Electronic Components

211 soru

Soru 141Soru →

A step-up transformer utilized in an industrial ignition system has 80 turns80\text{ turns} on its primary coil and 1200 turns1{}200\text{ turns} on its secondary coil. When the primary coil is energized by a 16 V AC16\text{ V AC} source, the secondary circuit delivers a current of 40 mA40\text{ mA} to its load. Assuming an ideal transformer, what is the current drawn by the primary coil?

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Cevap: 600 mA600\text{ mA}

Cevap

600 mA600\text{ mA}
In an ideal transformer, power is conserved (Pp=PsP_p = P_s), which means voltage and current vary inversely with the turns ratio. For this transformer, the secondary-to-primary turns ratio is 120080=15\frac{1{}200}{80} = 15. While voltage steps up by a factor of 15, current steps down by a factor of 15. Therefore, the primary current must be 15 times the secondary current: Ip=40 mA×15=600 mAI_p = 40\text{ mA} \times 15 = 600\text{ mA} (or 0.6 A0.6\text{ A}).

Adım Adım Çözüm

1
Calculate the turns ratio of the transformer
Turns ratio NsNp=120080=15\frac{N_s}{N_p} = \frac{1{}200}{80} = 15
The turns ratio establishes how voltage and current scale between the primary and secondary windings.
2
Apply the inverse current transformation formula for ideal transformers
Ip=Is×(NsNp)=40 mA×15=600 mAI_p = I_s \times \left(\frac{N_s}{N_p}\right) = 40\text{ mA} \times 15 = 600\text{ mA}
Because transformers conserve power (VpIp=VsIsV_p I_p = V_s I_s), a step-up in voltage results in a proportional step-down in current, meaning the primary current must be greater than the secondary current by the turns ratio.

Anahtar Kavram

Transformer Current and Turns Ratio Relationship
Tahmini Süre:1m 0s
Soru 142Soru →

Match each electronic component or resistor specification on the left with its correct circuit function or definition on the right.

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Öğeler

Potentiometer
Rheostat
Fixed Resistor
Tolerance Band

Eşleşmeler

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Cevap

Potentiometer matches with the three-terminal variable resistor used as an adjustable voltage divider; Rheostat matches with the two-terminal variable resistor used to adjust current flow; Fixed Resistor matches with the component providing a single, constant resistance value; Tolerance Band matches with the color-coded marking indicating allowable percentage variation from nominal resistance.
Each item is correctly matched to its standard electrical engineering and ASVAB definition: potentiometers provide variable three-terminal voltage division, rheostats provide two-terminal variable current control, fixed resistors supply a steady ohmic value, and tolerance bands indicate the precision percentage.

Adım Adım Çözüm

1
Identify the distinction between the two variable resistor types based on terminal count and primary function.
A potentiometer has 3 terminals and functions as a voltage divider; a rheostat has 2 active terminals in series to regulate current.
Variable resistors differ in wiring configuration: voltage division requires three connections across a potential difference, whereas current regulation requires two inline connections.
2
Identify the definition of a fixed resistor.
A fixed resistor provides a constant, unchangeable ohmic value.
Unlike potentiometers or rheostats, fixed resistors have no moving wiper mechanism.
3
Identify the purpose of the resistor tolerance band.
The tolerance band gives the allowable percentage deviation from the nominal value.
Manufacturing variations mean actual resistance lies within a percentage range defined by the 4th band.

Anahtar Kavram

Classification, terminal configurations, and functional specifications of fixed and variable resistors.
Tahmini Süre:40s
Soru 143Soru →

An electronics technician is designing a control panel that uses a single DC supply line to illuminate either a green 'Standby' lamp or a red 'Run' lamp, ensuring that only one lamp can receive power at any given time. Which type of switch configuration is required to route power between these two alternate loads?

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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 consists of a single common movable contact (one pole) that can connect to either of two stationary output contacts (two throws). Moving the switch actuator connects the single supply line to either the Standby lamp or the Run lamp, preventing both from being energized simultaneously.

