Basic Electrical Concepts

189 soru

Soru 121Soru →

An industrial temperature sensor module with a calibrated internal resistance of 4.5 kΩ4.5\text{ k}\Omega draws a steady direct current of 8 mA8\text{ mA} during normal operation. What is the voltage drop across the sensor module in volts (V\text{V})?

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Cevap: 36

Cevap

The voltage drop across the sensor module is 36 V36\text{ V}.
According to Ohm's Law, voltage (VV) equals current (II) multiplied by resistance (RR). Given I=8 mA=0.008 AI = 8\text{ mA} = 0.008\text{ A} and R=4.5 kΩ=4500 ΩR = 4.5\text{ k}\Omega = 4500\text{ }\Omega, the calculation is V=0.008 A×4500 Ω=36 VV = 0.008\text{ A} \times 4500\text{ }\Omega = 36\text{ V}.

Adım Adım Çözüm

1
Convert the given metric prefixes to standard base units (amperes and ohms).
I=8 mA=8×10−3 A=0.008 AI = 8\text{ mA} = 8 \times 10^{-3}\text{ A} = 0.008\text{ A} and R=4.5 kΩ=4.5×103 Ω=4500 ΩR = 4.5\text{ k}\Omega = 4.5 \times 10^{3}\text{ }\Omega = 4500\text{ }\Omega
Ohm's law requires fundamental SI units (volts, amperes, and ohms) for accurate mathematical evaluation.
2
Select the appropriate form of Ohm's Law to calculate voltage (VV).
V=I×RV = I \times R
Voltage across a resistive load equals the product of the current flowing through it and its resistance.
3
Substitute the converted values into the formula and solve.
V=0.008 A×4500 Ω=36 VV = 0.008\text{ A} \times 4500\text{ }\Omega = 36\text{ V}
Multiplying 8×10−3 A8 \times 10^{-3}\text{ A} by 4.5×103 Ω4.5 \times 10^{3}\text{ }\Omega simplifies directly to 8×4.5=36 V8 \times 4.5 = 36\text{ V} because the metric prefixes milli (10−310^{-3}) and kilo (10310^{3}) cancel each other.

Anahtar Kavram

Ohm's Law: Voltage Calculation (V=I×RV = I \times R)

Alternatif Yöntem

Recognize that the prefix multiplier milli- (10−310^{-3}) and kilo- (10310^{3}) cancel out when multiplied together: (8 mA)×(4.5 kΩ)=8×4.5 V=36 V(8\text{ mA}) \times (4.5\text{ k}\Omega) = 8 \times 4.5\text{ V} = 36\text{ V}.
Tahmini Süre:1m 0s
Soru 122Soru →

In basic electronics, distinguishing between charge behavior, measurement units, and current flow conventions is essential. Match each electrical concept on the left with its correct definition or governing rule on the right.

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Öğeler

Electron Flow Theory
Conventional Current Theory
Electric Current (II)
One Ampere (1 A1\text{ A})

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Cevap

Electron Flow Theory matches the movement of negative charge carriers from negative to positive; Conventional Current Theory matches the historical assumption of flow from positive to negative; Electric Current matches the rate of charge flow over time; One Ampere matches one coulomb of charge per second.
Each electrical term accurately maps to its standard scientific definition: Electron Flow describes negative electrons drifting toward the positive terminal; Conventional Current describes the standard positive-to-negative circuit model; Electric Current represents the rate of charge flow (I=Q/tI = Q/t); and One Ampere specifically equals one coulomb passing a given point in one second (1 C/s1\text{ C/s}).

Adım Adım Çözüm

1
Identify the physical reality of electron drift versus historical convention.
Electrons are negatively charged, so they drift from negative to positive (Electron Flow). Conventional current historically assumes flow from positive to negative.
Clear differentiation between historical convention and physical subatomic movement is foundational in electrical fundamentals.
2
Define the quantity of electric current.
Current (II) represents the rate of charge (QQ) flow per unit time (tt), yielding I=QtI = \frac{Q}{t}.
Current measures the flow rate of charge through a conductor, not electrical pressure or potential difference.
3
Relate the ampere to fundamental SI electrical units.
1 ampere=1 coulomb per second1\text{ ampere} = 1\text{ coulomb per second} (1 A=1 C/s1\text{ A} = 1\text{ C/s}).
The ampere quantifies current flow as a unit of coulombs transferred over time in seconds.

