Electronic Components

211 soru

Soru 121Soru →

An electronics technician is assembling an energy-storage circuit for an industrial sensor module. The circuit consists of two 100 μF100\text{ }\mu\text{F} capacitors connected in series, and this entire combination is wired in parallel with a single 150 μF150\text{ }\mu\text{F} capacitor. What is the total equivalent capacitance of this circuit in microfarads (μF\mu\text{F})?

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Cevap: 200

Cevap

The total equivalent capacitance of the circuit is 200 μF200\text{ }\mu\text{F}.
For capacitors in series, total capacitance decreases and is calculated as 1Cseries=1100 μF+1100 μF=2100 μF\frac{1}{C_{\text{series}}} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{100\text{ }\mu\text{F}} = \frac{2}{100\text{ }\mu\text{F}}, giving Cseries=50 μFC_{\text{series}} = 50\text{ }\mu\text{F}. When this combination is connected in parallel with the 150 μF150\text{ }\mu\text{F} capacitor, parallel capacitances add directly: Ctotal=50 μF+150 μF=200 μFC_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}.

Adım Adım Çözüm

1
Calculate the equivalent capacitance of the two 100 μF100\text{ }\mu\text{F} capacitors connected in series using the reciprocal formula 1Cseries=1C1+1C2\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}.
Cseries=50 μFC_{\text{series}} = 50\text{ }\mu\text{F}
Capacitors in series combine inversely, meaning two identical capacitors in series have half the capacitance of one capacitor.
2
Add the capacitance of the parallel branch to the series equivalent: Ctotal=Cseries+C3C_{\text{total}} = C_{\text{series}} + C_3.
Ctotal=50 μF+150 μF=200 μFC_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}
Capacitors connected in parallel add directly because their effective plate areas combine, increasing overall charge storage capacity.

Anahtar Kavram

Series and parallel equivalent capacitance rules (series capacitors combine inversely; parallel capacitors add directly).
Tahmini Süre:1m 0s
Soru 122Soru →

A 10 kΩ10\text{ k}\Omega linear potentiometer is wired as a variable rheostat by tying the center wiper (terminal 2) directly to outside terminal 3. If the control shaft is set to 25%25\% of its total travel starting from terminal 1 toward terminal 3, what is the effective resistance measured between terminal 1 and the combined terminal 2/3 connection?

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Cevap: 2.5 kΩ2.5\text{ k}\Omega

Cevap

2.5 kΩ2.5\text{ k}\Omega
A potentiometer wired with its wiper connected to one outer terminal functions as a two-terminal variable resistor (rheostat). Setting the shaft to 25%25\% rotation from terminal 1 creates a 2.5 kΩ2.5\text{ k}\Omega resistance path between terminal 1 and terminal 2. Because terminal 2 is strapped to terminal 3, the remaining 7.5 kΩ7.5\text{ k}\Omega section of the resistive element is completely bypassed by the short, leaving an effective resistance of 2.5 kΩ2.5\text{ k}\Omega.

Adım Adım Çözüm

1
Determine the resistance of the active resistive track between terminal 1 and the wiper (terminal 2).
R1−2=0.25×10 kΩ=2.5 kΩR_{1-2} = 0.25 \times 10\text{ k}\Omega = 2.5\text{ k}\Omega
Linear potentiometers divide resistance directly in proportion to shaft travel percentage from the reference end.
2
Analyze the state of the remaining track segment between the wiper (terminal 2) and terminal 3.
R2−3=7.5 kΩR_{2-3} = 7.5\text{ k}\Omega is short-circuited.
Connecting terminal 2 directly to terminal 3 creates a zero-resistance path across this section, bypassing it entirely.
3
Calculate the total effective resistance between terminal 1 and the joined 2/3 terminal.
Reff=2.5 kΩR_{\text{eff}} = 2.5\text{ k}\Omega
Current flows through the unshorted R1−2R_{1-2} segment and exits via the combined wiper/terminal 3 connection.

Anahtar Kavram

Rheostat Wiring and Potentiometer Track Division
Tahmini Süre:45s
Soru 123Soru →

A semiconductor diode is a two-terminal electronic component that acts as a one-way valve for electric current. For current to conduct freely through a standard diode in a direct current (DC) circuit, how must the diode be connected relative to the power source?

