Electronic Components

211 soru

Soru 81Soru →

Match each magnetic component type to its primary operational characteristic and circuit application.

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Öğeler

Air-core inductor
Ferrite-core toroidal inductor
Center-tapped step-down transformer
Variable autotransformer

Eşleşmeler

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Cevap

Air-core inductor matches with eliminating core saturation and hysteresis losses; Ferrite-core toroidal inductor matches with confining magnetic flux within a closed circular path to minimize EMI; Center-tapped step-down transformer matches with supplying two equal-magnitude, out-of-phase AC voltages; Variable autotransformer matches with utilizing a single continuous tapped winding without galvanic isolation.
Each component is correctly matched according to its electromagnetic construction: air-core inductors avoid high-frequency magnetic core losses; toroidal ferrite inductors contain magnetic flux to limit EMI; center-tapped transformers yield balanced dual out-of-phase secondary voltages; and autotransformers perform voltage adjustments via a single shared winding without galvanic isolation.

Adım Adım Çözüm

1
Analyze core material and geometry impacts on the inductor types.
Air lacks ferromagnetic material, meaning it cannot saturate or exhibit core losses at RF frequencies (identifying the air-core inductor). A toroid geometry forms a closed magnetic loop that keeps magnetic fields contained to prevent EMI (identifying the toroidal ferrite inductor).
Inductor core selection balances permeability, saturation limits, operating frequency, and stray field radiation.
2
Analyze winding arrangements and isolation properties for the transformer types.
A center-tapped secondary winding produces two equal-amplitude outputs that are 180^\\circ out of phase relative to the center tap (identifying the center-tapped transformer). An autotransformer uses a single tapped winding shared between input and output, which eliminates electrical isolation (identifying the variable autotransformer).
Winding configuration determines output phase symmetry and whether galvanic isolation exists between input and output circuits.

Anahtar Kavram

Operational characteristics, core geometries, and winding configurations of inductors and transformers
Tahmini Süre:1m 30s
Soru 82Soru →

Which of the following best describes the primary function of an inductor in an electronic circuit?

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Cevap: Opposing changes in electric current by storing energy in a magnetic field

Cevap

Opposing changes in electric current by storing energy in a magnetic field
An inductor is an electrical component designed to store energy within an electromagnetic field generated by current flowing through its coils. Due to self-inductance, it produces an opposing electromotive force whenever current changes, thereby resisting rapid current variations.

Adım Adım Çözüm

1
Identify the fundamental physical construction and behavior of an inductor.
An inductor is made of wire coiled around a core (air, iron, or ferrite).
When electric current passes through the coil, Ampere's law dictates that a magnetic field develops around the conductor.
2
Determine how the component reacts to dynamic circuit conditions.
According to Lenz's Law and Faraday's Law of Induction, any change in current produces a counter-electromotive force (back EMF) that opposes that change.
This property gives inductors their key application as chokes and energy storage devices in magnetic fields.

Anahtar Kavram

Inductors store energy in a magnetic field and oppose changes in electric current.
Tahmini Süre:40s
Soru 83Soru →

A technician needs to wire a single DC power source so that an operator can toggle power between an amber standby indicator and a green run indicator, ensuring only one light is energized at any given time. Which switch type should be selected?

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

Cevap

Single-Pole, Double-Throw (SPDT) switch
A Single-Pole, Double-Throw (SPDT) switch features one common movable contact (the pole) that connects to either of two fixed output contacts (the throws). When flipped, it breaks contact with the first load and makes contact with the second, allowing selective operation between two indicators.

Adım Adım Çözüm

1
Determine the number of input circuits (poles) required.
One input circuit (Single-Pole)
The circuit uses a single DC power feed connected to the switch's common terminal.
2
Determine the number of output positions (throws) required.
Two output paths (Double-Throw)
The operator must alternate power between two distinct loads (amber lamp and green lamp).
3
Combine the pole and throw specifications to identify the proper component.
SPDT switch
A single-pole, double-throw switch routes one common line to one of two selectable output terminals.

Anahtar Kavram

Switch Poles and Throws Nomenclature
Tahmini Süre:45s
Soru 84Soru →

Match each capacitor circuit application with its primary functional role in electronic systems.