Anahtar Kavram

Fundamental definitions of electric current, charge rate, ampere units, and directional flow models.

Alternatif Yöntem

Recall the fundamental dimensional relationships: Current is I=ΔQΔtI = \frac{\Delta Q}{\Delta t}, so its unit (1 A1\text{ A}) must be equal to 1 C1 s\frac{1\text{ C}}{1\text{ s}}. Then remember that because electrons possess a negative charge, like charges repel and opposite charges attract, forcing them to move away from the negative terminal toward the positive terminal.
Tahmini Süre:45s
Soru 123Soru →

In electrical and electronic systems, understanding the fundamental physical properties governing current flow is essential. Match each resistance-related concept on the left with its correct physical description on the right.

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Öğeler

Electrical Resistance (RR)
Material Resistivity (ρ\rho)
American Wire Gauge (AWG)
Positive Temperature Coefficient

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Cevap

Electrical Resistance matches the total opposition of a specific component in ohms; Material Resistivity matches the intrinsic geometry-independent material property; American Wire Gauge matches the inverse sizing scale where smaller numbers mean thicker wire; and Positive Temperature Coefficient matches the increase in resistance as temperature rises.
Each electrical concept matches its core definition: Electrical Resistance (RR) is the total opposition to current flow in ohms; Material Resistivity (ρ\rho) is the intrinsic material property; American Wire Gauge (AWG) is the inverse sizing standard where smaller numbers indicate thicker conductors; and a Positive Temperature Coefficient defines the increase of resistance with rising temperature.

Adım Adım Çözüm

1
Distinguish between total electrical resistance and intrinsic material resistivity.
Electrical resistance (RR) is the overall opposition in a specific conductor measured in ohms (Ω\Omega), while resistivity (ρ\rho) is the inherent property of the material itself.
Total resistance depends on both the material's resistivity and its physical dimensions according to R=ρLAR = \rho \frac{L}{A}.
2
Analyze standard conductor sizing and thermal behavior principles.
AWG numbers inversely scale with conductor diameter (smaller number = thicker wire), and a positive temperature coefficient means resistance rises with heat.
Wire gauge dictates cross-sectional area, and thermal excitation in metals increases electron collisions, raising resistance.

Anahtar Kavram

Core physical factors and definitions governing electrical resistance and resistivity in conductors
Tahmini Süre:45s
Soru 124Soru →

In direct-current (DC) circuit analysis, different forms of Watt's Law are used depending on which electrical quantities are known. Match each set of known circuit parameters on the left with the correct formula on the right used to calculate the target value directly.

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Öğeler

Known voltage (VV) and current (II) to calculate total power (PP)
Known current (II) and resistance (RR) to calculate power dissipation (PP)
Known voltage (VV) and resistance (RR) to calculate power dissipation (PP)
Known power rating (PP) and operating voltage (VV) to calculate operating current (II)

Eşleşmeler

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Cevap

Each given set of circuit parameters matches the direct algebraic form of Watt's Law designed for those specific variables: Voltage and Current to P=V×IP = V \times I; Current and Resistance to P=I2×RP = I^2 \times R; Voltage and Resistance to P=V2RP = \frac{V^2}{R}; and Power and Voltage to I=PVI = \frac{P}{V}.
Each match correctly pairs the given known variables with the exact Watt's Law equation variant required to solve directly for the unknown quantity without needing extraneous calculation steps.

Adım Adım Çözüm

1
Identify the base definition of electrical power in terms of voltage and current.
P=V×IP = V \times I
Power in watts equals electromotive force in volts multiplied by electron current flow in amperes.
2
Derive the power formula when only current and resistance are given using Ohm's Law (V=I×RV = I \times R).
P=(I×R)×I=I2×RP = (I \times R) \times I = I^2 \times R
Replacing voltage with the product of current and resistance calculates power without needing voltage explicitly.
3
Derive the power formula when only voltage and resistance are given using Ohm's Law (I=VRI = \frac{V}{R}).
P=V×(VR)=V2RP = V \times \left(\frac{V}{R}\right) = \frac{V^2}{R}
Replacing current with voltage divided by resistance calculates power without needing current explicitly.
4
Rearrange the primary power formula to solve for current when power and voltage are known.
I=PVI = \frac{P}{V}
Dividing both sides of P=V×IP = V \times I by voltage isolates the current parameter.