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Cevap: The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

Cevap

The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.
A diode conducts electric current when it is placed in forward bias. In a direct current circuit, forward bias is established by connecting the anode toward the positive terminal of the power supply and the cathode toward the negative terminal.

Adım Adım Çözüm

1
Identify the two terminals of a standard diode and the condition needed for conduction.
A diode has an anode (p-type material) and a cathode (n-type material), and it conducts current only when forward-biased.
Understanding diode terminal polarity is necessary to determine proper circuit orientation.
2
Determine the voltage orientation required to forward-bias the diode.
The anode must be at a higher electrical potential than the cathode.
Applying a positive voltage to the anode and a negative voltage to the cathode overcomes the internal barrier potential, narrowing the depletion region and permitting current flow.

Anahtar Kavram

A semiconductor diode allows current to flow when it is forward-biased, which occurs when the anode is connected toward the positive supply voltage and the cathode toward the negative side.
Tahmini Süre:45s
Soru 124Soru →

An AC step-down transformer has 600 turns600\text{ turns} on its primary winding and 50 turns50\text{ turns} on its secondary winding. The primary winding is connected to a 120 V120\text{ V} AC source, and the secondary circuit delivers current to a load drawing 2.4 A2.4\text{ A}. What is the current flowing in the primary winding?

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Cevap: 0.2 A0.2\text{ A}

Cevap

0.2 A0.2\text{ A}
In an ideal transformer, the ratio of primary current to secondary current is inversely proportional to the ratio of primary turns to secondary turns (Ip/Is=Ns/NpI_p / I_s = N_s / N_p). Since the transformer has a 12:112:1 step-down turns ratio (600/50=12600/50 = 12), the secondary voltage decreases by a factor of 12 while the primary current must be 12 times smaller than the secondary current: Ip=2.4 A/12=0.2 AI_p = 2.4\text{ A} / 12 = 0.2\text{ A}.

Adım Adım Çözüm

1
Determine the turns ratio between the primary and secondary windings
NpNs=60050=12\frac{N_p}{N_s} = \frac{600}{50} = 12
The turns ratio determines both the voltage transformation ratio and the inverse current transformation ratio.
2
Apply the transformer current relationship to solve for primary current (IpI_p)
IpIs=NsNp  ⟹  Ip=2.4 A×50600=2.4 A12=0.2 A\frac{I_p}{I_s} = \frac{N_s}{N_p} \implies I_p = 2.4\text{ A} \times \frac{50}{600} = \frac{2.4\text{ A}}{12} = 0.2\text{ A}
In an ideal transformer, power in equals power out (VpIp=VsIsV_p I_p = V_s I_s), which means current is inversely proportional to the number of turns.

Anahtar Kavram

Transformer Current and Turns Ratio Relationship
Tahmini Süre:1m 0s
Soru 125Soru →

Match each specialized switching or relay device to its corresponding operational characteristic and actuation method.

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Öğeler

Reed switch
Rotary switch
Contactor
Thermal overload relay

Eşleşmeler

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Cevap

Reed switch matches ferromagnetic flexible metal reeds sealed in glass; Rotary switch matches rotating shaft and wiper contact selecting between three or more positions; Contactor matches heavy-duty electromechanical device for high-current power loads; Thermal overload relay matches monitoring sustained overcurrent using a bimetallic strip.
Each switching component is matched according to its fundamental physics of operation: the reed switch relies on magnetic field induction on ferrous contacts; the rotary switch provides multi-position manual circuit routing; the contactor uses an electromagnetic solenoid for high-amperage switching; and the thermal overload relay uses thermal deflection of a bimetallic strip for motor overload protection.