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Öğeler

Coupling Capacitor
Filter Capacitor
Bypass Capacitor
Tuning Capacitor

Eşleşmeler

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Cevap

Coupling Capacitor matches transferring AC signal voltages between amplifier stages while blocking DC bias levels; Filter Capacitor matches smoothing pulsating direct current output from a rectifier; Bypass Capacitor matches diverting AC noise or signals directly to ground around a component; Tuning Capacitor matches providing adjustable capacitance to set resonant frequency in RF circuits.
Each application corresponds directly to fundamental capacitive behaviors: coupling capacitors leverage DC-blocking and AC-passing in series with the signal path; filter capacitors utilize charge storage to smooth rectified DC voltage ripple; bypass capacitors create a low-reactance parallel path to ground for AC signals; and tuning capacitors provide variable capacitance for frequency selection in resonant LC circuits.

Adım Adım Çözüm

1
Determine the function of a coupling capacitor in multi-stage signal amplification.
Coupling capacitors block DC bias voltages while passing AC information signals between amplifier stages.
Capacitors provide infinite opposition (reactance) to zero-frequency DC while exhibiting low capacitive reactance to high-frequency AC signals.
2
Determine the function of a filter capacitor in a DC power supply.
Filter capacitors reduce voltage ripple by charging at peak rectified voltages and discharging during valleys.
This energy storage smooths pulsating DC output into a steady DC voltage for electronic circuits.
3
Determine the function of a bypass capacitor across a circuit element.
Bypass capacitors shunt alternating current signals to ground around a component without disturbing the DC operating point.
The capacitor provides an alternate low-reactance path for AC currents.
4
Determine the function of a tuning capacitor in radio frequency systems.
Tuning capacitors vary circuit capacitance to alter the resonant frequency of an LC circuit.
Resonant frequency varies inversely with the square root of capacitance, allowing station or frequency selection.

Anahtar Kavram

Operational roles and circuit applications of capacitors across DC power filtering, signal coupling, AC bypassing, and RF tuning
Tahmini Süre:1m 0s
Soru 85Soru →

A 12 V12\text{ V} DC power source is connected in series with a 100 Ω100\ \Omega load resistor and a parallel pair of silicon diodes (D1D_1 and D2D_2). In the parallel branch, the anode of D1D_1 is connected toward the positive supply terminal, while the cathode of D2D_2 is connected toward the positive supply terminal. Assuming each silicon diode has a standard forward voltage drop of 0.7 V0.7\text{ V} and zero reverse leakage current, what is the voltage drop across the 100 Ω100\ \Omega load resistor?

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Cevap: 11.3 V11.3\text{ V}

Cevap

11.3 V11.3\text{ V}
Because Diode D1D_1 has its anode connected toward the positive terminal of the DC supply, it is forward-biased and conducts current, dropping its forward barrier potential of 0.7 V0.7\text{ V}. Diode D2D_2 has its cathode toward the positive supply, making it reverse-biased, so it acts as an open branch and conducts no current. According to Kirchhoff's Voltage Law, the total source voltage equals the sum of the series voltage drops. Subtracting the 0.7 V0.7\text{ V} diode drop from the 12 V12\text{ V} source gives a remaining 11.3 V11.3\text{ V} across the 100 Ω100\ \Omega resistor.

Adım Adım Çözüm

1
Analyze the biasing condition of each diode in the parallel branch
Diode D1D_1 is forward-biased (anode toward positive terminal) and Diode D2D_2 is reverse-biased (cathode toward positive terminal)
Diodes conduct conventional current only when the anode is positive relative to the cathode (forward bias).
2
Determine the effective voltage drop across the parallel diode combination
The parallel branch voltage is clamped at Vdiodes=0.7 VV_{\text{diodes}} = 0.7\text{ V}
Since D1D_1 is forward-biased, current flows through D1D_1, establishing its characteristic 0.7 V0.7\text{ V} forward barrier drop. D2D_2 is reverse-biased and behaves as an open circuit.
3
Apply Kirchhoff's Voltage Law (KVL) around the single series loop
Vresistor=Vsource−Vdiodes=12 V−0.7 V=11.3 VV_{\text{resistor}} = V_{\text{source}} - V_{\text{diodes}} = 12\text{ V} - 0.7\text{ V} = 11.3\text{ V}
The sum of the voltage drops across the series elements must equal the source voltage.

Anahtar Kavram

Diode biasing state determination and junction voltage drop in series-parallel DC circuits
Tahmini Süre:1m 30s
Soru 86Soru →

In electronics, inductors and transformers rely on magnetic fields to store energy, filter signals, and modify alternating current voltages. Match each component term to its correct definition or operational characteristic.