Anahtar Kavram

Watt's Law and its derived variations relate electrical power (PP), voltage (VV), current (II), and resistance (RR).
Tahmini Süre:45s
Soru 125Soru →

An avionics technician tests a direct current (DC) relay control circuit connected to a 36 V36\text{ V} power supply. An in-line ammeter indicates that the relay coil draws a steady current of 75 mA75\text{ mA}. What is the total electrical resistance of the relay coil?

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Cevap: 480 Ω480\text{ }\Omega

Cevap

The total electrical resistance of the relay coil is 480 Ω480\text{ }\Omega.
According to Ohm's Law, resistance is determined by dividing voltage by current (R=VIR = \frac{V}{I}). First, converting the current from milliamperes to amperes gives 75 mA=0.075 A75\text{ mA} = 0.075\text{ A}. Dividing the 36 V36\text{ V} supply by 0.075 A0.075\text{ A} yields a resistance of 480 Ω480\text{ }\Omega.

Adım Adım Çözüm

1
Convert the measured current from milliamperes to base amperes.
I=75 mA=75×10−3 A=0.075 AI = 75\text{ mA} = 75 \times 10^{-3}\text{ A} = 0.075\text{ A}
Standard electrical calculations using Ohm's law require values to be expressed in base SI units (volts, amperes, and ohms).
2
Apply Ohm's law rearranged to solve for resistance (R=VIR = \frac{V}{I}).
R=36 V0.075 A=480 ΩR = \frac{36\text{ V}}{0.075\text{ A}} = 480\text{ }\Omega
Resistance equals electromotive force (voltage) divided by current flow.

Anahtar Kavram

Calculating resistance in a DC circuit using Ohm's Law with metric prefix conversion.
Tahmini Süre:1m 0s
Soru 126Soru →

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

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

Cevap

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

Adım Adım Çözüm

1
Identify the nature of the power source and charge carriers
The circuit is powered by a direct-current (DC) source where charge carriers in metallic conductors are negatively charged electrons.
Understanding the carrier polarity is essential for determining movement under an electric field.
2
Apply electrostatic attraction and repulsion principles
Electrons carry a negative charge and are therefore repelled by the negative battery terminal and attracted toward the positive terminal.
Like charges repel and opposite charges attract.
3
Determine the physical direction in the external path
In the external circuit outside the power supply, electron flow proceeds from the negative terminal through the load to the positive terminal.
Electron flow theory describes the actual physical movement of electrons.

Anahtar Kavram

Electron flow theory dictates that in a DC circuit, electrons drift through the external conductor from the negative terminal to the positive terminal.
Tahmini Süre:35s
Soru 127Soru →

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

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Cevap: Increasing the electric current flowing through the coil

Cevap

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

Adım Adım Çözüm

1
Identify the factors determining the magnetic field strength of an electromagnet
The field strength of a solenoid or electromagnet depends directly on the electric current, the number of coil turns, and the magnetic permeability of the core.
Electromagnetic flux density is directly proportional to ampere-turns (I×NI \times N) and core permeability.
2
Evaluate the proposed modification against the operational formula
Raising the current (II) increases the total magnetomotive force (MMF=N⋅IMMF = N \cdot I), thereby producing a stronger magnetic field.
Greater electron movement generates more concentric magnetic lines of force around each loop of wire.

Anahtar Kavram

The strength of an electromagnet's magnetic field is directly proportional to the magnitude of current flowing through its windings.
Tahmini Süre:45s
Soru 128Soru →

Which of the following factors causes an increase in the electrical resistance of a solid metallic wire?

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Cevap: Increasing the operating temperature of the wire

Cevap

Increasing the operating temperature of the wire causes an increase in its electrical resistance.
In metallic conductors, electrical resistance increases as temperature rises. The added thermal energy causes the atoms in the metal lattice to vibrate more vigorously, resulting in more frequent collisions with moving free electrons and hindering the flow of electric current.

Adım Adım Çözüm

1
Identify the relationship between temperature and electrical resistance in metallic conductors.
Standard metals have a positive temperature coefficient of resistance, meaning resistance rises as temperature increases.
Thermal energy causes atoms in the metal lattice to vibrate more intensely, increasing collisions with flowing electrons.
2
Evaluate the effects of wire geometry and material properties using the resistance formula.
Electrical resistance is given by R=ρLAR = \rho \frac{L}{A}. Resistance increases with greater length (LL) and higher resistivity (ρ\rho), but decreases with larger cross-sectional area (AA) or higher conductivity.
Analyzing all choices confirms that among the options provided, only increasing the temperature results in higher resistance.

