Basic Electrical Concepts Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 5

Soru 81Soru →

A technician connects a diagnostic meter into a vehicle wire harness to measure the quantity of electric charge passing through a conductor per second. Which electrical property is the technician measuring?

Electromotive force

Electric current

Electrical resistance

Alternating frequency

Cevabı ve açıklamayı göster

Cevap: Electric current

Cevap

Electric current represents the measured rate of electrical charge passing through a conductor per second.

Electric current is defined as the rate at which electrical charge passes a given cross-section of a conductor, mathematically represented by

I = \frac{Q}{t}

, where one ampere equals one coulomb of charge per second.

Adım Adım Çözüm

1

Analyze the physical measurement described in the question.

The measurement monitors charge (

Q

) moving past a point over elapsed time (

t

).

Identifying the relationship between electrical charge and time determines the correct physical quantity.

2

Match the formula

I = \frac{Q}{t}

to its standard electrical property.

The rate of charge flow per second defines electric current, measured in amperes (

1\text{ A} = 1\text{ C/s}

).

Electric current is the formal definition of moving electric charge carriers through a conductor.

Anahtar Kavram

Electric current is the rate of flow of electric charge over time, expressed as

I = \frac{Q}{t}

and measured in amperes.

Tahmini Süre:35s

Soru 82Soru →

An electric heating element is rated to dissipate $1.8\text{ kW}$ of power when connected across a standard $120\text{ V}$ DC supply line. If a control circuit reduces the direct current flowing through the heating element to $5\text{ A}$ while its internal resistance remains constant, what is the new power dissipated by the element?

600\text{ W}

200\text{ W}

40\text{ W}

0.2\text{ W}

Cevabı ve açıklamayı göster

Cevap: $200\text{ W}$

Cevap

200\text{ W}

The correct answer of

200\text{ W}

is found by first converting

1.8\text{ kW}

to

1800\text{ W}

. Using the initial ratings, the element's resistance is

R = \frac{V^2}{P} = \frac{120^2}{1800} = 8\text{ }\Omega

. With the current controlled to

5\text{ A}

, the dissipated power is

P = I^2 R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

.

Adım Adım Çözüm

1

Convert the rated power from kilowatts to watts

P_1 = 1.8\text{ kW} = 1800\text{ W}

Metric prefixes must be converted to base SI units before performing Watt's Law calculations.

2

Calculate the constant internal resistance (

R

) of the heating element using the initial rating

R = \frac{V_1^2}{P_1} = \frac{(120\text{ V})^2}{1800\text{ W}} = \frac{14400}{1800} = 8\text{ }\Omega

The heating element's resistance remains unchanged and serves as the bridge to calculate power under the new operating current.

3

Calculate the new power dissipation with the reduced current (

I_2 = 5\text{ A}

)

P_2 = I_2^2 \times R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

Watt's Law specifies that power dissipation across a known resistance is directly proportional to the square of the current.

Anahtar Kavram

Watt's Law Power and Resistance Relationships (

P = \frac{V^2}{R}

and

P = I^2 R

)

Alternatif Yöntem

Determine the initial current

I_1 = \frac{P_1}{V_1} = \frac{1800\text{ W}}{120\text{ V}} = 15\text{ A}

. Since

P \propto I^2

for a constant resistance, scaling the power by the ratio of the currents squared gives

P_2 = P_1 \times \left(\frac{I_2}{I_1}\right)^2 = 1800\text{ W} \times \left(\frac{5\text{ A}}{15\text{ A}}\right)^2 = 1800 \times \left(\frac{1}{3}\right)^2 = \frac{1800}{9} = 200\text{ W}

.

Tahmini Süre:1m 30s

Soru 83Soru →

An avionics technician tests a $72\text{ V}$ direct current (DC) instrument circuit. If the circuit's total resistance is measured at $1.2\text{ k}\Omega$ , what is the current flowing through the circuit?

16.7\text{ mA}

60\text{ mA}

60\text{ A}

86.4\text{ A}

Cevabı ve açıklamayı göster

Cevap: $60\text{ mA}$

Cevap

60\text{ mA}

According to Ohm's law, current equals voltage divided by resistance (

I = V / R

). Converting

1.2\text{ k}\Omega

to base units gives

1200\text{ }\Omega

. Dividing

72\text{ V}

by

1200\text{ }\Omega

produces

0.06\text{ A}

, which is equivalent to

60\text{ mA}

.