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Öğeler

Henry (H)
Ferrite Core
Step-up Transformer
Inductive Reactance

Eşleşmeler

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Cevap

Henry (H) matches the standard unit of electrical inductance in a coil; Ferrite Core matches a magnetic ceramic material used in high-frequency coils to minimize energy losses; Step-up Transformer matches a device with more secondary turns than primary turns that increases output AC voltage; Inductive Reactance matches the opposition an inductor offers to the flow of alternating current, measured in ohms.
Each concept accurately pairs with its standard electronic definition: Henry (H) is the unit of inductance; Ferrite Core is the high-frequency ceramic magnetic material; Step-up Transformer increases output AC voltage via greater secondary turns; and Inductive Reactance is the coil's opposition to AC current measured in ohms.

Adım Adım Çözüm

1
Identify the standard unit of measurement for inductance
The Henry (H) is the standard unit of inductance.
Inductance measures a coil's ability to induce back-EMF opposing current changes, measured in henries.
2
Identify the specialized core material used for high frequencies
Ferrite core matches the ceramic magnetic material.
Ferrite reduces eddy current losses at radio and high frequencies due to its high electrical resistivity.
3
Identify the transformer type that raises alternating voltage
A step-up transformer increases output voltage.
By the turns ratio principle, having more secondary turns than primary turns increases secondary AC voltage.
4
Identify the property representing opposition to AC current in ohms
Inductive reactance represents AC opposition in ohms.
Inductive reactance (XL=2πfLX_L = 2\pi f L) quantifies how much an inductor restricts alternating current at a given frequency.

Anahtar Kavram

Basic terminology, units, materials, and operating principles of inductors and transformers.

Alternatif Yöntem

Recall standard electronic abbreviations and formulas: LL is measured in Henries (H), XLX_L is measured in Ohms (Ω\Omega), and Ns>NpN_s > N_p signifies a step-up voltage transformer.
Tahmini Süre:45s
Soru 87Soru →

An electronics technician is testing a bipolar junction transistor (BJT) configured as an electronic switch. When the base receives sufficient input voltage and current to drive the transistor into full conduction with minimum resistance between the collector and emitter, in which operating region is the transistor functioning?

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Cevap: Saturation region

Cevap

Saturation region
When a bipolar junction transistor is driven with sufficient base current to achieve maximum conduction, it operates in the saturation region. In saturation, both internal PN junctions are forward-biased, reducing the collector-to-emitter voltage (VCEV_{CE}) to a fraction of a volt (typically around 0.2 V0.2\text{ V}), effectively behaving like a closed mechanical switch.

Adım Adım Çözüm

1
Identify the circuit role and desired state of the transistor
The transistor operates as an electronic switch driven into full conduction (fully ON).
Transistors in switching circuits toggle between two primary binary states: fully OFF and fully ON.
2
Relate the fully ON switching condition to the characteristic BJT operating regions
Full conduction with minimal collector-to-emitter voltage drop occurs in the saturation region.
When both the base-emitter and base-collector junctions become forward-biased due to sufficient base drive, the transistor enters saturation, maximizing current flow through the load.

Anahtar Kavram

BJT Operating Regions and Switching States
Tahmini Süre:1m 0s
Soru 88Soru →

An electromechanical relay provides electrical isolation between the control circuit powering its coil and the separate circuit switched by its contacts.

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Cevap: True

Cevap

True
Electromechanical relays operate by using a magnetic field from a coil to mechanically open or close contacts, providing complete electrical isolation between the low-power control circuit and the switched load circuit.

Adım Adım Çözüm

1
Identify the operating mechanism of an electromechanical relay.
Current through the relay coil generates a magnetic field that physically attracts an armature to open or close contacts.
Understanding the physical separation between the electromagnetic coil and the contact terminals determines whether an electrical connection exists between them.
2
Evaluate the electrical connectivity between the coil circuit and the switched load circuit.
The two circuits are linked magnetically and mechanically, not electrically.
Because there is no conductive path between the coil terminals and the contact terminals, the relay provides complete electrical isolation.

Anahtar Kavram

Electrical isolation in electromechanical relays
Tahmini Süre:30s
Soru 89Soru →

In a linear DC power supply circuit, a technician places a smoothing filter capacitor in parallel with the load resistor following a bridge rectifier. What primary function does this capacitor perform in this circuit?

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Cevap: It charges during peak rectified voltage and releases stored electrical energy when voltage falls, reducing ripple voltage.