Anahtar Kavram

Physical factors governing conductor resistance (temperature coefficient, geometry, and material resistivity)
Tahmini Süre:45s
Soru 129Soru →

A technician is servicing an off-grid solar telemetry station and measures the operating parameters of four distinct direct current (DC) load circuits. Match each circuit description on the left with its corresponding calculated electrical quantity on the right using Ohm's law.

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Öğeler

A 24 V24\text{ V} cooling fan circuit with a measured internal resistance of 120 Ω120\text{ }\Omega
A 48 V48\text{ V} remote sensor beacon drawing a steady current of 16 mA16\text{ mA}
An environmental heating strip drawing 2.5 A2.5\text{ A} when supplied with 15 V15\text{ V}
A backup LED indicator branch drawing 50 mA50\text{ mA} through a 180 Ω180\text{ }\Omega resistor

Eşleşmeler

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Cevap

The 24 V24\text{ V} fan at 120 Ω120\text{ }\Omega matches 0.20 A0.20\text{ A}; the 48 V48\text{ V} beacon at 16 mA16\text{ mA} matches 3.0 kΩ3.0\text{ k}\Omega; the 15 V15\text{ V} heater drawing 2.5 A2.5\text{ A} matches 6.0 Ω6.0\text{ }\Omega; and the 50 mA50\text{ mA} indicator across 180 Ω180\text{ }\Omega matches 9.0 V9.0\text{ V}.
Each match correctly applies the appropriate algebraic variation of Ohm's law (V=I×RV = I \times R, I=VRI = \frac{V}{R}, or R=VIR = \frac{V}{I}) while properly converting metric units (1 mA=0.001 A1\text{ mA} = 0.001\text{ A} and 1 kΩ=1000 Ω1\text{ k}\Omega = 1000\text{ }\Omega).

Adım Adım Çözüm

1
Calculate the current for the cooling fan circuit
I=VR=24 V120 Ω=0.20 AI = \frac{V}{R} = \frac{24\text{ V}}{120\text{ }\Omega} = 0.20\text{ A}
Ohm's law defines current as voltage divided by resistance.
2
Calculate the resistance for the remote sensor beacon
R=VI=48 V0.016 A=3000 Ω=3.0 kΩR = \frac{V}{I} = \frac{48\text{ V}}{0.016\text{ A}} = 3000\text{ }\Omega = 3.0\text{ k}\Omega
Convert milliamperes to amperes (16 mA=0.016 A16\text{ mA} = 0.016\text{ A}) and solve for resistance by dividing voltage by current.
3
Calculate the resistance for the environmental heating strip
R=VI=15 V2.5 A=6.0 ΩR = \frac{V}{I} = \frac{15\text{ V}}{2.5\text{ A}} = 6.0\text{ }\Omega
Ohm's law defines resistance as voltage divided by current.
4
Calculate the voltage for the backup LED indicator branch
V=I×R=0.050 A×180 Ω=9.0 VV = I \times R = 0.050\text{ A} \times 180\text{ }\Omega = 9.0\text{ V}
Convert milliamperes to amperes (50 mA=0.050 A50\text{ mA} = 0.050\text{ A}) and calculate voltage by multiplying current by resistance.

Anahtar Kavram

Ohm's Law states the fundamental relationship between voltage, current, and resistance in a direct current circuit: V=I×RV = I \times R, I=VRI = \frac{V}{R}, and R=VIR = \frac{V}{I}.
Tahmini Süre:1m 15s
Soru 130Soru →

Electromagnetism and induction are core principles governing how electric circuits interact with magnetic fields. Match each electromagnetic concept with its correct definition or operational description.

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Öğeler

Electromagnet
Electromagnetic induction
Mutual induction
Magnetic field lines

Eşleşmeler

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Cevap

Electromagnet matches the temporary magnet created by electric current through a coil; Electromagnetic induction matches the generation of voltage from a changing magnetic field; Mutual induction matches voltage induced in a nearby coil by varying current; Magnetic field lines match imaginary loops representing flux direction and strength.
Each electromagnetic term is matched to its precise definition: an electromagnet creates a magnetic field from electric current; electromagnetic induction creates voltage from a changing magnetic field; mutual induction describes two inductively coupled coils; and magnetic field lines visually represent flux patterns.