Adım Adım Çözüm

1
Identify the number of input circuits (poles) required by the scenario.
One incoming DC supply line means 1 pole (Single-Pole, SP).
The 'pole' count defines how many independent input circuits or movable contact arms are operated by the switch mechanism.
2
Identify the number of output positions (throws) needed for switching between loads.
Two distinct output paths (Standby lamp and Run lamp) means 2 throws (Double-Throw, DT).
The 'throw' count defines how many individual output positions or contacts each pole can connect to.
3
Combine pole and throw nomenclature to select the appropriate component.
Single-Pole, Double-Throw (SPDT).
An SPDT switch directs current from one common terminal to either of two branch terminals in a mutually exclusive manner.

Anahtar Kavram

Switch pole and throw configurations (SPST, SPDT, DPST, DPDT)
Tahmini Süre:1m 0s
Soru 144Soru →

In a standard electronic schematic diagram, what does the triangular arrowhead of a semiconductor diode symbol indicate?

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Cevap: The permitted direction of conventional current flow

Cevap

The permitted direction of conventional current flow
In standard electronic schematic symbols, the triangular arrowhead points from the anode toward the cathode, designating the direction that conventional electric current (from positive to negative) is allowed to flow during forward bias.

Adım Adım Çözüm

1
Analyze the anatomy of a standard semiconductor diode schematic symbol.
The symbol features a triangle pointing directly toward a vertical barrier line.
Recognizing the individual elements of schematic drafting symbols is necessary to read circuit diagrams.
2
Correlate the triangular symbol element with electrical flow conventions.
The triangle represents an arrowhead pointing from the anode (positive terminal) to the cathode (negative terminal), matching conventional current direction.
Standard electronics schematics use conventional current (positive-to-negative flow) to define component polarity arrows.

Anahtar Kavram

A semiconductor diode schematic symbol uses a triangular arrowhead pointing from anode to cathode to indicate the permitted direction of conventional current flow.
Tahmini Süre:45s
Soru 145Soru →

A four-band carbon-film resistor has the color bands in the following sequence: Blue, Gray, Brown, and Gold. What is the nominal resistance value and tolerance rating of this resistor?

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Cevap: 680 Ω with ±5% tolerance680\ \Omega\text{ with } \pm 5\%\text{ tolerance}

Cevap

680 Ω with ±5% tolerance680\ \Omega\text{ with } \pm 5\%\text{ tolerance}
Decoding the four color bands in order yields: first band (Blue) = 6, second band (Gray) = 8, third band (Brown) = ×101\times 10^1 (1010), and fourth band (Gold) = ±5%\pm 5\%. Calculating 68×1068 \times 10 gives a nominal resistance of 680 Ω680\ \Omega with ±5%\pm 5\% tolerance.

Adım Adım Çözüm

1
Identify the significant digits from the first two color bands
Blue corresponds to the digit 6 and Gray corresponds to the digit 8, forming the significant number 68.
In a standard four-band resistor, the first two bands represent the first and second significant digits of the resistance value.
2
Determine the multiplier from the third color band and calculate the nominal resistance
Brown corresponds to a multiplier of 10110^1 (1010). Multiplying 68×10=680 Ω68 \times 10 = 680\ \Omega.
The third band indicates the power of ten by which the significant digits are multiplied.
3
Identify the tolerance percentage from the fourth color band
Gold corresponds to a tolerance of ±5%\pm 5\%.
The fourth band indicates the manufacturing tolerance of the resistor.

Anahtar Kavram

Four-band resistor color code decoding
Tahmini Süre:35s
Soru 146Soru →

An electronics technician is assembling a tuning filter network. The circuit consists of two inductors connected in series with values of 18 mH18\text{ mH} and 42 mH42\text{ mH}. This series combination is then connected in parallel with a third inductor rated at 40 mH40\text{ mH}. Assuming there is no mutual inductance between the coils, what is the total equivalent inductance of the circuit in millihenries (mH\text{mH})?

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Cevap: 24

Cevap

The total equivalent inductance of the circuit is 24 mH24\text{ mH}.
Combining the two series inductors (18 mH+42 mH18\text{ mH} + 42\text{ mH}) yields an equivalent branch inductance of 60 mH60\text{ mH}. Connecting this branch in parallel with the 40 mH40\text{ mH} inductor gives a total inductance of 60×4060+40=2400100=24 mH\frac{60 \times 40}{60 + 40} = \frac{2400}{100} = 24\text{ mH}.

