF) is the potential generated by a power source, whereas Voltage Drop is the potential difference consumed across a passive circuit element.
Sources provide energy to charges, while loads dissipate or convert that energy.
2
Identify the standard unit definition for electric potential.
The volt (V) is the SI unit representing work per unit charge, equal to 1 J/C1\text{ J/C}.
Potential difference fundamentally measures work needed to move charge between two points.
3
Determine the operational properties of the instrument used to measure voltage.
A voltmeter possesses high internal resistance and connects in parallel across components.
Parallel placement reads potential difference without disturbing the current flow through the monitored circuit.

Anahtar Kavram

Distinction between electromotive force, potential difference across loads, the physical definition of the volt, and voltmeter configuration.
Tahmini Süre:1m 0s
Soru 31Soru →

Which of the following statements correctly describes how charge carrier (electron) movement differs between an alternating current (AC) circuit and a direct current (DC) circuit?

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Cevap: Electrons in an AC circuit oscillate back and forth periodically, whereas electrons in a DC circuit drift continuously in a single direction.

Cevap

Electrons in an AC circuit oscillate back and forth periodically, whereas electrons in a DC circuit drift continuously in a single direction.
In direct current (DC), the electromotive force maintains a constant polarity, causing charge carriers (electrons) to drift steadily in one direction along the conductor. In alternating current (AC), the polarity of the voltage source reverses periodically at a given frequency, causing the electrons to oscillate rapidly back and forth around relatively fixed positions without traveling continuously from one end to the other.

Adım Adım Çözüm

1
Identify the fundamental physical behavior of electric charge in Direct Current (DC).
In a DC circuit, a constant voltage polarity creates an electric field that causes electrons to drift steadily in one continuous direction.
Direct current is characterized by unidirectional flow over time.
2
Identify the fundamental physical behavior of electric charge in Alternating Current (AC).
In an AC circuit, the source voltage periodically reverses polarity, causing the electric field to reverse direction and the electrons to oscillate back and forth in place.
Alternating current changes direction at a specific frequency (e.g., 60 Hz60\text{ Hz}), resulting in zero net displacement over complete cycles.

Anahtar Kavram

Direct Current (DC) involves unidirectional electron drift, while Alternating Current (AC) involves periodic bidirectional oscillation of electrons.
Tahmini Süre:45s
Soru 32Soru →

A technician is constructing a series battery bank using standard 1.5 V1.5\text{ V} cells. During the assembly process, they accidentally reverse the polarity of exactly two cells. When they measure the total electromotive force (EMF) across the completed battery bank, the multimeter reads 15 V15\text{ V}. How many total cells are in the battery bank?

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Cevap: 14 cells

Cevap

The battery bank contains 14 cells.
The total electromotive force (EMF) of a series circuit is the algebraic sum of the individual cell voltages. A correctly oriented cell adds 1.5 V1.5\text{ V}, while a reversed cell subtracts 1.5 V1.5\text{ V}. Each reversed cell not only fails to add its voltage but actively opposes the other cells, effectively reducing the total maximum potential by 3.0 V3.0\text{ V} per reversed cell. To find the total number of cells (NN), we use the equation: Vtotal=(Nforward×1.5 V)−(Nreversed×1.5 V)V_{\text{total}} = (N_{\text{forward}} \times 1.5\text{ V}) - (N_{\text{reversed}} \times 1.5\text{ V}). Given 15 V15\text{ V} total and 2 reversed cells, the equation becomes 15=(N−2)(1.5)−(2×1.5)15 = (N - 2)(1.5) - (2 \times 1.5). Solving this gives 15=1.5N−615 = 1.5N - 6, which results in 1.5N=211.5N = 21, meaning N=14N = 14 cells.