Adım Adım Çözüm

1

Convert the given resistance from kilo-ohms to base ohms

R = 1.2\text{ k}\Omega = 1.2 \times 10^3\text{ }\Omega = 1200\text{ }\Omega

Standard electrical calculations using Ohm's law require values to be expressed in SI base units (volts, amperes, and ohms).

2

Apply Ohm's law formula to solve for current (

I

)

I = \frac{V}{R} = \frac{72\text{ V}}{1200\text{ }\Omega} = 0.06\text{ A}

Current is directly proportional to voltage and inversely proportional to resistance.

3

Convert the current from amperes to milliamperes

I = 0.06\text{ A} \times 1000\text{ mA/A} = 60\text{ mA}

Expressing the small current value in milliamperes matches standard electronic measurement conventions.

Anahtar Kavram

Ohm's Law Current Calculation (

I = \frac{V}{R}

) with Metric Prefix Conversion

Alternatif Yöntem

Convert the calculation directly using milliamperes:

1\text{ V} / 1\text{ k}\Omega = 1\text{ mA}

. Therefore,

72\text{ V} / 1.2\text{ k}\Omega = 60\text{ mA}

.

Tahmini Süre:1m 0s

Soru 84Soru →

An avionics technician replaces a solid copper grounding wire of length $L$ and circular diameter $d$ with a replacement wire made of a resistive alloy. The alloy has $4$ times the resistivity of copper ( $\rho_{\text{alloy}} = 4\rho_{\text{copper}}$ ), and the replacement wire is cut to half the original length ( $\frac{1}{2}L$ ). If the replacement wire must have the exact same electrical resistance as the original copper wire, what must be the diameter of the replacement wire in terms of $d$ ?

2\,d

\sqrt{2}\,d

\frac{1}{2}\,d

4\,d

Cevabı ve açıklamayı göster

Cevap: $\sqrt{2}\,d$

Cevap

\sqrt{2}\,d

Resistance depends on material resistivity and physical dimensions according to

R = \rho \frac{L}{A} = \frac{4\rho L}{\pi d^2}

. Multiplying resistivity by

4

and multiplying length by

\frac{1}{2}

produces a numerator factor of

4 \times \frac{1}{2} = 2

. To keep the overall resistance constant, the denominator area must also increase by a factor of

2

. Because cross-sectional area is proportional to the square of diameter (

A \propto d^2

), the diameter must increase by a factor of

\sqrt{2}

, giving

\sqrt{2}\,d

.

Adım Adım Çözüm

1

Express the electrical resistance of the original copper wire using its physical dimensions and resistivity.

R_{\text{orig}} = \rho_{\text{copper}} \frac{L}{A_1} = \rho_{\text{copper}} \frac{L}{\frac{\pi d^2}{4}} = \frac{4\rho_{\text{copper}}L}{\pi d^2}

The resistance of a cylindrical conductor is governed by

R = \rho \frac{L}{A}

, where circular cross-sectional area is

A = \frac{\pi d^2}{4}

.

2

Express the electrical resistance of the replacement alloy wire in terms of its new parameters

\rho_{\text{alloy}} = 4\rho_{\text{copper}}

,

L_{\text{new}} = 0.5L

, and unknown diameter

d_{\text{new}}

.

R_{\text{new}} = (4\rho_{\text{copper}}) \frac{0.5L}{\frac{\pi d_{\text{new}}^2}{4}} = \frac{4(2\rho_{\text{copper}}L)}{\pi d_{\text{new}}^2} = \frac{8\rho_{\text{copper}}L}{\pi d_{\text{new}}^2}

Substitute the scaled values of resistivity and length into the conductor resistance equation.

3

Equate the original resistance to the replacement resistance and solve for

d_{\text{new}}

.

\frac{4\rho_{\text{copper}}L}{\pi d^2} = \frac{8\rho_{\text{copper}}L}{\pi d_{\text{new}}^2} \implies \frac{1}{d^2} = \frac{2}{d_{\text{new}}^2} \implies d_{\text{new}}^2 = 2d^2 \implies d_{\text{new}} = \sqrt{2}\,d

The replacement specification mandates identical resistance (

R_{\text{orig}} = R_{\text{new}}

).