Adım Adım Çözüm

1
Calculate the equivalent inductance of the series branch (LseriesL_{\text{series}}).
Lseries=18 mH+42 mH=60 mHL_{\text{series}} = 18\text{ mH} + 42\text{ mH} = 60\text{ mH}
Inductors in series combine by direct addition, similar to series resistors (Lseries=L1+L2L_{\text{series}} = L_1 + L_2).
2
Calculate the total inductance of the series branch in parallel with the third inductor.
Ltotal=60 mH×40 mH60 mH+40 mH=2400100=24 mHL_{\text{total}} = \frac{60\text{ mH} \times 40\text{ mH}}{60\text{ mH} + 40\text{ mH}} = \frac{2400}{100} = 24\text{ mH}
Inductors in parallel combine using the reciprocal or product-over-sum formula (Ltotal=LA×LBLA+LBL_{\text{total}} = \frac{L_A \times L_B}{L_A + L_B}).

Anahtar Kavram

Calculation of total inductance for series-parallel inductor combinations without mutual coupling.
Tahmini Süre:1m 15s
Soru 147Soru →

Different semiconductor diode designs are tailored for specific roles in electronic systems. Match each specialized diode type on the left with its primary operating function on the right.

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Öğeler

Zener Diode
Light-Emitting Diode (LED)
Photodiode
Schottky Diode

Eşleşmeler

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Cevap

Zener Diode matches maintaining a steady reference voltage in reverse breakdown; Light-Emitting Diode matches emitting light when forward biased; Photodiode matches acting as an optical sensor in reverse bias; Schottky Diode matches providing high switching speeds and low forward voltage drop.
Each diode type utilizes semiconductor properties for a targeted function: Zener diodes maintain a stable clamping voltage in reverse breakdown, LEDs emit photons when forward-biased, photodiodes sense light through light-induced reverse current, and Schottky diodes allow high-speed rectification with a lower forward voltage drop.

Adım Adım Çözüm

1
Identify the primary electrical behavior of a Zener diode
Zener diodes uniquely conduct in reverse breakdown to regulate voltage across loads
They provide constant reference voltages in DC power supplies
2
Differentiate optoelectronic diodes (LED vs. Photodiode)
The LED emits light when forward-biased, while the photodiode detects light by conducting reverse current
One transforms electrical energy to light, whereas the other transforms light into an electrical signal
3
Identify the metal-semiconductor barrier characteristic of the Schottky diode
Schottky diodes feature low barrier voltage (~0.2 V to 0.3 V) and rapid switching capability
They are selected for high-frequency clamping and power rectification

Anahtar Kavram

Specialized Diode Types and Functional Applications
Tahmini Süre:45s
Soru 148Soru →

When selecting and replacing overcurrent protective devices in electrical and electronic equipment, technicians must verify both operating specifications stamped on the component body. What two fundamental ratings define the safe operational limits of a standard cartridge fuse?

Aşağıdaki boşlukları doldurun

Every standard cartridge fuse has two critical ratings: the rating, which defines the maximum continuous electrical flow the element can carry before melting, and the rating, which indicates the maximum circuit potential the fuse can safely interrupt without allowing an electrical arc to sustain across the blown element.
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Cevap

The first blank is current (or amperage/ampere), and the second blank is voltage (or working voltage/volt).
A fuse is an overcurrent protective device whose primary function is dictated by its current rating (in amperes), which determines the current magnitude required to melt the fusible element. Its secondary rating is its voltage rating (in volts), which specifies the maximum operating circuit voltage under which the fuse can safely extinguish and prevent an electrical arc once blown.

Adım Adım Çözüm

1
Analyze the description for the first blank regarding continuous electrical flow and melting threshold.
Identified as current rating (measured in amperes).
The internal fusible metal link is designed to heat up and melt (I2RI^2R heating) when current exceeds the rated continuous amperage.
2
Analyze the description for the second blank regarding maximum circuit potential and arc suppression.
Identified as voltage rating (measured in volts).
The voltage rating ensures that after the fuse element melts and opens the circuit, the open gap has sufficient dielectric breakdown resistance to prevent arcing.