Adım Adım Çözüm

1
Identify the basic construction and function of an electromagnet.
An electromagnet requires electric current flowing through a coil wound on a ferromagnetic core to produce magnetism.
This pairs 'Electromagnet' directly with the definition describing a temporary magnet formed by current through a wire coil around a core.
2
Define the fundamental principle of electromagnetic induction.
Discovered by Faraday, moving a conductor through a magnetic field or exposing it to changing flux creates an induced voltage.
This matches 'Electromagnetic induction' with the generation of voltage across a conductor experiencing changing magnetic flux.
3
Distinguish mutual induction from general induction.
Mutual induction specifically requires interaction between two distinct coils (primary and secondary).
This links 'Mutual induction' to the definition involving a varying current in one coil inducing voltage across a separate nearby coil.
4
Identify the visual concept used for magnetic flux.
Field lines illustrate lines of magnetic force and field intensity.
This pairs 'Magnetic field lines' with the representation of flux strength and direction.

Anahtar Kavram

Core terminology and physical principles of electromagnetism and induction in basic electrical systems.
Tahmini Süre:45s
Soru 131Soru →

An electrical circuit that transfers 180 coulombs180\text{ coulombs} of charge through a conductor over a span of 3 minutes3\text{ minutes} maintains an electric current of 60 amperes60\text{ amperes}.

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Cevap: False

Cevap

False. The actual current is 1 A1\text{ A} because time must be expressed in seconds (3 minutes=180 seconds3\text{ minutes} = 180\text{ seconds}) when calculating current in amperes (I=QtI = \frac{Q}{t}).
The statement is false because the ampere represents charge transfer per second (1 A=1 C/s1\text{ A} = 1\text{ C/s}). A charge of 180 coulombs180\text{ coulombs} flowing over 3 minutes3\text{ minutes} (180 seconds180\text{ seconds}) yields an electric current of exactly 1 ampere1\text{ ampere}, not 60 amperes60\text{ amperes}.

Adım Adım Çözüm

1
Convert the elapsed time from minutes to seconds
3 minutes×60 seconds/minute=180 seconds3\text{ minutes} \times 60\text{ seconds/minute} = 180\text{ seconds}
The base unit of electric current (ampere) is defined as coulombs per second (1 A=1 C/s1\text{ A} = 1\text{ C/s}).
2
Calculate the electric current using the charge rate formula I=QtI = \frac{Q}{t}
I=180 C180 s=1 AI = \frac{180\text{ C}}{180\text{ s}} = 1\text{ A}
Electric current is the rate of charge flow through a cross-section of a conductor.
3
Compare the calculated value with the statement's value
The calculated current is 1 A1\text{ A}, whereas the statement asserts 60 A60\text{ A}.
The statement erroneously divides charge by time in minutes rather than seconds.

Anahtar Kavram

Electric current is defined as the rate of charge flow per unit time, where one ampere equals one coulomb per second (I=QtI = \frac{Q}{t}).
Tahmini Süre:45s
Soru 132Soru →

A continuous spool of aluminum hookup wire has a total length of 80 meters80\text{ meters} and a measured electrical resistance of 2.4 Ω2.4\ \Omega. If an electronics technician cuts a 20-meter20\text{-meter} piece from this spool for a project, what is the electrical resistance of the cut piece in ohms (Ω\Omega)?

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Cevap: 0.6

Cevap

0.6 Ω
The electrical resistance of a uniform wire is directly proportional to its length (R=ρLAR = \rho \frac{L}{A}). Since 20 m20\text{ m} is one-fourth (14\frac{1}{4} or 0.250.25) of the original 80 m80\text{ m} length, the resistance of the cut piece is one-fourth of the total resistance: 2.4 Ω×0.25=0.6 Ω2.4\ \Omega \times 0.25 = 0.6\ \Omega.