Anahtar Kavram

Conductor resistance is directly proportional to resistivity and length, and inversely proportional to the square of conductor diameter (

R \propto \frac{\rho L}{d^2}

).

Tahmini Süre:1m 30s

Soru 85Soru →

Match each circuit modification on the left with its resulting effect on circuit current according to Ohm's law on the right.

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Cevabı ve açıklamayı göster

Cevap

Doubling voltage with constant resistance doubles current; doubling resistance with constant voltage halves current; doubling both voltage and resistance keeps current unchanged; halving voltage while doubling resistance reduces current to one-fourth.

Ohm's law states

I = \frac{V}{R}

. Analyzing each case:

V \times 2 \implies I \times 2

;

R \times 2 \implies I \times 0.5

; both

\times 2 \implies I \times 1

; and

V \times 0.5

combined with

R \times 2 \implies I \times 0.25

.

Adım Adım Çözüm

1

Recall Ohm's law relating current, voltage, and resistance

I = \frac{V}{R}

Current equals voltage divided by resistance.

2

Apply proportional reasoning for each condition

Direct proportionality for voltage, inverse proportionality for resistance.

Changes in numerator directly multiply the result, while changes in denominator inversely divide the result.

Anahtar Kavram

Ohm's law governs the relationship between voltage, current, and resistance (

I = \frac{V}{R}

), establishing that current is directly proportional to voltage and inversely proportional to resistance.

Tahmini Süre:45s

Soru 86Soru →

In direct current (DC) electronics, which of the following statements correctly describes the fundamental distinction between electromotive force (EMF) and electrical current?

Electromotive force is the rate of electron flow through a conductor, whereas electrical current is the potential energy stored per unit charge.

Electromotive force is the total resistance opposing charge movement, whereas electrical current is the mathematical product of voltage and resistance.

Electromotive force is the electrical potential difference that drives charge carriers, whereas electrical current is the rate at which charge flows through the circuit.

Electromotive force is an oscillating electrical wave present only in alternating circuits, whereas electrical current is the steady flow found only in direct current circuits.

Cevabı ve açıklamayı göster

Cevap: Electromotive force is the electrical potential difference that drives charge carriers, whereas electrical current is the rate at which charge flows through the circuit.

Cevap

Electromotive force is the electrical potential difference that drives charge carriers, whereas electrical current is the rate at which charge flows through the circuit.

Electromotive force (EMF) is the electric potential difference produced by a source (such as a battery or generator) that provides the energy and pressure needed to push free electrons through a closed circuit. Electrical current is the rate of flow of that electric charge, typically measured in amperes (coulombs per second).

Adım Adım Çözüm

1

Identify the definition and physical role of electromotive force (EMF).

EMF represents the electric potential difference (measured in volts) generated by a source, acting as the electrical pressure that drives charges.

Establishing the source of electrical push is required to distinguish cause from effect in an electrical circuit.

2

Identify the definition and physical role of electrical current.

Current represents the flow rate of electric charge (measured in amperes, or coulombs per second) passing through a conductor.

Differentiating the driving force from the resulting flow clarifies the core relationship between voltage and current.

3

Evaluate the statements to select the one that accurately pairs both concepts.

The statement describing EMF as the driving potential difference and current as the rate of charge flow is correct.

This matches standard electrical definitions without confusing units, formulas, or circuit types.

Anahtar Kavram

Electromotive Force (EMF) vs. Current

Tahmini Süre:1m 0s

Soru 87Soru →

Match each electromagnetic component or induction phenomenon with its precise operational principle.

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Cevabı ve açıklamayı göster

Cevap

Electromagnetic Relay pairs with low-current coil energization pulling isolated contacts; Inductive Back-EMF pairs with counter-EMF resisting instantaneous current changes; Step-Down Transformer pairs with mutual induction reducing voltage while increasing current capacity; Ferromagnetic Core Solenoid pairs with concentrating flux lines via high permeability to maximize mechanical force.

Each electromagnetic concept directly reflects its physical mechanism: relays isolate circuits through electromagnetically actuated mechanical contacts, back-EMF generates an opposing counter-voltage during current shifts, step-down transformers exchange voltage for current via mutual AC induction, and iron-core solenoids maximize pull force by concentrating magnetic flux density.

Adım Adım Çözüm

1

Analyze the electrical isolation and mechanical switching mechanism of an electromagnetic relay.