Cevap

It charges during peak rectified voltage and releases stored electrical energy when voltage falls, reducing ripple voltage.
A parallel filter capacitor functions as an electrostatic energy reservoir across the load. During peak outputs from the rectifier, the capacitor absorbs electrical charge. As the pulsating rectified input falls toward zero, the capacitor supplies stored energy to the load, filling the troughs between pulses and significantly attenuating the unwanted AC ripple voltage.

Adım Adım Çözüm

1
Identify the component and its circuit role after the rectifier stage.
The component is a capacitor placed in parallel with the DC load resistor as a power supply filter.
The output of a rectifier consists of pulsating DC with substantial AC ripple that must be smoothed into a steady DC voltage level.
2
Analyze the electrostatic charge and discharge mechanism of the capacitor across the voltage cycle.
The capacitor quickly charges to peak output voltage during the rising edge of each rectified pulse, then discharges its stored charge into the load when the rectifier output drops below the capacitor voltage.
This continuous charge-discharge cycle fills in the valleys between rectified pulses, stabilizing the output voltage and minimizing ripple.

Anahtar Kavram

Capacitive filtering and ripple voltage reduction in DC power supplies
Tahmini Süre:1m 0s
Soru 90Soru →

Specialized semiconductor diodes are engineered with distinct physical and electrical properties to perform specific functions in electronic circuits. Match each specialized diode and operating condition on the left with its corresponding functional characteristic or application on the right.

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Öğeler

Zener diode operated beyond its reverse breakdown voltage (VZV_Z)
Varactor (Varicap) diode operated under variable reverse bias
Schottky barrier diode in forward conduction
Light-Emitting Diode (LED) in forward bias

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Cevap

The Zener diode in reverse breakdown matches stable voltage regulation; the Varactor diode under reverse bias matches voltage-variable capacitance for RF tuning; the Schottky diode matches fast switching with low forward voltage drop; and the Light-Emitting Diode matches electroluminescent photon emission.
Each diode type corresponds to its unique semiconductor junction physics: Zener diodes maintain stable reverse breakdown voltages for regulation; varactor diodes provide voltage-dependent junction capacitance; Schottky diodes utilize metal-semiconductor junctions for high-speed switching and low forward drop; and LEDs release photons through forward-biased electron-hole recombination.

Adım Adım Çözüm

1
Analyze the operational principle of the Zener diode in reverse bias.
Identified as maintaining a constant voltage (VZV_Z) across terminals.
Zener diodes are specifically doped to operate safely in the avalanche or Zener breakdown region to provide voltage regulation.
2
Analyze the operational principle of the Varactor diode under reverse bias.
Identified as a voltage-controlled variable capacitor.
The reverse bias modulates the width of the dielectric depletion region, changing junction capacitance proportionally to the applied reverse voltage.
3
Analyze the construction and forward characteristics of the Schottky barrier diode.
Identified by its metal-semiconductor junction providing low forward drop (0.2–0.3 V0.2\text{--}0.3\text{ V}) and negligible reverse recovery time.
Because majority carriers dominate conduction, there is no minority carrier recombination delay, enabling high-frequency switching.
4
Analyze the forward bias behavior of the Light-Emitting Diode (LED).
Identified as emitting light via electroluminescence.
Direct bandgap semiconductor recombination converts injected forward current into light.

Anahtar Kavram

Operating characteristics and specialized semiconductor applications of Zener, Varactor, Schottky, and Light-Emitting Diodes.
Tahmini Süre:1m 30s
Soru 91Soru →

A technician installs an inductor (choke) in series with the output line of a direct current (DC) power supply to filter out residual alternating current (AC) ripple. What is the primary effect of the inductor on this electrical current?

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Cevap: It opposes rapid changes in current, resisting the AC ripple while allowing steady DC to pass with minimal resistance.

Cevap

The inductor opposes rapid current variations (the AC ripple) due to inductive reactance while allowing steady direct current (DC) to pass freely.
An inductor resists changes in current through self-induction. When placed in series with a DC power supply line, its inductive reactance (XL=2πfLX_L = 2\pi f L) creates significant opposition to the fluctuating AC ripple component while providing near-zero opposition to the constant DC current.