Adım Adım Çözüm

1
Identify the actuation mechanism of a reed switch.
Recognize that reed switches operate magnetically via enclosed flexible ferrous blades.
Reed switches rely on external magnetic flux rather than manual toggling or internal thermal deflection.
2
Analyze the mechanical multi-position selection function of a rotary switch.
Associate the rotary switch with angular shaft rotation routing a signal to multiple switch poles/throws.
Rotary switches provide multi-throw circuit selection through rotational movement.
3
Distinguish between standard low-power control relays and contactors.
Connect contactors to high-current power switching controlled via electromagnetic coils.
Contactors feature heavy-duty contacts built specifically to handle high current without contact welding.
4
Determine the protective operation of a thermal overload relay.
Pair the thermal overload relay with bimetallic thermal expansion sensing excessive motor current.
Thermal overload relays protect motor windings by converting persistent current into heat that mechanically triggers auxiliary contacts.

Anahtar Kavram

Specialized switch and relay actuation mechanisms and circuit roles
Tahmini Süre:1m 0s
Soru 126Soru →

In order for a standard PN junction diode to conduct conventional electric current, its cathode must be connected to a higher positive electrical potential than its anode.

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Cevap: False

Cevap

False
The statement is false because forward biasing a PN junction diode requires the anode (P-type material) to be at a higher positive potential than the cathode (N-type material). Applying a higher positive potential to the cathode places the diode in reverse bias, which widens the depletion layer and prevents current flow.

Adım Adım Çözüm

1
Identify the relationship between diode terminals and semiconductor materials.
The anode is connected to the P-type semiconductor region, and the cathode is connected to the N-type semiconductor region.
Understanding terminal orientation is essential to determining how external voltage affects the internal PN junction barrier.
2
Evaluate the biasing condition described in the statement.
Connecting the cathode to a more positive potential than the anode creates a reverse-bias condition, widening the depletion region and preventing conventional current conduction.
For conduction to occur, the diode must be forward-biased by applying a more positive potential to the anode relative to the cathode.

Anahtar Kavram

Diode Biasing and Polarity
Tahmini Süre:45s
Soru 127Soru →

An electronics technician connects three inductors in parallel across an AC filter circuit. The inductors have inductance values of 60 mH60\text{ mH}, 30 mH30\text{ mH}, and 20 mH20\text{ mH}, with negligible mutual inductance between them. What is the total equivalent inductance of the circuit in millihenries (mH)?

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Cevap: 10

Cevap

The total equivalent inductance of the parallel circuit is 10 mH10\text{ mH}.
Inductors in parallel without mutual coupling combine according to the reciprocal formula 1Ltotal=1L1+1L2+1L3\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}. Computing 160+130+120=660=110\frac{1}{60} + \frac{1}{30} + \frac{1}{20} = \frac{6}{60} = \frac{1}{10}, and inverting yields 10 mH10\text{ mH}.

Adım Adım Çözüm

1
State the reciprocal formula for inductors connected in parallel
1Ltotal=1L1+1L2+1L3\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}
When inductors are in parallel and uncoupled, the total rate of current change is the sum of the individual branch rates, resulting in a reciprocal combination rule identical to parallel resistors.
2
Substitute the individual inductance values
1Ltotal=160+130+120\frac{1}{L_{\text{total}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}
Direct substitution of the known branch values: L1=60 mHL_1 = 60\text{ mH}, L2=30 mHL_2 = 30\text{ mH}, and L3=20 mHL_3 = 20\text{ mH}.
3
Find the common denominator and sum the terms
1Ltotal=1+2+360=660=110 mH−1\frac{1}{L_{\text{total}}} = \frac{1 + 2 + 3}{60} = \frac{6}{60} = \frac{1}{10}\text{ mH}^{-1}
Using the lowest common denominator (6060) to perform fractional addition.
4
Invert the reciprocal to obtain equivalent inductance
Ltotal=10 mHL_{\text{total}} = 10\text{ mH}
Taking the reciprocal of 110 mH−1\frac{1}{10}\text{ mH}^{-1} gives 10 mH10\text{ mH}.

Anahtar Kavram

Parallel Inductance Calculation
Tahmini Süre:1m 0s
Soru 128Soru →

A quality assurance technician is inspecting a four-band carbon-composition resistor with the color bands in order: Yellow, Violet, Red, and Silver. To meet quality standards, the resistor must operate within its specified manufacturer tolerance. What is the minimum acceptable resistance value, in ohms (Ω\Omega), for this component?