Anahtar Kavram

Dual Ratings of Circuit Protection Devices (Current Rating vs. Voltage Rating)
Tahmini Süre:1m 0s
Soru 149Soru →

An auxiliary power transformer on a vessel steps down an input voltage of 440 V AC440\text{ V AC} to 110 V AC110\text{ V AC}. If the secondary winding is connected to a lighting circuit drawing 14 A14\text{ A}, what is the primary current in amperes (A\text{A}), assuming an ideal transformer with zero power losses?

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Cevap: 3.5

Cevap

The primary current is 3.5 A3.5\text{ A}.
In an ideal transformer with negligible losses, the primary apparent power equals the secondary apparent power (Vp×Ip=Vs×IsV_p \times I_p = V_s \times I_s). Because the transformer reduces voltage from 440 V AC440\text{ V AC} to 110 V AC110\text{ V AC} (a 4:14:1 step-down ratio), the primary winding requires only one-fourth of the secondary current. Multiplying 14 A14\text{ A} by 110440\frac{110}{440} yields 3.5 A3.5\text{ A}.

Adım Adım Çözüm

1
State the ideal transformer power balance equation
Vp×Ip=Vs×IsV_p \times I_p = V_s \times I_s
An ideal transformer operates at 100% efficiency, meaning input power (PpP_p) equals output power (PsP_s).
2
Isolate the unknown primary current variable (IpI_p)
Ip=Is×(VsVp)I_p = I_s \times \left(\frac{V_s}{V_p}\right)
Dividing both sides by the primary voltage (VpV_p) provides the direct formula for primary current.
3
Substitute the given numerical values and calculate
Ip=14 A×(110 V440 V)=14×14=3.5 AI_p = 14\text{ A} \times \left(\frac{110\text{ V}}{440\text{ V}}\right) = 14 \times \frac{1}{4} = 3.5\text{ A}
The voltage is stepped down by a factor of 44, which requires the current to be stepped down by the inverse factor of 44 in the primary circuit relative to the secondary.

Anahtar Kavram

Transformer Voltage and Current Transformation Ratio
Tahmini Süre:1m 0s
Soru 150Soru →

A linear potentiometer has a total resistance of 20 kΩ20\text{ k}\Omega between its two outer terminals (Terminal 1 and Terminal 3). If the center wiper (Terminal 2) is adjusted so that the resistance between Terminal 1 and Terminal 2 is 6.5 kΩ6.5\text{ k}\Omega, what is the resistance in kilohms (kΩ\text{k}\Omega) between Terminal 2 and Terminal 3?

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Cevap: 13.5

Cevap

The resistance between Terminal 2 and Terminal 3 is 13.5 kΩ13.5\text{ k}\Omega.
In a three-terminal potentiometer, the resistive element connects between the two outer terminals (1 and 3), providing a fixed total resistance of 20 kΩ20\text{ k}\Omega. The adjustable wiper (terminal 2) contacts this element and divides it into two series portions such that R1−2+R2−3=R1−3R_{1-2} + R_{2-3} = R_{1-3}. Subtracting the known section (6.5 kΩ6.5\text{ k}\Omega) from the total (20 kΩ20\text{ k}\Omega) gives 13.5 kΩ13.5\text{ k}\Omega for the remaining section between Terminal 2 and Terminal 3.

Adım Adım Çözüm

1
Set up the resistance relationship for the three terminals of a potentiometer.
Rtotal=R1−2+R2−3R_{\text{total}} = R_{1-2} + R_{2-3}
The wiper divides the continuous internal resistive strip into two series sections whose resistances add up to the total end-to-end resistance.
2
Subtract the resistance of the first section from the total end-to-end resistance.
R2−3=20 kΩ−6.5 kΩ=13.5 kΩR_{2-3} = 20\text{ k}\Omega - 6.5\text{ k}\Omega = 13.5\text{ k}\Omega
Isolating the unknown segment gives the remaining resistance along the track to the opposite outer terminal.

Anahtar Kavram

Potentiometer Terminal Resistance Division
Tahmini Süre:30s
Soru 151Soru →

An air-core inductor is connected in series with an alternating current (AC) voltage source. If a solid soft-iron core with high magnetic permeability is fully inserted into the center of the coil, how does this modification affect the coil's inductance and the overall circuit current?