Adım Adım Çözüm

1
Identify the physical relationship governing wire resistance and length.
Resistance is directly proportional to length: R=ρLAR = \rho \frac{L}{A}, which means R2=R1×L2L1R_2 = R_1 \times \frac{L_2}{L_1}.
Because the material (resistivity ρ\rho) and cross-sectional area (AA) remain constant, resistance scales directly with length.
2
Find the proportion of the new length relative to the original length.
LcutLtotal=20 m80 m=0.25\frac{L_{\text{cut}}}{L_{\text{total}}} = \frac{20\text{ m}}{80\text{ m}} = 0.25
This establishes what fraction of the total opposition to electron flow the shorter segment represents.
3
Calculate the resistance of the 20-meter wire segment.
Rcut=2.4 Ω×0.25=0.6 ΩR_{\text{cut}} = 2.4\ \Omega \times 0.25 = 0.6\ \Omega
Multiplying the total resistance by the length ratio yields the exact resistance of the cut section.

Anahtar Kavram

The electrical resistance of a conductor is directly proportional to its length (R∝LR \propto L) when cross-sectional area and material remain constant.
Tahmini Süre:45s
Soru 133Soru →

An HVAC technician is testing a motorized zone damper actuator powered by a 24 V24\text{ V} direct current (DC) power supply. An ammeter connected in series measures a current draw of 400 mA400\text{ mA} through the actuator coil. What is the electrical resistance of the actuator coil?

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Cevap: 60 Ω60\text{ }\Omega

Cevap

The electrical resistance of the damper actuator coil is 60 Ω60\text{ }\Omega.
According to Ohm's Law, resistance is calculated by dividing the voltage across the component by the current flowing through it (R=VIR = \frac{V}{I}). Converting 400 mA400\text{ mA} to amperes yields 0.4 A0.4\text{ A}. Dividing 24 V24\text{ V} by 0.4 A0.4\text{ A} gives 60 Ω60\text{ }\Omega.

Adım Adım Çözüm

1
Convert the measured current from milliamperes (mA) to base amperes (A)
I=400 mA=4001000 A=0.4 AI = 400\text{ mA} = \frac{400}{1000}\text{ A} = 0.4\text{ A}
Standard electrical calculations using Ohm's Law require current to be expressed in amperes.
2
State and apply the Ohm's Law formula solved for resistance (RR)
R=VIR = \frac{V}{I}
Resistance equals electromotive force (voltage) divided by current.
3
Substitute the known values and calculate the resistance
R=24 V0.4 A=60 ΩR = \frac{24\text{ V}}{0.4\text{ A}} = 60\text{ }\Omega
Dividing 2424 by 0.40.4 gives the coil resistance in ohms.

Anahtar Kavram

Ohm's Law (R=VIR = \frac{V}{I}) and metric prefix conversion
Tahmini Süre:1m 0s
Soru 134Soru →

A technician is testing a direct current (DC) cooling fan motor installed in an equipment rack. When powered by a 48 V48\text{ V} DC supply, the motor draws a steady current of 250 mA250\text{ mA}. What is the electrical resistance of the cooling fan motor in ohms (Ω\Omega)?

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Cevap: 192

Cevap

The electrical resistance of the cooling fan motor is 192 Ω192\text{ }\Omega.
According to Ohm's law, resistance is found by dividing voltage by current (R=VIR = \frac{V}{I}). Converting 250 mA250\text{ mA} to 0.25 A0.25\text{ A} and dividing 48 V48\text{ V} by 0.25 A0.25\text{ A} gives exactly 192 Ω192\text{ }\Omega.

Adım Adım Çözüm

1
Convert the current from milliamperes to amperes
I=250 mA=0.25 AI = 250\text{ mA} = 0.25\text{ A}
Ohm's law calculations require current to be expressed in the base SI unit of amperes.
2
Calculate resistance using Ohm's law formula (R=VIR = \frac{V}{I})
R=48 V0.25 A=192 ΩR = \frac{48\text{ V}}{0.25\text{ A}} = 192\text{ }\Omega
Dividing the supply voltage by the total circuit current yields the resistance in ohms.

Anahtar Kavram

Calculating resistance using Ohm's law (R=VIR = \frac{V}{I}) combined with metric prefix conversion from milliamperes to amperes.
Tahmini Süre:1m 0s
Soru 135Soru →

According to the principles of electromagnetic induction, what occurs in a closed wire loop when the surrounding magnetic field changes?

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Cevap: An electric current is induced to flow through the wire.

Cevap

An electric current is induced to flow through the wire.
Electromagnetic induction occurs when a changing magnetic field cuts across a conductor, creating an electromotive force (voltage). In a closed loop, this induced voltage causes an electric current to flow.