Identify that a relay enables a low-power circuit to safely switch a separate heavy-duty circuit via magnetic pull on an armature.

Relays function as electrically operated mechanical switches where the control coil is isolated from the load contacts.

2

Apply Lenz's Law and Faraday's Law to evaluate inductive back-EMF behavior.

Determine that the self-induced voltage always acts in the opposite polarity to the change in circuit current (

V = -L \frac{di}{dt}

).

Inductors store energy in magnetic fields and resist rapid current fluctuations by generating counter-electromotive force.

3

Examine power conservation and mutual induction relationships in a step-down transformer.

Confirm that fewer secondary turns reduce secondary voltage while stepping up current proportionally under AC operation.

Assuming ideal transformer efficiency (

P_{in} = P_{out}

), a decrease in voltage requires an inverse, proportional increase in current.

4

Assess the effect of magnetic core material on solenoid pulling performance.

Recognize that a ferromagnetic core lowers magnetic reluctance and concentrates flux lines, amplifying linear physical force.

High-permeability materials channel magnetic flux far more effectively than air, boosting magnetic field density.

Anahtar Kavram

Fundamental principles of electromagnetism, self-induction, mutual induction, and magnetic flux concentration

Tahmini Süre:2m 0s

Soru 88Soru →

When a direct-current (DC) circuit is energized, which of the following statements correctly describes the physical movement of electrical charge through a solid metallic conductor?

Electrons move from the positive terminal toward the negative terminal.

Electrons move from the negative terminal toward the positive terminal.

Voltage particles move from the negative terminal toward the positive terminal.

Protons move freely through the metallic lattice toward the positive terminal.

Cevabı ve açıklamayı göster

Cevap: Electrons move from the negative terminal toward the positive terminal.

Cevap

Electrons move from the negative terminal toward the positive terminal.

In solid conductors like copper wire, electrical current is physically composed of negatively charged free electrons drifting away from the negative power source terminal and toward the positive terminal.

Adım Adım Çözüm

1

Identify the charge carriers responsible for current in solid metal conductors.

Free (valence) electrons carry electrical charge in metals, while positive atomic nuclei remain stationary in the lattice.

Establishing the physical particle involved distinguishes electron flow from atomic or ionic movement.

2

Determine the direction of electrostatic force exerted on negative charges by the power source.

Negatively charged electrons are repelled by the negative terminal and attracted to the positive terminal.

Like charges repel and opposite charges attract, establishing the direction of electron drift.

Anahtar Kavram

Physical current in metallic conductors consists of electron flow traveling from negative potential to positive potential.

Tahmini Süre:30s

Soru 89Soru →

Match each voltage and electromotive force (EMF) concept on the left with its correct technical definition or operating behavior on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Electromotive Force (EMF) pairs with total electrical energy generated per unit charge; Terminal Voltage pairs with actual electrical potential delivered across the load under current; Internal Voltage Drop pairs with potential difference consumed within the source due to internal resistance; Open-Circuit Voltage pairs with potential difference measured across terminals when no current is drawn.

Electromotive Force (EMF) represents the total energy supplied per unit charge inside an electrical generator or chemical cell. Under no-load conditions (zero current), the open-circuit voltage equals the EMF. When connected to a load, current produces an internal voltage drop across the internal resistance, resulting in a lower measured terminal voltage available to the circuit.

Adım Adım Çözüm

1

Identify the fundamental energy conversion definition of source potential

Electromotive Force (EMF) corresponds to the total work converted to electrical energy per unit charge (

1\text{ Volt} = 1\text{ Joule per Coulomb}

).

EMF defines the driving force created by non-electrical mechanisms before any internal losses occur.

2

Differentiate between loaded operating potential and no-load terminal potential

Terminal Voltage is the operational output under load, whereas Open-Circuit Voltage is the static terminal measurement when current is zero.

When an external circuit is closed, current flowing through internal resistance reduces terminal voltage below the generated EMF.

3

Isolate the effect of internal resistance on source performance

Internal Voltage Drop represents the product of load current and internal resistance (

I \cdot r_{int}

).

Internal losses account for the discrepancy between total EMF and measured terminal voltage under load.