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Cevap: 4230

Cevap

The minimum acceptable resistance value is 4230 Ω4230\text{ }\Omega.
The first two bands (Yellow = 4, Violet = 7) establish the significant figures of 47. The third band (Red) indicates a multiplier of 102=10010^2 = 100, giving a nominal rating of 4700 Ω4700\text{ }\Omega. The fourth band (Silver) specifies a ±10%\pm 10\% tolerance. Subtracting 10%10\% of 4700 Ω4700\text{ }\Omega (470 Ω470\text{ }\Omega) from the nominal rating gives the minimum allowable threshold of 4230 Ω4230\text{ }\Omega.

Adım Adım Çözüm

1
Decode the first two color bands to obtain the significant numerical digits.
Yellow represents the digit 44, and Violet represents the digit 77, forming the base value of 4747.
Standard EIA four-band resistor color coding assigns the first two bands as significant figures of the resistance value.
2
Multiply the base significant figures by the power of ten indicated by the third band.
Red indicates a multiplier of 10210^2 (100100), so the nominal resistance is 47×100=4700 Ω47 \times 100 = 4700\text{ }\Omega.
The third band functions as the decimal multiplier for four-band fixed resistors.
3
Identify the tolerance rating from the fourth band.
Silver represents a tolerance of ±10%\pm 10\%.
The fourth band specifies the allowable deviation range from the nominal resistance value.
4
Calculate the lower allowable resistance threshold.
Tolerance amount is 4700 Ω×0.10=470 Ω4700\text{ }\Omega \times 0.10 = 470\text{ }\Omega. Minimum acceptable value is 4700 Ω−470 Ω=4230 Ω4700\text{ }\Omega - 470\text{ }\Omega = 4230\text{ }\Omega.
The minimum acceptable value represents the nominal value minus the total tolerance deviation.

Anahtar Kavram

Four-Band Resistor Color Code Decoding and Tolerance Range Calculation

Alternatif Yöntem

Multiply the nominal resistance directly by the minimum percentage factor: 4700 Ω×(1.00−0.10)=4700 Ω×0.90=4230 Ω4700\text{ }\Omega \times (1.00 - 0.10) = 4700\text{ }\Omega \times 0.90 = 4230\text{ }\Omega.
Tahmini Süre:45s
Soru 129Soru →

Monolithic integrated circuits contain microscopic electronic components such as transistors, diodes, and resistors fabricated directly onto a single thin piece of semiconductor material. What is this semiconductor base commonly called?

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Cevap: Silicon wafer

Cevap

Silicon wafer
A monolithic integrated circuit is produced by creating all necessary circuit elements—including transistors, diodes, and resistors—on a single thin slice of crystalline semiconductor material (most commonly silicon). In microelectronics manufacturing, this foundation slice is called a silicon wafer.

Adım Adım Çözüm

1
Identify the structural foundation of a monolithic integrated circuit
Monolithic ICs are manufactured as a single, solid physical unit on a continuous semiconductor slice
The word 'monolithic' indicates that all circuit components share a single, unified semiconductor base rather than separate discrete parts
2
Apply standard electronics manufacturing terminology to the semiconductor base
A thin, polished slice of crystalline silicon used as the substrate is termed a wafer
Wafers provide the crystal foundation into which microscopic active and passive components are diffused, etched, and interconnected

Anahtar Kavram

Monolithic integrated circuit construction and semiconductor wafer substrate
Soru 130Soru →

In an electronic power supply circuit, what is the primary function of a rectifier diode?

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Cevap: To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction

Cevap

To convert alternating current (AC) into direct current (DC) by allowing current to flow in only one direction
A semiconductor diode permits electrical current to flow easily in only one direction (when forward-biased) and blocks current flow in the reverse direction (when reverse-biased). In power supply circuits, this unidirectional property is utilized in rectification to convert alternating current (AC) into direct current (DC).

Adım Adım Çözüm

1
Identify the fundamental operational characteristic of a semiconductor diode.
A diode acts as a one-way electrical valve, exhibiting very low resistance in the forward direction and extremely high resistance in the reverse direction.
Understanding the unidirectional conduction property of the PN junction is essential for determining its circuit application.
2
Connect this one-way conduction property to its purpose in a power supply circuit.
When alternating current (AC) is applied, the diode conducts only during the positive half-cycles, producing a pulsating direct current (DC) output.
Rectification is the process of converting bidirectional AC into unidirectional DC.