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Cevap: Inductance increases, causing the circuit current to decrease.

Cevap

Inductance increases, causing the circuit current to decrease.
Inserting a soft-iron core with high magnetic permeability concentrates the magnetic lines of force inside the inductor. Because inductance is directly proportional to permeability (L=μN2AlL = \frac{\mu N^2 A}{l}), the inductance increases. In an AC circuit, inductive reactance is given by XL=2πfLX_L = 2\pi f L. As inductance increases, inductive reactance increases, providing greater opposition to alternating current. By Ohm's law (I=VXLI = \frac{V}{X_L}), the higher opposition causes the circuit current to decrease.

Adım Adım Çözüm

1
Determine the effect of inserting an iron core on coil inductance.
The inductance (LL) increases significantly because the magnetic permeability (μ\mu) of soft iron is hundreds of times greater than that of air (L=μN2AlL = \frac{\mu N^2 A}{l}).
Inductance is directly proportional to the magnetic permeability of the core material.
2
Determine the relationship between inductance and inductive reactance in an AC circuit.
Inductive reactance (XLX_L) increases proportionally with inductance (XL=2πfLX_L = 2\pi f L).
Inductive reactance represents the total counter-electromotive force opposition to alternating current.
3
Apply Ohm's law for AC circuits to determine current change.
Circuit current (II) decreases (I=VXLI = \frac{V}{X_L}).
Under constant source voltage, an increase in circuit reactance reduces total current flow.

Anahtar Kavram

Magnetic core permeability increases inductance and inductive reactance, which reduces AC circuit current.
Tahmini Süre:1m 0s
Soru 152Soru →

Match each integrated circuit (IC) package type or physical feature with its correct operational or structural description.

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Öğeler

Dual In-line Package (DIP)
Surface-Mount Device (SMD / SOIC)
Single In-line Package (SIP)
Index Notch / Pin 1 Dot

Eşleşmeler

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Cevap

Dual In-line Package (DIP) matches the rectangular housing with two parallel rows of through-hole pins; Surface-Mount Device (SMD) matches the low-profile housing soldered directly onto surface pads; Single In-line Package (SIP) matches the single row of connecting pins; and Index Notch / Pin 1 Dot matches the physical alignment marker used to identify the starting reference pin.
Dual In-line Packages (DIP) are through-hole components with two parallel pin rows. Surface-Mount Devices (SMD) attach directly to surface pads on the circuit board. Single In-line Packages (SIP) contain a single row of leads along one edge. The index notch or dot is the industry-standard visual alignment marker used to identify Pin 1 and determine the proper pin count orientation.

Adım Adım Çözüm

1
Identify the pin layout configurations for standard through-hole IC packages
DIP designates two parallel rows of pins; SIP designates a single inline row of pins.
The names 'Dual' and 'Single' refer directly to the number of parallel pin rows on the package.
2
Differentiate surface-mount packaging from through-hole construction
SMD/SOIC packages solder directly to surface traces/pads without extending through drilled holes in the PCB.
Surface-mount technology is characterized by direct surface soldering, reducing board thickness and component footprint.
3
Identify the orientation and indexing convention for integrated circuits
A notch or dot on the package casing indicates the top of the chip and locates Pin 1 immediately to its left.
Physical indexing marks are necessary to prevent backwards installation and correctly map schematic pins to physical leads.

Anahtar Kavram

Integrated circuit package classifications and physical indexing standards
Tahmini Süre:1m 0s
Soru 153Soru →

A metal oxide varistor (MOV) is installed in parallel across the incoming AC power lines of an industrial power supply to protect sensitive electronics against transient overvoltage spikes. When a high-voltage surge exceeds the clamping voltage rating of the MOV, how does this protective device respond?

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Cevap: Its electrical resistance drops sharply, shunting excess surge current away from downstream components and clamping line voltage.

Cevap

Its electrical resistance drops sharply, shunting excess surge current away from downstream components and clamping line voltage.
Metal oxide varistors (MOVs) are voltage-dependent protective devices that exhibit high electrical resistance under normal line voltages. When exposed to a transient voltage surge exceeding their clamping voltage rating, their resistance drops almost instantaneously to a very low value. This low-resistance path safely shunts the excess surge current away from sensitive downstream electronics, clamping the voltage across the load to a safe level.