Adım Adım Çözüm

1
Identify the fundamental operating principle of an inductor.
Inductors store energy in a magnetic field and inherently oppose any change in electrical current (VL=LdidtV_L = L \frac{di}{dt}).
According to Lenz's Law, when current changes through a coil, a self-induced counter-electromotive force (CEMF) is produced that resists that change.
2
Analyze the inductor's frequency-dependent opposition (inductive reactance, XLX_L).
For steady DC (f=0 Hzf = 0\text{ Hz}), XL=2πfL=0 ΩX_L = 2\pi f L = 0\,\Omega, so DC passes with negligible opposition. For AC ripple (f>0 Hzf > 0\text{ Hz}), XL>0 ΩX_L > 0\,\Omega, which impedes the fluctuation.
Higher frequency signals encounter significant inductive reactance, attenuating the unwanted AC ripple while letting constant DC through to the load.

Anahtar Kavram

Inductors oppose changes in alternating current via inductive reactance while allowing steady direct current to flow with minimal opposition.
Tahmini Süre:45s
Soru 92Soru →

In standard electronic schematic symbols for bipolar junction transistors (BJTs), the arrow is located on the emitter terminal and points outward away from the base for an NPN transistor, indicating the direction of conventional current flow.

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Cevap: True

Cevap

True
The statement is true because electronic schematic standards place the directional arrow exclusively on the emitter lead. For an NPN transistor, this arrow points outward (away from the base), which reflects the direction of conventional current flow across the forward-biased base-emitter junction.

Adım Adım Çözüm

1
Identify the specific terminal that carries the directional arrow on a BJT schematic symbol.
The arrow is always placed on the emitter terminal, distinguishing it from the collector and base leads.
Standard electronic schematic conventions designate the emitter lead with an arrow to define transistor polarity and terminal identity.
2
Determine the orientation of the arrow for an NPN transistor configuration.
In an NPN transistor, the arrow points outward, away from the base.
The arrow represents the direction of forward-biased conventional current flow, moving from the P-type base material out through the N-type emitter material.
3
Evaluate the accuracy of the full statement.
The statement correctly identifies both the terminal location (emitter) and the outward direction for an NPN transistor.
Both structural elements conform exactly to standard electrical schematic rules.

Anahtar Kavram

BJT schematic symbol identification and terminal arrow direction
Tahmini Süre:45s
Soru 93Soru →

In an electromechanical relay, which internal component creates the magnetic field needed to pull the movable armature and actuate the electrical contacts?

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Cevap: An electromagnetic wire coil

Cevap

An electromagnetic wire coil creates the magnetic field that pulls the armature to actuate the contacts in an electromechanical relay.
An electromechanical relay consists of an insulated wire coil wound around a ferromagnetic core. When a control current passes through this coil, it produces an electromagnetic field that pulls the hinged iron armature, physically shifting the contact positions to open or close the circuit.

Adım Adım Çözüm

1
Identify the operating mechanism of an electromechanical relay.
An electromechanical relay uses an electrical control signal to produce mechanical motion via magnetic force.
Relays use electromagnetism to allow a low-power circuit to open or close a separate electrical circuit.
2
Determine which component generates the magnetic force within the relay.
An insulated wire wrapped around an iron core forms an electromagnetic coil that produces magnetic flux when energized.
When current flows through the coil, the resulting magnetic attraction pulls the movable iron armature toward the core, changing the position of the contacts.

Anahtar Kavram

Electromechanical relay coil operation and magnetic contact actuation
Tahmini Süre:45s
Soru 94Soru →

An avionics technician is assembling a filtering network using three capacitors connected in series. The capacitors have ratings of 30 μF30\text{ }\mu\text{F}, 20 μF20\text{ }\mu\text{F}, and 60 μF60\text{ }\mu\text{F}. What is the total equivalent capacitance of this series network in microfarads (μF\mu\text{F})?

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Cevap: 10

Cevap

The total equivalent capacitance of the series circuit is 10 μF10\text{ }\mu\text{F}.
For capacitors connected in series, the total capacitance is found using the formula 1Ctotal=1C1+1C2+1C3\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}. Substituting 30 μF30\text{ }\mu\text{F}, 20 μF20\text{ }\mu\text{F}, and 60 μF60\text{ }\mu\text{F} gives 260+360+160=660=110 μF\frac{2}{60} + \frac{3}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10\text{ }\mu\text{F}}. Inverting this fraction yields the correct total capacitance of 10 μF10\text{ }\mu\text{F}, which is less than the smallest individual capacitor (20 μF20\text{ }\mu\text{F}).

Adım Adım Çözüm

1
State the series capacitance relationship
1Ctotal=1C1+1C2+1C3\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
Capacitors in series combine inversely because the total dielectric separation increases, reducing total capacitance.
2
Calculate the sum of the reciprocals using a common denominator
130+120+160=2+3+160=660=110\frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}
Finding a least common multiple of 6060 allows direct fractional addition.
3
Invert the result to find CtotalC_{\text{total}}
Ctotal=10 μFC_{\text{total}} = 10\text{ }\mu\text{F}
The final equivalent capacitance is the inverse of the sum of the reciprocals.