Anahtar Kavram

Unidirectional current conduction and AC-to-DC rectification by semiconductor diodes
Tahmini Süre:45s
Soru 131Soru →

A technician is installing a master disconnect switch for a 240 V AC equipment circuit supplied by two active 120 V lines (L1L_1 and L2L_2). To completely isolate the load by opening both hot conductors simultaneously using a single mechanical toggle, which type of switch is required?

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Cevap: Double-Pole, Single-Throw (DPST) switch

Cevap

Double-Pole, Single-Throw (DPST) switch
A Double-Pole, Single-Throw (DPST) switch features two independent internal contact sets (poles) mechanically linked to move together under a single toggle. This enables simultaneous opening and closing of both L1L_1 and L2L_2 hot conductors to achieve complete electrical isolation of the equipment.

Adım Adım Çözüm

1
Identify the number of isolated conductive paths (poles) required by the circuit.
Two distinct live conductors (L1L_1 and L2L_2) must be controlled at the same time, which requires a Double-Pole (DP) configuration.
The term 'pole' defines the number of separate, electrically isolated circuits or lines the switch can open or close concurrently.
2
Determine the number of active output positions (throws) required for each pole.
The application requires simple ON/OFF isolation, which corresponds to a Single-Throw (ST) contact arrangement.
The term 'throw' defines the number of closed contact positions available per pole; a single-throw switch has one closed state (ON) and one open state (OFF).

Anahtar Kavram

Switch Pole and Throw Configurations (DPST vs. SPST/SPDT)
Tahmini Süre:1m 0s
Soru 132Soru →

A semiconductor diode is a fundamental component that controls current flow in electronic circuits. Match each diode term on the left with its corresponding characteristic or operational definition on the right.

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Öğeler

Anode
Cathode
Forward bias
Reverse bias

Eşleşmeler

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Cevap

The Anode pairs with the P-type terminal represented by the triangle base, the Cathode pairs with the N-type terminal identified by the painted stripe, Forward bias pairs with the state allowing conduction when the P-type region is positive, and Reverse bias pairs with the state blocking conduction when the N-type region is positive.
A semiconductor diode consists of an anode (P-type side, represented by the base of the triangle symbol) and a cathode (N-type side, indicated by a band on the physical package and a vertical bar on the symbol). Connecting the positive supply to the anode creates forward bias, collapsing the depletion layer and allowing current conduction. Connecting the positive supply to the cathode creates reverse bias, expanding the depletion layer and insulating the circuit.

Adım Adım Çözüm

1
Identify the two terminals of a PN junction diode
The anode represents the P-type material (triangle base on schematics), while the cathode represents the N-type material (vertical line on schematics and painted ring on physical components).
Correctly identifying component leads is essential for analyzing circuit orientation and polarity.
2
Determine the operating conditions for current conduction and cutoff
Applying positive voltage to the anode creates forward bias and allows conduction; applying positive voltage to the cathode creates reverse bias and halts conduction.
Forward bias shrinks the internal depletion layer, whereas reverse bias widens it to prevent electron movement.
3
Match each term to its exact operational and physical descriptor
All four terms are unambiguously paired with their respective terminal characteristics and biasing behaviors.
Verifies complete conceptual understanding of diode fundamentals.

Anahtar Kavram

Diode terminal identification (anode vs. cathode) and junction biasing states (forward vs. reverse bias)
Tahmini Süre:45s
Soru 133Soru →

A double-pole, single-throw (DPST) switch is designed to simultaneously open or close two electrically independent circuits using a single mechanical actuator.

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Cevap: True

Cevap

True. A DPST switch controls two isolated circuit paths simultaneously with a single actuator mechanism.
The statement accurately defines a DPST switch. In electrical engineering and electronics, 'pole' refers to the number of separate circuits controlled by the switch, and 'throw' refers to the number of closed contact positions available per pole. A DPST switch therefore controls two separate circuits with a single ON/OFF throw action.