Adım Adım Çözüm

1
Identify the relationship between changing magnetic fields and conductors
A changing magnetic field cutting across a conductor generates an electromotive force (EMF).
Faraday's Law of Induction establishes that voltage is generated whenever magnetic flux lines cross a conductive path.
2
Apply the generated electromotive force to a closed circuit
The induced EMF causes electric current to circulate through the complete loop.
In any closed conductive loop, an induced potential difference creates electron flow.

Anahtar Kavram

Electromagnetic induction (generation of voltage and current via changing magnetic flux)
Tahmini Süre:40s
Soru 136Soru →

A continuous direct-current (DC) power feed delivers a steady current of 5 amperes5\text{ amperes} through a heavy-duty copper busbar. Which statement correctly describes the physical flow of charge representing this 5 A5\text{ A} current?

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Cevap: Five coulombs of electric charge pass through a cross-section of the busbar each second.

Cevap

Five coulombs of electric charge pass through a cross-section of the busbar each second.
Electric current (II) measures the rate at which charge passes through a conductor. One ampere (1 A1\text{ A}) is defined as one coulomb (1 C1\text{ C}) of electrical charge passing through a cross-sectional area in one second (1 s1\text{ s}). Therefore, a steady current of 5 A5\text{ A} means that 5 coulombs5\text{ coulombs} of charge pass through the cross-section of the busbar every second.

Adım Adım Çözüm

1
Identify the fundamental definition of electric current and its standard unit.
Current (II) is defined as the rate of transfer of electric charge (QQ) per unit of time (tt), expressed by the formula I=QtI = \frac{Q}{t}.
The ampere is the base SI unit of electrical current, defined as one coulomb of charge passing a point per second (1 A=1 C/s1\text{ A} = 1\text{ C/s}).
2
Apply the given current value of 5 A5\text{ A} to the rate formula.
I=5 A=5 coulombs1 secondI = 5\text{ A} = \frac{5\text{ coulombs}}{1\text{ second}}.
A steady current of 5 A5\text{ A} directly translates to a charge flow rate of 5 coulombs per second5\text{ coulombs per second} through the conductor.

Anahtar Kavram

Electric current is the time rate of charge flow through a conductor, where one ampere equals one coulomb per second (I=Q/tI = Q/t).
Tahmini Süre:40s
Soru 137Soru →

A magnetic field is generated around a conductor only when alternating current (AC) flows through it, whereas direct current (DC) produces no magnetic field.

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Cevap: False

Cevap

False
The statement is false because any flow of electric current through a conductor creates a magnetic field. Direct current (DC) produces a steady, non-fluctuating magnetic field around the wire, while alternating current (AC) produces an expanding and collapsing magnetic field.

Adım Adım Çözüm

1
Identify the fundamental relationship between electric current and magnetic fields.
Electric current is the movement of electric charge, and any moving charge produces a magnetic field around its path.
This is the core principle of electromagnetism discovered by Hans Christian Oersted.
2
Evaluate the behavior of direct current (DC) in a conductor.
A steady direct current produces a steady, stationary magnetic field forming concentric circular lines of flux around the conductor.
A constant current still consists of moving electrical charges, which continuously maintain a magnetic field.
3
Distinguish between magnetic field generation and electromagnetic induction.
While electromagnetic induction (generating a voltage in another conductor) requires a changing magnetic field (such as from AC), generating a magnetic field itself requires only the presence of electric current.
Confusing field generation with transformer induction leads to the false belief that DC cannot produce magnetism.

Anahtar Kavram

Magnetic Field Generation by Electric Current
Tahmini Süre:35s
Soru 138Soru →

In direct-current (DC) circuits, understanding the fundamental definitions, units, and directional models of electric charge is essential for troubleshooting. Match each electrical current concept on the left with its correct physical description or definition on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Electron Flow Theory
Conventional Current Theory
Electric Current Rate (1 A1\text{ A})
Total Electric Charge (1 C1\text{ C})

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Electron Flow Theory pairs with the physical drift of negative charge carriers from negative to positive; Conventional Current Theory pairs with the classical convention of positive charge moving from positive to negative; Electric Current Rate (1 A1\text{ A}) pairs with one coulomb passing per second; and Total Electric Charge (1 C1\text{ C}) pairs with the quantity of charge from 6.24×10186.24 \times 10^{18} electrons.
Each concept accurately reflects its physical and scientific definition: electron flow describes negative carriers moving negative-to-positive, conventional flow describes positive-to-negative modeling, an ampere defines charge rate per second, and a coulomb defines total accumulated electron charge.