Adım Adım Çözüm

1
Identify the formula for the total voltage of cells in series when some are reversed.
Vtotal=(Nforward×1.5 V)−(Nreversed×1.5 V)V_{\text{total}} = (N_{\text{forward}} \times 1.5\text{ V}) - (N_{\text{reversed}} \times 1.5\text{ V})
Reversed cells subtract their voltage from the total EMF instead of adding to it.
2
Express the number of forward cells in terms of the total number of cells (NN).
Nforward=N−2N_{\text{forward}} = N - 2
The total number of cells is the sum of forward-facing and reversed cells, and we know exactly 2 are reversed.
3
Substitute the known values into the voltage equation and solve for NN.
15=(N−2)(1.5)−(2×1.5)15 = (N - 2)(1.5) - (2 \times 1.5)
This sets up the algebraic equation to find the total number of cells.
4
Simplify and solve the equation for NN.
15=1.5N−3−3  ⟹  15=1.5N−6  ⟹  21=1.5N  ⟹  N=1415 = 1.5N - 3 - 3 \implies 15 = 1.5N - 6 \implies 21 = 1.5N \implies N = 14
Solving the equation reveals that 14 total cells are required to produce a 15 V15\text{ V} output with 2 reversed cells.

Anahtar Kavram

Series Voltage Summation and Polarity
Soru 33Soru →

An automotive technician measures an electromotive force (EMF) of 12.6 V12.6\text{ V} across the terminals of a vehicle battery with no load connected. What does this electromotive force represent?

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Cevap: The electrical potential difference available to move electrons through a circuit

Cevap

The electrical potential difference available to move electrons through a circuit
Electromotive force (EMF), commonly referred to as voltage, is the electrical potential difference established between two points. It acts as the electrical driving pressure that causes free electrons to move through a conductor when a complete circuit is formed.

Adım Adım Çözüm

1
Identify the electrical quantity being measured and its unit
The measurement is 12.6 V12.6\text{ V}, which represents electromotive force (EMF) or voltage.
Volts (V\text{V}) are the standard unit of measure for electrical potential difference and electromotive force.
2
Relate electromotive force to fundamental circuit principles
EMF is the electrical pressure or work per unit charge created by a source to move electrons through an external pathway.
Voltage creates the electric field necessary to establish and maintain current in a closed circuit.

Anahtar Kavram

Electromotive Force and Potential Difference
Tahmini Süre:45s
Soru 34Soru →

A standard portable flashlight is powered by two alkaline dry cell batteries connected in series. What type of electrical current flows through the flashlight bulb when it is turned on?

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Cevap: Direct current (DC), because electric charge flows continuously in a single, constant direction

Cevap

Direct current (DC), because electric charge flows continuously in a single, constant direction
Batteries produce direct current (DC) because their chemical reactions establish fixed positive and negative terminals, causing electrical charge to flow continuously in a single direction through the closed circuit.

Adım Adım Çözüm

1
Identify the power source and its characteristic electrical output
Alkaline dry cells are electrochemical batteries, which produce a steady electromotive force with fixed positive and negative terminals.
Determining the power source type indicates whether the electrical output is unidirectional (DC) or oscillating (AC).
2
Classify the resulting current flow in the closed circuit
Since the battery polarity remains constant, electrons move continuously in one direction through the circuit, which is the definition of direct current (DC).
Direct current is defined by unidirectional charge flow, whereas alternating current regularly reverses direction.

Anahtar Kavram

Direct Current (DC) vs. Alternating Current (AC)
Tahmini Süre:30s
Soru 35Soru →

Unlike a steady DC power system where individual electrons physically drift through the entirety of the closed conductive path over time, the electrons in a standard 60 Hz60\text{ Hz} AC household system only vibrate back and forth within a localized region of the wire and never complete a full lap around the circuit.

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Cevap: True

Cevap

True. DC causes a continuous physical progression of electrons, while AC causes localized oscillation.
The statement is true because the constant unidirectional electric field in a DC system allows for a continuous physical progression of electrons around the circuit loop. In an AC system, the rapidly alternating electric field causes the electrons to simply wiggle back and forth locally, transferring energy without requiring a full physical transit of the circuit.