Adım Adım Çözüm

1
Analyze the switch terminology for poles and throws.
'Double-pole' (DP) signifies two separate, electrically isolated circuits or lines being controlled, while 'single-throw' (ST) indicates that each pole has only one active closed position (ON/OFF).
Understanding standard electrical switch nomenclature is necessary to identify contact arrangements.
2
Evaluate the mechanical and electrical operation of the switch.
The two internal sets of contacts move in unison via one toggle or lever, switching both independent circuits at the same instant.
A single mechanical linkage operates both poles together to ensure simultaneous opening or closing.

Anahtar Kavram

Switch Pole and Throw Nomenclature (DPST Operation)
Tahmini Süre:45s
Soru 134Soru →

A step-down transformer has a 10:110:1 turns ratio between its primary and secondary windings. If the primary winding is connected across a steady 12 V12\text{ V} direct current (DC) power supply, what voltage is measured across the secondary winding under steady-state conditions?

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Cevap: 0 V0\text{ V}

Cevap

0 V0\text{ V}
Transformers operate through mutual electromagnetic induction, requiring an alternating current (AC) to generate an expanding and collapsing magnetic field. A steady direct current (DC) produces a static, unvarying magnetic field once the circuit reaches steady state. Because Faraday's Law requires a changing magnetic field (ΔΦΔt≠0\frac{\Delta\Phi}{\Delta t} \neq 0) to induce an electromotive force, the secondary winding produces exactly 0 V0\text{ V}.

Adım Adım Çözüm

1
Identify the type of input voltage applied to the transformer primary winding.
The input source is a steady 12 V12\text{ V} direct current (DC) supply.
Transformers rely on mutual electromagnetic induction, which fundamentally requires a time-varying magnetic field.
2
Analyze the magnetic flux created in the core during steady-state DC operation.
A steady DC current produces a stationary, unchanging magnetic flux (ΔΦΔt=0\frac{\Delta\Phi}{\Delta t} = 0).
According to Faraday's Law of Induction, voltage is only induced in a conductor when magnetic flux lines expand or collapse across the winding turns.
3
Determine the resulting secondary output voltage.
The induced secondary voltage is 0 V0\text{ V}.
With zero rate of change of magnetic flux in steady state, no electromotive force (EMF) is induced in the secondary winding.

Anahtar Kavram

Transformers require alternating current (AC) to induce a secondary voltage via changing magnetic flux; steady direct current (DC) produces no secondary voltage.
Tahmini Süre:1m 0s
Soru 135Soru →

In an automotive electrical system, an electromechanical relay is installed between a dashboard switch and a high-current cooling fan. What is the primary purpose of using the relay in this circuit configuration?

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Cevap: It allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.

Cevap

The correct answer states that it allows a low-current switch to control a high-current load while reducing voltage drop across long wiring runs.
An electromechanical relay utilizes a low current passing through an internal coil to generate a magnetic field. This magnetic field pulls an armature to close heavy-duty contacts, safely powering high-current loads like cooling fans without subjecting delicate dashboard switches to high current or requiring thick, heavy wiring runs through the passenger compartment.

Adım Adım Çözüm

1
Identify the distinct circuit sections established by the relay.
The dashboard switch operates within a low-current control circuit connected to the relay coil, while the cooling fan operates within a separate high-current load circuit connected through the relay contacts.
Relays provide electrical isolation and remote switching between control circuits and power circuits.
2
Evaluate the technical necessity of isolating high-current loads from cabin switches.
Routing substantial current directly through the dashboard switch would require thick gauge wiring into the vehicle interior, cause significant voltage drop over long distances, and degrade manual switch contacts due to arcing.
Electromechanical relays permit small control currents to safely actuate high-amperage contacts placed near the battery and load.
3
Match the identified operational principle to the corresponding functional description.
The relay serves as a remote electromechanical switch allowing low-current input to control high-current output.
Relays do not act as fuses, voltage transformers, or DC-to-AC inverters.

Anahtar Kavram

Electromechanical relay function in high-current DC switching
Tahmini Süre:1m 0s
Soru 136Soru →

On the cylindrical casing of a standard axial-lead semiconductor diode, what feature is identified by the painted stripe or band near one of the ends?