Adım Adım Çözüm

1
Differentiate between the two current flow directional models
Electron flow represents actual negative charge carriers moving from negative to positive terminals, while conventional flow designates positive charge moving from positive to negative terminals.
Clear separation of physical electron movement from historical circuit conventions prevents directional polarity errors.
2
Distinguish rate of flow (current) from static quantity of charge
Current measured in amperes (1 A=1 C/s1\text{ A} = 1\text{ C/s}) represents the rate of charge transfer, whereas charge measured in coulombs (1 C≈6.24×1018 electrons1\text{ C} \approx 6.24 \times 10^{18}\text{ electrons}) represents the accumulated charge quantity.
Differentiating rate (I=Q/tI = Q/t) from total charge (QQ) is fundamental to quantitative circuit analysis.
3
Verify all four concept-to-definition pairings
Each electrical concept on the left uniquely and unambiguously aligns with its physical definition on the right.
Ensures complete matching accuracy without conceptual overlap.

Anahtar Kavram

Electric current is the time rate of charge flow through a conductor (I=Q/tI = Q/t), where electrons physically drift from negative to positive potentials while conventional current is modeled from positive to negative.
Tahmini Süre:45s
Soru 139Soru →

An electronics technician is sorting materials for a repair job based on their electrical properties. Arrange the following materials in order from the highest electrical conductivity (best conductor) to the lowest electrical conductivity (best insulator).

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

Copper, Germanium, Rubber
Metals like copper have a high concentration of free electrons, making them excellent conductors. Group IV elements such as germanium act as semiconductors with intermediate conductivity. Non-metallic compounds like rubber hold their electrons tightly in covalent structures, acting as insulators with the lowest conductivity.

Adım Adım Çözüm

1
Identify the conductor among the materials
Copper is identified as a conductor with high electrical conductivity.
Conductors have loosely bound valence electrons that move freely under an applied voltage.
2
Identify the semiconductor among the materials
Germanium is identified as a semiconductor with moderate electrical conductivity.
Semiconductors have four valence electrons and exhibit conductivity between conductors and insulators.
3
Identify the insulator among the materials
Rubber is identified as an insulator with extremely low electrical conductivity.
Insulators hold valence electrons tightly, offering very high resistance to current flow.
4
Sequence the materials from highest to lowest electrical conductivity
Copper (conductor) →\rightarrow Germanium (semiconductor) →\rightarrow Rubber (insulator)
The ranking follows the natural continuum of charge carrier availability in solids.

Anahtar Kavram

Classification and relative electrical conductivity of conductors, semiconductors, and insulators.
Tahmini Süre:45s
Soru 140Soru →

An automated assembly fixture uses an industrial direct current (DC) solenoid valve to actuate a pneumatic clamp. The solenoid coil has an internal resistance of 600 Ω600\text{ }\Omega and requires a steady current of 35 mA35\text{ mA} to operate reliably. What DC voltage must the power supply deliver to the solenoid valve?

Cevabı ve açıklamayı göster

Cevap: 21 V21\text{ V}

Cevap

21 V21\text{ V}
According to Ohm's Law, voltage is calculated by multiplying current by resistance (V=I×RV = I \times R). Converting the current from milliamperes to amperes gives 35 mA=0.035 A35\text{ mA} = 0.035\text{ A}. Multiplying 0.035 A×600 Ω0.035\text{ A} \times 600\text{ }\Omega results in 21 V21\text{ V}.

Adım Adım Çözüm

1
Convert the operating current from milliamperes (mA\text{mA}) to base amperes (A\text{A}).
35 mA=351,000 A=0.035 A35\text{ mA} = \frac{35}{1,000}\text{ A} = 0.035\text{ A}
Ohm's Law calculations require electrical quantities in standard SI base units.
2
Apply Ohm's Law (V=I×RV = I \times R) to find the required supply voltage.
V=0.035 A×600 Ω=21 VV = 0.035\text{ A} \times 600\text{ }\Omega = 21\text{ V}
Voltage equals the product of current and resistance.

Anahtar Kavram

Ohm's Law (V=I×RV = I \times R) with metric prefix conversion
Tahmini Süre:1m 0s
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