Anahtar Kavram

Series capacitance follows the reciprocal sum rule, resulting in a total capacitance smaller than any individual capacitor in the string.
Tahmini Süre:1m 0s
Soru 95Soru →

A control transformer in an HVAC system steps down an input line voltage of 240 V240\text{ V} AC to an output voltage of 24 V24\text{ V} AC to safely power a thermostat circuit. If the primary winding consists of 800800 turns of wire, what is the number of turns on the secondary winding?

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Cevap: 80

Cevap

80 turns
The secondary winding requires 80 turns. According to the transformer turns ratio relationship, VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}. Since the secondary voltage (24 V24\text{ V}) is one-tenth of the primary input voltage (240 V240\text{ V}), the secondary coil must have one-tenth the number of turns of the primary coil: 800×0.10=80800 \times 0.10 = 80 turns.

Adım Adım Çözüm

1
Identify known electrical parameters from the problem statement.
Vp=240 VV_p = 240\text{ V}, Vs=24 VV_s = 24\text{ V}, and Np=800 turnsN_p = 800\text{ turns}.
Establishing the input and output parameters is necessary to set up the transformer relationship.
2
Set up the ideal transformer turns ratio equation relating voltage and turns.
VpVs=NpNs\frac{V_p}{V_s} = \frac{N_p}{N_s}
The voltage induced across transformer windings is directly proportional to the number of turns in each coil.
3
Rearrange the equation to isolate the unknown secondary winding turns (NsN_s).
Ns=Np×(VsVp)N_s = N_p \times \left(\frac{V_s}{V_p}\right)
Algebraic isolation allows direct computation of secondary turns.
4
Substitute the known numerical values into the formula and calculate the result.
Ns=800×(24 V240 V)=800×0.10=80 turnsN_s = 800 \times \left(\frac{24\text{ V}}{240\text{ V}}\right) = 800 \times 0.10 = 80\text{ turns}
Carrying out the arithmetic yields the exact number of secondary turns required.

Anahtar Kavram

Transformer Turns Ratio and Voltage Transformation

Alternatif Yöntem

Calculate the turns-per-volt ratio of the primary coil: 800 turns240 V=103 turns/V\frac{800\text{ turns}}{240\text{ V}} = \frac{10}{3}\text{ turns/V}. Then multiply by the desired secondary voltage: 24 V×103 turns/V=80 turns24\text{ V} \times \frac{10}{3}\text{ turns/V} = 80\text{ turns}.
Tahmini Süre:45s
Soru 96Soru →

In an electronic power system, two ideal transformers are cascaded (connected in sequence).

Transformer 1 receives an input voltage of 1200 V AC1{}200\text{ V AC} on its primary coil, which has 3000 turns3{}000\text{ turns}. The secondary coil of Transformer 1 has 600 turns600\text{ turns}, and its output voltage is fed directly into the primary coil of Transformer 2.

Transformer 2 further steps down this voltage to supply exactly 48 V AC48\text{ V AC} to an electronic load.

If the secondary coil of Transformer 2 contains 50 turns50\text{ turns}, how many turns must the primary coil of Transformer 2 have?

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Cevap: 250

Cevap

250 turns
The correct answer is derived by sequentially applying the transformer turns ratio formula (Vp/Vs=Np/NsV_p/V_s = N_p/N_s) across the cascaded system. Transformer 1 operates with a 5:1 step-down ratio, reducing the 1200 V1{}200\text{ V} input to 240 V240\text{ V}. Transformer 2 receives this 240 V240\text{ V} and must further step it down to 48 V48\text{ V}, which also requires a 5:1 ratio. Multiplying the 5050 secondary turns by this ratio yields 250250 primary turns.