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Cevap: The cathode terminal

Cevap

The cathode terminal
On standard through-hole axial diodes, the painted stripe marks the cathode terminal. In schematic diagrams, this corresponds to the vertical bar toward which the triangle points.

Adım Adım Çözüm

1
Identify the standard packaging markings used on axial-lead semiconductor diodes.
Standard cylindrical diodes (such as the 1N4001 series) use a single painted band at one end to distinguish polarity.
Because diodes are polarized components that conduct current primarily in one direction, physical orientation must be identifiable.
2
Correlate the physical stripe with the corresponding electronic terminal.
The stripe aligns with the cathode (negative lead during forward conduction), matching the vertical bar in the standard schematic symbol.
Correct alignment ensures the diode is wired in forward bias rather than accidental reverse bias.

Anahtar Kavram

Physical diode lead identification and polarity marking
Tahmini Süre:45s
Soru 137Soru →

In electronic power and communication systems, various inductor and transformer configurations serve specialized functions. Match each inductive component on the left with its primary operational characteristic on the right.

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Öğeler

RF Choke
Isolation Transformer
Air-Core Inductor
Step-Up Transformer

Eşleşmeler

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Cevap

RF Choke matches with presenting high opposition to high-frequency AC signals while allowing DC to pass; Isolation Transformer matches with providing a 1:1 turns ratio to electrically decouple circuits for safety; Air-Core Inductor matches with eliminating magnetic core losses at ultra-high frequencies; Step-Up Transformer matches with increasing alternating voltage while decreasing output current proportionally.
Each inductive device is matched according to its fundamental electromagnetic function: an RF choke blocks AC via inductive reactance while passing DC; an isolation transformer provides 1:1 galvanic isolation; an air-core inductor eliminates ferromagnetic core losses at high frequencies; and a step-up transformer increases voltage while proportionally reducing current.

Adım Adım Çözüm

1
Identify the primary behavior of an RF choke
Recognized as a high-frequency AC blocking inductor that freely passes DC.
Inductive reactance (XL=2πfLX_L = 2\pi f L) increases directly with frequency, creating significant impedance at radio frequencies.
2
Identify the function of an isolation transformer
Recognized as a 1:1 turns ratio transformer used for safety and ground-loop isolation.
Equal winding counts maintain input voltage on the secondary side while eliminating direct conductive connections.
3
Determine the main design advantage of an air-core inductor
Recognized as eliminating ferromagnetic core losses.
Without a solid iron or magnetic core, there are zero eddy currents or magnetic hysteresis losses at high frequencies.
4
Evaluate the voltage and current relationship of a step-up transformer
Recognized as increasing voltage while decreasing current proportionally.
Due to the principle of power conservation (VpIp≈VsIsV_p I_p \approx V_s I_s), increasing secondary voltage requires a corresponding drop in secondary current.

Anahtar Kavram

Operational characteristics and applications of inductors and transformers, including frequency-dependent choke behavior, core losses, galvanic isolation, and turns ratio power relationships.
Tahmini Süre:1m 0s
Soru 138Soru →

An electronics technician selects a four-band carbon-film resistor with the color sequence Brown, Green, Red, and Gold. What is the nominal resistance and tolerance of this resistor?

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Cevap: 1.5 kΩ with ±5% tolerance1.5\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance}

Cevap

1.5 kΩ with ±5% tolerance1.5\text{ k}\Omega\text{ with } \pm 5\%\text{ tolerance} (1,500 Ω±5%1,500\ \Omega \pm 5\%)
Decoding the four color bands in standard order gives: Brown = 1 (first digit), Green = 5 (second digit), Red = 10210^2 or ×100\times 100 (multiplier), and Gold = ±5%\pm 5\% (tolerance). Calculating 15×100 Ω15 \times 100\ \Omega yields 1,500 Ω1,500\ \Omega, which converts to 1.5 kΩ1.5\text{ k}\Omega with a ±5%\pm 5\% tolerance.