Anahtar Kavram

Distinction between source electromotive force, internal resistance drop, and loaded versus unloaded terminal voltage

Tahmini Süre:1m 0s

Soru 90Soru →

A copper feed wire in a naval radar subsystem has a length of $50\text{ m}$ , a circular cross-sectional diameter of $2.0\text{ mm}$ , and a measured DC resistance of $0.27\ \Omega$ . During maintenance, this line is replaced with an alloy wire that has a resistivity $4.0$ times that of copper, a total length of $75\text{ m}$ , and a diameter of $3.0\text{ mm}$ . What is the electrical resistance of the replacement wire in ohms?

Cevabı ve açıklamayı göster

Cevap: 0.72

Cevap

The electrical resistance of the replacement wire is 0.72 ohms.

Electrical resistance is given by the formula

R = \rho \frac{L}{A} = \frac{4\rho L}{\pi d^2}

. Comparing the replacement wire to the original: resistivity increases by a factor of

4.0

, length increases by a factor of

\frac{75}{50} = 1.5

, and cross-sectional area increases by a factor of

\left(\frac{3.0}{2.0}\right)^2 = 2.25

. Combining these factors gives an overall resistance change factor of

\frac{4.0 \times 1.5}{2.25} = \frac{8}{3}

. Multiplying the baseline resistance of

0.27\ \Omega

by

\frac{8}{3}

gives exactly

0.72\ \Omega

.

Adım Adım Çözüm

1

Relate electrical resistance to material resistivity and conductor dimensions.

R = \rho \frac{L}{A} = \frac{4\rho L}{\pi d^2}

Resistance is directly proportional to resistivity and length, and inversely proportional to cross-sectional area (which depends on the square of the diameter).

2

Determine the relative scaling factors for resistivity, length, and diameter.

Resistivity factor

= 4.0

; Length factor

= \frac{75}{50} = 1.5

; Diameter factor

= \frac{3.0}{2.0} = 1.5

Proportional scaling allows direct calculation without calculating intermediate microscopic values.

3

Calculate the combined scaling factor for resistance.

\frac{R_2}{R_1} = \frac{\rho_2}{\rho_1} \times \frac{L_2}{L_1} \times \left(\frac{d_1}{d_2}\right)^2 = 4.0 \times 1.5 \times \left(\frac{1}{1.5}\right)^2 = \frac{4.0}{1.5} = \frac{8}{3}

Area scales quadratically with diameter, so diameter increase reduces resistance by a factor of

1.5^2 = 2.25

.

4

Calculate the final resistance of the replacement wire.

R_2 = 0.27\ \Omega \times \frac{8}{3} = 0.72\ \Omega

Multiplying the original resistance by the overall scaling factor yields the new total resistance.

Anahtar Kavram

Dependence of electrical resistance on resistivity, conductor length, and cross-sectional area

Alternatif Yöntem

Calculate the resistivity of copper first using

\rho_{\text{Cu}} = \frac{R_1 A_1}{L_1} = \frac{0.27 \times \pi (0.001)^2}{50} \approx 1.696 \times 10^{-8}\ \Omega\cdot\text{m}

. Then calculate

\rho_{\text{alloy}} = 4.0 \times \rho_{\text{Cu}} \approx 6.786 \times 10^{-8}\ \Omega\cdot\text{m}

, and compute

R_2 = \frac{\rho_{\text{alloy}} \times 75}{\pi (0.0015)^2} = 0.72\ \Omega

.

Tahmini Süre:1m 30s

Soru 91Soru →

A direct current (DC) heating element is connected to a $120\text{ V}$ power supply. If an ammeter measures a current of $4\text{ A}$ flowing through the element, what is its electrical resistance?

480\text{ }\Omega

30\text{ }\Omega

0.033\text{ }\Omega

116\text{ }\Omega

Cevabı ve açıklamayı göster

Cevap: $30\text{ }\Omega$

Cevap

30\text{ }\Omega

According to Ohm's law, electrical resistance (

R

) is defined as the ratio of voltage (

V

) across a component to the current (

I

) passing through it:

R = \frac{V}{I}

. Substituting

120\text{ V}

and

4\text{ A}

gives

R = \frac{120}{4} = 30\text{ }\Omega

.

Adım Adım Çözüm

1

Identify the given values and the required quantity

Voltage

V = 120\text{ V}

, Current

I = 4\text{ A}

, Resistance

R

is unknown

Establishing the known circuit parameters determines which Ohm's law formula variation to use.