Adım Adım Çözüm

1
Calculate the secondary voltage of Transformer 1.
Vs1=240 V ACV_{s1} = 240\text{ V AC}
Using the turns ratio formula Vs1/Vp1=Ns1/Np1V_{s1} / V_{p1} = N_{s1} / N_{p1}, we find Vs1=1200×(600/3000)=1200×0.2=240 V ACV_{s1} = 1{}200 \times (600 / 3{}000) = 1{}200 \times 0.2 = 240\text{ V AC}.
2
Identify the primary voltage of Transformer 2.
Vp2=240 V ACV_{p2} = 240\text{ V AC}
Because the transformers are cascaded in series, the output voltage of Transformer 1 serves directly as the input voltage for Transformer 2.
3
Set up the turns ratio equation for Transformer 2.
Np2/50=240/48N_{p2} / 50 = 240 / 48
Applying the turns ratio formula Np2/Ns2=Vp2/Vs2N_{p2} / N_{s2} = V_{p2} / V_{s2} to the second transformer using its specific input and output voltages.
4
Solve for the primary turns of Transformer 2.
Np2=250 turnsN_{p2} = 250\text{ turns}
Np2=50×(240/48)=50×5=250 turnsN_{p2} = 50 \times (240 / 48) = 50 \times 5 = 250\text{ turns}.

Anahtar Kavram

Cascaded Transformer Turns Ratios
Soru 97Soru →

An electronics technician is testing a power supply circuit featuring a four-diode full-wave bridge rectifier driven by a 60 Hz60\text{ Hz} AC secondary winding. During testing under load, one of the four rectifier diodes fails open. Which of the following best describes the electrical output observed across the load resistor?

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Cevap: The circuit delivers pulsating DC with a ripple frequency of 60 Hz60\text{ Hz} because only one half-cycle path remains functional.

Cevap

The circuit operates as a half-wave rectifier, supplying pulsating DC with a ripple frequency of 60 Hz60\text{ Hz} because only one alternation pair conducts.
A standard bridge rectifier uses two diodes per half-cycle to route current through the load in the same direction. When one diode opens, the series loop for that half-cycle is broken, producing zero output during that alternation. The other pair continues to conduct normally during the opposite half-cycle, resulting in half-wave rectification where pulses occur once per AC period at 60 Hz60\text{ Hz}.

Adım Adım Çözüm

1
Analyze normal full-wave bridge rectifier conduction paths.
In normal operation, two diodes conduct in series on the positive half-cycle, and the opposite two diodes conduct in series on the negative half-cycle, yielding full-wave pulsating DC at twice the input line frequency (2×60 Hz=120 Hz2 \times 60\text{ Hz} = 120\text{ Hz}).
Understanding normal operation establishes the baseline for analyzing component failure.
2
Determine the impact of one diode failing open on both AC half-cycles.
The half-cycle requiring the open diode cannot complete its series loop, resulting in zero conduction during that alternation. The opposite half-cycle still has both diodes forward-biased and fully operational.
Diodes operate as one-way directional switches in specific series pairs per AC polarity.
3
Identify the resulting waveform characteristics across the load.
The circuit delivers unidirectional current pulses during only one half-cycle of the input AC sine wave, which defines half-wave rectification with a fundamental ripple frequency equal to the input line frequency (60 Hz60\text{ Hz}).
A single pulse per full AC cycle corresponds directly to the source frequency of 60 Hz.

Anahtar Kavram

Diode bridge rectification mechanics and single-diode open fault behavior
Tahmini Süre:1m 30s
Soru 98Soru →

A technician is configuring a PNP bipolar junction transistor to operate as a switch. To forward-bias the base-emitter junction and turn the transistor ON, what electrical condition must be established at the base terminal relative to the emitter?

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Cevap: The base voltage must be lower (more negative) than the emitter voltage.

Cevap

The base voltage must be lower (more negative) than the emitter voltage.
For a PNP bipolar junction transistor to turn ON, its base-emitter P-N junction must be forward-biased. Because the emitter is P-type semiconductor material and the base is N-type material, the base terminal must be held at a lower (more negative) electrical potential than the emitter terminal (typically by at least 0.7 V0.7\text{ V} in standard silicon transistors).

Adım Adım Çözüm

1
Identify the transistor polarity and internal semiconductor structure.
A PNP transistor consists of an N-type base sandwiched between a P-type emitter and a P-type collector.
Understanding layer material types determines the required voltage polarities across the junctions.
2
Determine the biasing requirement for the base-emitter junction to allow conduction.
To forward-bias a P-N junction (P-type emitter to N-type base), the P-type region must be more positive than the N-type region.
Current can only flow across a forward-biased junction when the appropriate potential difference overcomes the barrier voltage (typically 0.7 V0.7\text{ V} for silicon).
3
Formulate the necessary base terminal condition relative to the emitter.
The base voltage must be pulled lower (more negative) relative to the emitter voltage, allowing conventional current to flow out of the base.
This establishes forward bias at the base-emitter junction, switching the PNP transistor ON.