Adım Adım Çözüm

1
Identify the function and circuit configuration of a metal oxide varistor (MOV).
An MOV is a voltage-dependent, non-linear resistor connected in parallel across the power lines to protect against transient overvoltage spikes.
Recognizing that the MOV is wired in parallel confirms it serves as a voltage clamp rather than a series disconnect.
2
Determine the electrical behavior of the MOV when line voltage exceeds its clamping threshold.
The internal resistance of the MOV drops rapidly from very high resistance (megohms) to very low resistance (near zero ohms).
The semiconductor zinc-oxide material conducts heavily once the breakdown voltage is reached.
3
Evaluate the resulting protection provided to downstream components.
The low-resistance state diverts high surge currents away from the load, holding the voltage across the circuit to a safe clamped level.
By shunting the transient energy across the parallel branch, sensitive downstream circuits are shielded from damage.

Anahtar Kavram

Metal Oxide Varistor (MOV) transient overvoltage clamping mechanism
Tahmini Süre:1m 0s
Soru 154Soru →

A technician installs a standard semiconductor diode in series with a direct current (DC) motor circuit to provide reverse polarity protection. Which of the following best describes the operating characteristic of the diode that enables this protection?

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Cevap: It allows current to flow easily in only one direction while blocking current flow in the reverse direction.

Cevap

The diode allows current to flow easily in only one direction while blocking current flow in the reverse direction.
A semiconductor diode functions as an electrical one-way check valve. When forward-biased (anode more positive than cathode), it allows current to pass with minimal resistance. When reverse-biased (voltage polarity reversed), the diode exhibits extremely high resistance and blocks current, safeguarding downstream components from incorrect power connections.

Adım Adım Çözüm

1
Identify the primary electrical function of a semiconductor diode
A diode is designed to allow electric current to pass easily in one direction (forward bias) while presenting very high resistance to current flowing in the opposite direction (reverse bias).
Understanding the fundamental one-way conduction property explains how it functions as a circuit safeguard.
2
Apply the diode characteristic to the reverse polarity protection scenario
When power is connected with normal polarity, the diode is forward-biased and allows normal circuit operation. If the battery is connected backward, the diode becomes reverse-biased and blocks the reverse current, preventing component damage.
This directly demonstrates why unidirectional conductivity provides reverse polarity protection.

Anahtar Kavram

Unidirectional current flow and biasing in semiconductor diodes
Tahmini Süre:45s
Soru 155Soru →

An electronics technician is assembling a circuit that requires a 10 kΩ10\text{ k}\Omega fixed resistor. If the first two color bands on the resistor are Brown (1) and Black (0), which color must the third (multiplier) band be?

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Cevap: Orange

Cevap

Orange
Orange represents a decimal multiplier of 10310^3 (10001{}000). Multiplying the significant digits 10 (from Brown and Black) by 10001{}000 yields 10000 Ω10{}000\ \Omega, which equals 10 kΩ10\text{ k}\Omega.

Adım Adım Çözüm

1
Convert the target resistance from kilohms to ohms.
10 kΩ=10000 Ω10\text{ k}\Omega = 10{}000\ \Omega
Standard color code multipliers apply directly to resistance values expressed in ohms.
2
Determine the two-digit base number from the first two color bands.
Brown (1) and Black (0) form the significant digits 10.
In a 4-band resistor, the first two bands represent the leading significant digits.
3
Calculate the necessary multiplier value and identify its corresponding color.
10×103=10000 Ω10 \times 10^3 = 10{}000\ \Omega; 10310^3 (10001{}000) corresponds to Orange.
The third band indicates the power of ten (number of trailing zeros) to multiply by the significant digits.

Anahtar Kavram

Four-Band Resistor Color Code Multipliers
Tahmini Süre:30s
Soru 156Soru →

A standard silicon diode requires a forward-bias voltage of approximately 0.7 V0.7\text{ V} across its terminals to overcome its internal barrier potential and conduct current.