Adım Adım Çözüm

1
Identify the first two significant digit color bands.
Brown represents the digit 1, and Green represents the digit 5, forming the number 15.
In a standard four-band resistor, the first two bands establish the base significant figures.
2
Determine the multiplier from the third band.
Red represents a multiplier of 10210^2 (100100).
The third band indicates the power of ten by which the significant digits are multiplied.
3
Calculate the nominal resistance value.
15×100=1,500 Ω=1.5 kΩ15 \times 100 = 1,500\ \Omega = 1.5\text{ k}\Omega.
Multiplying the base digits by the multiplier yields the total nominal resistance.
4
Identify the tolerance from the fourth band.
Gold indicates a tolerance of ±5%\pm 5\%.
The fourth band specifies the acceptable manufacturing variation percentage.

Anahtar Kavram

Four-band resistor color code decoding and tolerance identification
Tahmini Süre:35s
Soru 139Soru →

In electrical and electronic control systems, electromechanical relays and multi-contact switches rely on specialized internal components and contact configurations to direct current and automate circuits. Match each switch or relay component to its primary mechanical or electrical operational function.

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Öğeler

Common (COM) Terminal
Relay Armature
Normally Closed (NC) Contacts
Relay Solenoid Coil

Eşleşmeler

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Cevap

Common (COM) Terminal matches the shared conductive connection point that routes current to selectable paths; Relay Armature matches the movable ferromagnetic component attracted by magnetic flux; Normally Closed (NC) Contacts match the contact pair that maintains a closed conductive path in the resting de-energized state and opens when actuated; Relay Solenoid Coil matches the insulated wire winding that produces an electromagnetic pulling force.
Each component corresponds directly to its mechanical or electrical role in relay operation: the Common terminal serves as the shared current pathway, the armature provides the physical motion via magnetic attraction, Normally Closed contacts provide resting continuity, and the solenoid coil generates the magnetic force from the control circuit.

Adım Adım Çözüm

1
Identify the function of the Common (COM) Terminal
Matched with the shared conductive connection point that routes current to one of multiple selectable contact paths.
In single-pole double-throw (SPDT) configurations and standard relays, the COM terminal is the central node connected internally to either the NO or NC contact.
2
Identify the function of the Relay Armature
Matched with the movable ferromagnetic component that physically transfers motion to open or close contacts.
The armature is the mechanical piece of iron that pivots when drawn toward the core by electromagnetic attraction.
3
Identify the function of the Normally Closed (NC) Contacts
Matched with the contact pair that maintains continuity during the resting de-energized state and opens upon actuation.
'Normally' designates the shelf or unpowered state, meaning NC contacts conduct electricity until an active force separates them.
4
Identify the function of the Relay Solenoid Coil
Matched with the insulated wire winding that produces an electromagnetic pulling force when control current flows through it.
The coil converts electrical input energy into a magnetic field via electromagnetic induction to drive the switching mechanism.

Anahtar Kavram

Electromechanical Relay Structure and Contact Nomenclature

Alternatif Yöntem

Categorize each item first by its physical domain: identify the electrical input terminal (COM), the mechanical moving element (Armature), the resting electrical state (NC contacts), and the electromagnetic actuator (Coil).
Tahmini Süre:1m 0s
Soru 140Soru →

A light-emitting diode (LED) produces light when it is connected in a reverse-bias orientation within a direct current (DC) circuit.

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Cevap: False

Cevap

False. A light-emitting diode (LED) produces light only when forward-biased; in a reverse-bias configuration, it blocks current flow and stays dark.
The statement is false because a light-emitting diode (LED) must be forward-biased (anode connected to the positive potential and cathode connected to the negative potential) to conduct electrical current and release energy in the form of light. In a reverse-bias orientation, the diode blocks current and produces no light.

Adım Adım Çözüm

1
Identify the operating requirement of a semiconductor diode and LED.
Diodes conduct electric current in only one direction (forward bias) and block current in the opposite direction (reverse bias).
An LED functions as a standard PN-junction diode with the specialized property of emitting light when electric current conducts through it.
2
Determine the state of the LED in a reverse-bias orientation.
Under reverse bias, the internal depletion region widens, preventing current flow across the junction.
Because no charge carriers cross the junction under reverse bias, no electron-hole recombination occurs, preventing any light emission.

Anahtar Kavram

Light-emitting diodes (LEDs) require forward biasing (anode connected to a higher positive potential than cathode) to conduct current and emit light.
Tahmini Süre:45s
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