Anahtar Kavram

PNP Transistor Biasing and Conduction Principles
Tahmini Süre:1m 0s
Soru 99Soru →

An electronics technician is organizing replacement components for various power supply and control circuits. Pair each protective device with its designated operational response or circuit function. What is the correct match for each component?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Fast-acting fuse
Slow-blow (time-delay) fuse
Metal-oxide varistor (MOV)
Bimetallic thermal circuit breaker

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Fast-acting fuse matches with melting an internal sacrificial element almost instantaneously during overcurrent; Slow-blow fuse matches with tolerating momentary startup inrush currents while melting under sustained overload; Metal-oxide varistor (MOV) matches with clamping transient high-voltage spikes by rapidly dropping resistance; Bimetallic thermal circuit breaker matches with using heat-induced mechanical deflection of bonded metals to trip a reusable switch.
Each protective component operates on a specific physical mechanism: fast-acting fuses melt rapidly to safeguard delicate electronics; slow-blow fuses incorporate intentional thermal delays to permit harmless inrush currents; MOVs clamp high-voltage spikes by conducting heavily above their rated clamping voltage; and thermal circuit breakers employ bimetallic strip bending to offer resettable overcurrent interruption.

Adım Adım Çözüm

1
Identify the primary fault type (overcurrent vs. overvoltage) addressed by each protective component.
Fast-acting fuses, slow-blow fuses, and thermal circuit breakers protect against excessive current, whereas metal-oxide varistors protect against transient high-voltage surges.
Categorizing devices by their primary protection role prevents confusing overvoltage suppressors with overcurrent interrupters.
2
Distinguish between the response speeds and mechanisms of the overcurrent protective devices.
Fast-acting fuses blow instantly; slow-blow fuses accommodate brief inrush surges; thermal breakers provide resettable mechanical tripping via bimetallic heating.
Different circuit loads, such as inductive motors versus sensitive solid-state electronics, require tailored opening times and reset capabilities.
3
Correlate each device with its matching functional description.
All four protective devices are correctly paired with their physical operating characteristics.
Accurate pairing verifies mastery of basic electronic protection principles.

Anahtar Kavram

Operational characteristics, response times, and functions of circuit protective devices
Tahmini Süre:1m 0s
Soru 100Soru →

When a freewheeling (flyback) diode is installed across an inductive DC relay coil to suppress transient voltage spikes, its cathode must be connected to the positive supply rail so that the diode remains reverse-biased during normal energized operation and becomes forward-biased only when the collapsing magnetic field inverts the coil's terminal polarity.

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Cevap: True

Cevap

True: In a flyback protection configuration, connecting the diode's cathode to the positive supply rail keeps it reverse-biased during steady-state energization, while allowing inductive collapse back-EMF to forward-bias the diode and clamp harmful voltage transients.
The statement is true because connecting the cathode of a freewheeling diode to the positive DC rail holds the diode in reverse bias during normal circuit operation, preventing unwanted conduction. When the control switch opens, the inductor's collapsing magnetic field inverts the terminal voltage polarity (back-EMF), which makes the anode positive relative to the cathode, placing the diode in forward bias to clamp voltage spikes and dissipate stored inductive energy.

Adım Adım Çözüm

1
Analyze the biasing condition of the diode during normal steady-state energization of the inductive relay.
With the cathode tied to the positive DC supply rail and the anode connected to the switched low side, the diode is reverse-biased and draws only negligible leakage current, allowing full operating voltage across the coil.
A semiconductor diode conducts only when its anode is positive relative to its cathode; applying a higher potential to the cathode establishes reverse bias.
2
Determine the electrical response of the relay coil when the control switch opens to de-energize the circuit.
In accordance with Lenz's Law, the collapsing magnetic field induces a high-voltage back-EMF across the coil that reverses the terminal polarity, driving the switched terminal significantly positive relative to the supply rail.
An inductor opposes instantaneous changes in current by reversing its induced terminal voltage.
3
Evaluate the diode's conduction state during the induced back-EMF transient.
The reversed polarity places a higher positive voltage at the diode's anode than at its cathode, forward-biasing the diode to create a circulating discharge path that clamps the voltage spike to approximately 0.7 V0.7\text{ V} above the supply rail.
Forward conduction through the diode safely dissipates the stored magnetic energy and protects switching components from overvoltage breakdown.

Anahtar Kavram

Flyback diode polarity and inductive transient voltage clamping in DC circuits
Tahmini Süre:1m 30s
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