2

Apply Ohm's law rearranged to solve for resistance:

R = \frac{V}{I}

R = \frac{120\text{ V}}{4\text{ A}} = 30\text{ }\Omega

Resistance represents the opposition to current flow and is calculated as the ratio of electromotive force to electrical current.

Anahtar Kavram

Ohm's Law Resistance Calculation (

R = \frac{V}{I}

)

Tahmini Süre:45s

Soru 92Soru →

A solid copper wire used in an industrial application has a length of $200\text{ meters}$ and a cross-sectional area of $0.85 \, mm^2$ . The resistivity of copper at $20^\circ C$ is $1.7 \times 10^{-8} \, \Omega\cdot m$ , and its temperature coefficient of resistance is $0.004 \text{ per } ^\circ C$ . What is the resistance of this wire, in ohms, when its operating temperature reaches $70^\circ C$ ?

Cevabı ve açıklamayı göster

Cevap: 4.8

Cevap

4.8

The solution requires a synthesis of two core resistance principles. First, the baseline resistance is found by dividing the length by the area (converted to square meters) and multiplying by the resistivity, yielding 4.0 ohms. Second, the 50°C temperature increase is multiplied by the coefficient (0.004) to find a 20% increase in total resistance, raising the final value to 4.8 ohms.

Adım Adım Çözüm

1

Convert the cross-sectional area from

mm^2

to

m^2

.

0.85 \, mm^2 = 0.85 \times 10^{-6} \, m^2

Standard SI units must be used to cancel with the meters in the resistivity value.

2

Calculate the resistance at

20^\circ C

(

R_{20}

) using Pouillet's law (

R = \rho \frac{L}{A}

).

R_{20} = (1.7 \times 10^{-8}) \times \frac{200}{0.85 \times 10^{-6}} = 4.0 \, \Omega

This establishes the baseline resistance of the wire before temperature effects are considered.

3

Calculate the change in temperature (

\Delta T

).

\Delta T = 70^\circ C - 20^\circ C = 50^\circ C

The temperature coefficient is applied to the difference between the operating temperature and the reference temperature.

4

Calculate the new resistance at

70^\circ C

using the temperature coefficient formula:

R_{new} = R_{ref}[1 + \alpha(\Delta T)]

.

R_{70} = 4.0 \times [1 + 0.004 \times 50] = 4.0 \times [1 + 0.2] = 4.0 \times 1.2 = 4.8 \, \Omega

As temperature increases, the resistance of a standard conductor like copper increases proportionally.

Anahtar Kavram

Calculating absolute resistance using dimensions and resistivity, followed by applying temperature coefficients.

Tahmini Süre:3m 0s

Soru 93Soru →

Match each direct current (DC) circuit scenario on the left with its corresponding calculated electrical quantity on the right using Ohm's law ( $V = I \times R$ ).

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Cevabı ve açıklamayı göster

Cevap

The

120\text{ V}

load with

2.4\text{ k}\Omega

matches the resulting current of

50\text{ mA}

; the

5\text{ V}

circuit drawing

20\text{ mA}

matches the circuit resistance of

250\text{ }\Omega

; the

400\text{ mA}

current through

50\text{ }\Omega

matches the voltage drop of

20\text{ V}

; and the

36\text{ V}

loop carrying

3\text{ mA}

matches the loop resistance of

12\text{ k}\Omega

.

Each circuit condition correctly maps to its target electrical parameter by converting metric prefixes (

1\text{ k}\Omega = 1000\text{ }\Omega

,

1\text{ mA} = 0.001\text{ A}

) and applying the appropriate algebraic form of Ohm's law:

I = \frac{V}{R}

,

R = \frac{V}{I}

, and

V = I \times R

.

Adım Adım Çözüm

1

Calculate current for the

120\text{ V}

,

2.4\text{ k}\Omega

circuit

I = \frac{120\text{ V}}{2400\text{ }\Omega} = 0.05\text{ A} = 50\text{ mA}

Ohm's law states

I = \frac{V}{R}

, requiring resistance in fundamental ohms (

1\text{ k}\Omega = 1000\text{ }\Omega

).

2

Calculate resistance for the

5\text{ V}

,

20\text{ mA}

circuit branch

R = \frac{5\text{ V}}{0.02\text{ A}} = 250\text{ }\Omega

Rearranging Ohm's law gives

R = \frac{V}{I}

, requiring current in amperes (

20\text{ mA} = 0.02\text{ A}

).