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Cevap: True

Cevap

The statement is true because standard silicon semiconductor diodes require approximately 0.7 V0.7\text{ V} of forward-bias voltage to overcome the internal barrier potential and allow current conduction.
Silicon semiconductor diodes have an internal PN junction barrier potential of approximately 0.7 V0.7\text{ V}. In a forward-biased circuit, current cannot flow freely until the applied external voltage meets or exceeds this 0.7 V0.7\text{ V} threshold.

Adım Adım Çözüm

1
Identify the semiconductor material and its characteristic barrier potential.
Silicon diodes possess an intrinsic barrier potential of approximately 0.7 V0.7\text{ V}.
The material properties of the silicon PN junction determine the electric field strength across the depletion layer.
2
Evaluate the condition required for current conduction.
Applying a forward voltage of at least 0.7 V0.7\text{ V} neutralizes the depletion region and allows charge carriers to cross the junction.
Voltages below 0.7 V0.7\text{ V} leave the diode in a non-conducting state, while voltages at or above 0.7 V0.7\text{ V} permit continuous current flow.

Anahtar Kavram

Silicon Diode Forward Voltage Drop and Barrier Potential
Soru 157Soru →

In power supply filtering and voltage transformation circuits, inductors and transformers operate through specific electromagnetic principles and structural designs. Match each electromagnetic concept or construction feature with its correct operational description.

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Öğeler

Counter Electromotive Force (Back EMF)
Mutual Inductance
Inductive Reactance (XLX_L)
Laminated Iron Core

Eşleşmeler

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Cevap

Counter Electromotive Force matches the generation of an opposing internal voltage; Mutual Inductance matches the transfer of energy between adjacent coils via alternating magnetic flux; Inductive Reactance matches frequency-dependent opposition to alternating current; Laminated Iron Core matches the structural method using insulated steel sheets to minimize eddy current losses.
Each concept correctly identifies a foundational operating mechanism in inductive components: self-induction produces Counter EMF opposing current changes; mutual inductance allows electrical isolation and power transfer across coils; inductive reactance describes frequency-dependent AC opposition; and core lamination mitigates destructive eddy current losses in iron-core devices.

Adım Adım Çözüm

1
Identify the definition of self-induction and Back EMF in single inductors.
Back EMF is the self-induced counter-voltage opposing any change in coil current.
Inductors store energy in a magnetic field; when current fluctuates, the changing field generates an opposing potential difference.
2
Identify the principle of mutual induction linking transformer windings.
Mutual inductance is the electromagnetic coupling where changing magnetic flux from one winding induces voltage in another.
Transformers rely on shared magnetic flux across electrically isolated primary and secondary coils.
3
Identify the dynamic AC behavior of inductors defined by inductive reactance.
Inductive reactance (XLX_L) represents the AC opposition given by XL=2πfLX_L = 2\pi f L.
Reactance is measured in ohms and increases proportionally with both frequency and inductance value.
4
Identify the purpose of laminations in magnetic transformer and inductor cores.
Laminations break up conductive loops to suppress eddy current power losses.
Solid conductive iron cores develop strong circular induced currents (eddy currents) that cause severe heating unless laminated.

Anahtar Kavram

Fundamental electromagnetic properties and structural principles governing inductors and transformers
Tahmini Süre:1m 15s
Soru 158Soru →

An electronics technician is servicing power supplies and control circuits containing various protection components. Match each circuit protective device or mechanism on the left with its primary operational principle on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Magnetic trip mechanism in a circuit breaker
Bimetallic strip in a thermal circuit breaker
Polymeric positive temperature coefficient (PPTC) resettable fuse
Transient voltage suppression (TVS) diode

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Magnetic trip mechanism matches with electromagnetic flux pulling an armature for instantaneous tripping; Bimetallic strip matches with progressive bending from thermal expansion during prolonged overcurrents; PPTC resettable fuse matches with transitioning to high resistance upon I2RI^2R heating and self-resetting; TVS diode matches with avalanche breakdown to clamp fast transient overvoltage spikes.
Each protective device is correctly paired with its operating physics: magnetic elements react instantly to short circuits via electromagnetic attraction; bimetallic strips bend slowly in response to resistive heat during continuous overloads; PPTC devices restrict current by increasing resistance when heated and recover when cool; and TVS diodes suppress voltage spikes through fast avalanche breakdown in parallel with the load.