Adım Adım Çözüm

1
Analyze the nature of electron movement in a steady DC system.
DC maintains a constant voltage polarity, which forces electrons to slowly drift in one continuous direction around the circuit.
Understanding DC electron drift establishes the baseline for comparison.
2
Analyze the nature of electron movement in an AC system.
AC voltage constantly reverses polarity, which in a 60 Hz60\text{ Hz} system occurs 120 times per second.
This rapid reversal determines how the electric field influences the electrons.
3
Evaluate the distance an electron can travel in an AC system before reversing.
Because electron drift velocity is incredibly slow, the rapid polarity reversal forces them to merely vibrate in place rather than completing a circuit lap.
Comparing the slow drift speed to the high AC frequency proves that AC electrons cannot traverse the entire circuit.

Anahtar Kavram

Electron Drift and Oscillation in AC vs. DC
Soru 36Soru →

A DC workbench circuit connected to a fixed-voltage power supply draws a current of 8 mA8\text{ mA} through a 1.5 kΩ1.5\text{ k}\Omega resistor. If the resistor is replaced with a 6.0 kΩ6.0\text{ k}\Omega resistor while the supply voltage remains unchanged, what is the new current flowing through the circuit?

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Cevap: 2 mA2\text{ mA}

Cevap

2 mA2\text{ mA}
The correct current is 2 mA2\text{ mA}. First, applying Ohm's Law to the initial state yields a constant supply voltage of V=I×R=(8 mA)×(1.5 kΩ)=12 VV = I \times R = (8\text{ mA}) \times (1.5\text{ k}\Omega) = 12\text{ V}. When the resistance increases to 6.0 kΩ6.0\text{ k}\Omega, the new current is calculated as I=VR=12 V6.0 kΩ=2 mAI = \frac{V}{R} = \frac{12\text{ V}}{6.0\text{ k}\Omega} = 2\text{ mA}. Proportional reasoning confirms this: quadrupling resistance at constant voltage reduces current to one-fourth of its initial value (8 mA÷4=2 mA8\text{ mA} \div 4 = 2\text{ mA}).

Adım Adım Çözüm

1
Calculate the fixed supply voltage (VV) using the initial current and resistance values.
V=I1×R1=(8 mA)×(1.5 kΩ)=(0.008 A)×(1500 Ω)=12 VV = I_1 \times R_1 = (8\text{ mA}) \times (1.5\text{ k}\Omega) = (0.008\text{ A}) \times (1500\ \Omega) = 12\text{ V}
Ohm's Law states that voltage is the product of current and resistance.
2
Calculate the new current (I2I_2) using the constant supply voltage and the new resistance value.
I2=VR2=12 V6.0 kΩ=12 V6000 Ω=0.002 A=2 mAI_2 = \frac{V}{R_2} = \frac{12\text{ V}}{6.0\text{ k}\Omega} = \frac{12\text{ V}}{6000\ \Omega} = 0.002\text{ A} = 2\text{ mA}
Current equals voltage divided by resistance. Alternatively, since resistance increased by a factor of 4 (1.5 kΩ1.5\text{ k}\Omega to 6.0 kΩ6.0\text{ k}\Omega) at constant voltage, current must decrease by a factor of 4 (8 mA/4=2 mA8\text{ mA} / 4 = 2\text{ mA}).

Anahtar Kavram

Ohm's Law and inverse proportionality between current and resistance at constant voltage
Tahmini Süre:1m 0s
Soru 37Soru →

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.

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Öğeler

Open-circuit cell voltage
Loaded cell terminal voltage
External battery charging voltage
Fundamental definition of the volt

Eşleşmeler

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Cevap

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, E\mathcal{E}) is the total energy per unit charge supplied by a source. Under no-load conditions (I=0I = 0), the terminal voltage equals EMF. When delivering current (I>0I > 0) to a load, internal resistance produces a drop (IrI r), lowering terminal voltage to E−Ir\mathcal{E} - I r. During charging, the external supply must exceed EMF by E+Ir\mathcal{E} + I r to force current inward. Fundamentally, one volt is defined as one joule of energy per coulomb of charge (1 V=1 J/C1\text{ V} = 1\text{ J/C}).

Adım Adım Çözüm

1
Analyze open-circuit state for a DC voltage source
Terminal voltage equals internal electromotive force (V=EV = \mathcal{E})
When no current flows (I=0 AI = 0\text{ A}), no voltage is dropped across internal resistance (Ir=0 VI r = 0\text{ V}).
2
Evaluate terminal voltage during active discharge under load
Terminal voltage drops to V=E−IrV = \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=E+IrV = \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 V=1 J/C1\text{ V} = 1\text{ J/C}
Voltage is defined as electrical potential energy (work in joules) per unit charge (coulombs).

Anahtar Kavram

Electromotive Force (EMF) vs. Terminal Voltage and Internal Resistance Relations
Tahmini Süre:1m 30s
Soru 38Soru →

An alternating current (AC) sinusoidal voltage with a peak value of 10 V10\text{ V} delivers the same electrical power to a resistive heating element as a steady 10 V10\text{ V} direct current (DC) supply.

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Cevap: False

Cevap

False. An alternating current (AC) sine wave with a peak voltage of 10 V10\text{ V} delivers an effective (RMS) voltage of approximately 7.07 V7.07\text{ V}, which provides less electrical power to a resistor than a constant 10 V10\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 (VRMS=0.707×VpeakV_{\text{RMS}} = 0.707 \times V_{\text{peak}}). For a 10 V10\text{ V} peak AC supply, the effective voltage is only 7.07 V7.07\text{ V}, resulting in half the power output of a steady 10 V10\text{ V} DC supply.

Adım Adım Çözüm

1
Identify the relationship between peak voltage and effective (RMS) voltage for alternating current.
VRMS=Vpeak2≈0.707×Vpeak=0.707×10 V=7.07 VV_{\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 V10\text{ V} DC source and the 10 V10\text{ V} peak AC source across a resistor RR.
PDC=(10 V)2R=100RP_{\text{DC}} = \frac{(10\text{ V})^2}{R} = \frac{100}{R}, whereas PAC=(VRMS)2R=(7.07 V)2R=50RP_{\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 V10\text{ V} peak AC source delivers only 50%50\% of the power delivered by the 10 V10\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.

Anahtar Kavram

AC RMS Voltage vs. DC Equivalent Heating Power
Tahmini Süre:35s
Soru 39Soru →

An electronics technician measures a potential difference of 3.5 kV3.5\text{ kV} across the secondary winding of an ignition transformer. What is this electromotive force expressed in volts (V\text{V})?

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Cevap: 3500

Cevap

3,500 V
The prefix 'kilo-' (abbreviated as 'k') indicates a factor of 10001{}000. Therefore, an electromotive force of 3.5 kV3.5\text{ kV} is equivalent to 3.5×1000=3500 V3.5 \times 1{}000 = 3{}500\text{ V}.

Adım Adım Çözüm

1
Identify the standard metric prefix multiplier for kilo (k)
The prefix kilo represents a multiplier of 10310^3, or 10001{}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
3.5×1000=3500 V3.5 \times 1{}000 = 3{}500\text{ V}
To convert from a larger prefixed unit (kilovolts) to the base electrical unit (volts), multiply by 10001{}000.

Anahtar Kavram

Metric prefix conversion of electrical potential difference and electromotive force (EMF)
Tahmini Süre:30s
Soru 40Soru →

An avionics technician is testing a dual-polarity DC power supply that provides +12 V+12\text{ V} on the positive rail and −12 V-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?

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Cevap: 24 V24\text{ V}

Cevap

24 V24\text{ V}
Voltage (electromotive force) is fundamentally a measure of electric potential difference between two specific points. By subtracting the negative rail potential (−12 V-12\text{ V}) from the positive rail potential (+12 V+12\text{ V}), the resulting potential difference is (+12 V)−(−12 V)=24 V(+12\text{ V}) - (-12\text{ V}) = 24\text{ V}.

Adım Adım Çözüm

1
Identify the reference potentials of both supply rails relative to common ground
Vpositive=+12 VV_{\text{positive}} = +12\text{ V} and Vnegative=−12 VV_{\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
Vload=Vpositive−Vnegative=12 V−(−12 V)=24 VV_{\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.

Anahtar Kavram

Voltage as Electric Potential Difference
Tahmini Süre:45s
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