Adım Adım Çözüm

1
Distinguish between overcurrent protective devices and transient overvoltage suppression devices.
Identify that circuit breaker elements and PPTC fuses protect against excessive current flow, whereas TVS diodes protect against voltage transients.
Protective devices function either as series current limiters/interrupts or parallel voltage clamps.
2
Differentiate between the dual trip mechanisms of thermal-magnetic circuit breakers.
The magnetic solenoid provides instantaneous response to large short-circuit currents, while the bimetal strip provides an inverse-time response to moderate sustained overloads.
Magnetic tripping prevents catastrophic equipment damage during direct shorts, whereas thermal tripping prevents wire insulation overheating from prolonged overloads.
3
Examine the solid-state behavior of PPTC fuses and TVS diodes.
PPTC devices rely on temperature-dependent polymer resistance expansion to limit overcurrent, while TVS diodes rely on p-n junction avalanche breakdown to shunt transient overvoltage.
Associating each solid-state component with its physical trigger establishes the correct pairs.

Anahtar Kavram

Operational Principles of Overcurrent and Overvoltage Protective Devices
Tahmini Süre:1m 15s
Soru 159Soru →

Fill in the blanks to complete the statement describing variable resistor terminology.

Aşağıdaki boşlukları doldurun

A three-terminal variable resistor commonly used to create an adjustable voltage divider is called a , and the internal adjustable contact that moves across the resistive track is called the .
Cevabı ve açıklamayı göster

Cevap

blank_1: potentiometer; blank_2: wiper
A potentiometer is a three-terminal variable resistor widely used to provide an adjustable voltage division in electronic circuits. The two outer terminals connect across the full resistive element, while the center movable contact, called the wiper, adjusts the output voltage based on its mechanical position.

Adım Adım Çözüm

1
Identify the standard three-terminal variable resistor component
A potentiometer (often abbreviated 'pot') is a three-terminal variable resistor configured to divide voltage across its resistive track.
By definition in basic electronics, a three-terminal variable resistor used for voltage division is designated as a potentiometer.
2
Identify the movable mechanical contact mechanism
The sliding contact that traverses the resistive material to tap off a portion of the total resistance is called the wiper.
The standard technical term for the adjustable middle contact of a potentiometer or rheostat is the wiper.

Anahtar Kavram

Potentiometer structure and terminology
Tahmini Süre:30s
Soru 160Soru →

An audio technician is designing a passive crossover network and installs an inductor in series with a low-frequency woofer. Which of the following best explains why the inductor is used in this configuration?

Cevabı ve açıklamayı göster

Cevap: It opposes high-frequency alternating currents while allowing low-frequency signals to pass easily.

Cevap

The inductor opposes high-frequency alternating currents while allowing low-frequency signals to pass easily.
An inductor generates an opposing electromotive force (back EMF) that increases with the rate of change of current. Because inductive reactance is calculated as XL=2πfLX_L = 2\pi f L, opposition rises with higher signal frequencies. Consequently, placing an inductor in series with a woofer permits low-frequency audio to pass with minimal attenuation while attenuating unwanted high frequencies.

Adım Adım Çözüm

1
Identify the relationship between frequency and inductor behavior.
Inductive reactance is defined by the formula XL=2πfLX_L = 2\pi f L, where ff is frequency and LL is inductance.
Understanding how an inductor's opposition changes with frequency determines which signals it blocks or passes.
2
Analyze circuit operation at low versus high audio frequencies.
At low audio frequencies (ff is small), XLX_L is low, presenting almost zero opposition. At high audio frequencies (ff is large), XLX_L is high, presenting strong opposition to current flow.
This behavior creates a low-pass filter when placed in series with a speaker load.
3
Match the electrical characteristics to the speaker application.
The series inductor acts as a low-pass filter (choke), directing bass signals to the woofer while blocking treble frequencies.
Woofers are designed strictly for low-frequency sound reproduction.

Anahtar Kavram

Inductive reactance (XL=2πfLX_L = 2\pi f L) increases with frequency, allowing an inductor to block high-frequency AC signals while allowing low-frequency signals to pass.
Tahmini Süre:1m 0s
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