3

Calculate voltage drop across the

50\text{ }\Omega

element carrying

400\text{ mA}

V = 0.4\text{ A} \times 50\text{ }\Omega = 20\text{ V}

Ohm's law formula

V = I \times R

computes potential difference when current and resistance are known.

4

Calculate total resistance for the

36\text{ V}

,

3\text{ mA}

sensor loop

R = \frac{36\text{ V}}{0.003\text{ A}} = 12000\text{ }\Omega = 12\text{ k}\Omega

Applying

R = \frac{V}{I}

with

3\text{ mA} = 0.003\text{ A}

yields

12000\text{ }\Omega

, or

12\text{ k}\Omega

.

Anahtar Kavram

Ohm's law mathematically relates voltage (

V

), current (

I

), and resistance (

R

) via

V = I \times R

,

I = \frac{V}{R}

, and

R = \frac{V}{I}

, requiring consistent base metric units.

Tahmini Süre:1m 15s

Soru 94Soru →

A technician wraps two separate copper wire coils, Coil 1 and Coil 2, around a single soft iron core. Coil 1 is connected to a steady 12V DC power supply, and the current has reached a steady state. Coil 2 is connected only to a sensitive digital voltmeter. If the technician rapidly pulls the soft iron core completely out of the coils while Coil 1 remains connected to the power supply, which of the following best describes the reading on the voltmeter during the exact moment the core is moving?

A voltage reading that spikes momentarily and then returns to zero, because the sudden decrease in magnetic permeability causes a rapid change in magnetic flux.

A continuous reading of zero volts, because a direct current (DC) power supply cannot produce the alternating magnetic field required for electromagnetic induction.

A steady 12V reading, because removing the iron core eliminates the magnetic resistance between the coils, allowing the full electromotive force to transfer.

A permanently higher voltage reading, because the opposing magnetic poles of the iron core are no longer present to attract and restrict the electron flow.

Cevabı ve açıklamayı göster

Cevap: A voltage reading that spikes momentarily and then returns to zero, because the sudden decrease in magnetic permeability causes a rapid change in magnetic flux.

Cevap

The correct answer is a voltage reading that spikes momentarily and then returns to zero, because the sudden decrease in magnetic permeability causes a rapid change in magnetic flux.

Electromagnetic induction requires a changing magnetic flux over time. While a steady DC current normally produces a constant, unchanging magnetic field, the physical act of rapidly removing the highly permeable iron core drastically reduces the magnetic flux. This rapid change in flux induces a temporary voltage spike in the secondary coil. The induction ceases as soon as the core is completely removed, because the flux becomes constant again, returning the voltmeter reading to zero.

Adım Adım Çözüm

1

Analyze the initial steady state of the circuit.

Coil 1 produces a constant magnetic flux that is concentrated by the soft iron core. Because the flux is not changing over time, the induced voltage in Coil 2 is zero.

Faraday's Law of Induction requires a changing magnetic field to induce an electromotive force (EMF).

2

Evaluate the physical change caused by the technician's action.

The technician removes the soft iron core, replacing a material with high magnetic permeability with air, which has very low magnetic permeability.

Magnetic permeability directly determines how well a material can support the formation of a magnetic field within itself.

3

Determine the effect of this physical change on the magnetic flux.

The total magnetic flux passing through both coils drops rapidly as the highly permeable core is removed.

Less permeability means weaker magnetic field strength for the same amount of current.

4

Apply Faraday's Law to the changing magnetic flux.

The rapidly decreasing magnetic flux induces a temporary voltage (a spike) in Coil 2.

An EMF is induced in a coil whenever there is a change in magnetic flux passing through it, proportional to the rate of change (

E = -N \times \frac{\Delta\Phi}{\Delta t}

).

5

Determine the final state after the core is completely removed.

The magnetic flux stabilizes at a new, much lower constant value. Since the flux is no longer changing, the induced voltage in Coil 2 returns to zero.

Induction only occurs during the period of change.

Anahtar Kavram

Faraday's Law of Induction and Magnetic Permeability

Soru 95Soru →

In standard metallic conductors such as copper wire, electric current consists of the directed movement of free valence electrons drifting through the material.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True

The statement is true because the physical flow of electric current in metallic conductors like copper is carried exclusively by free valence electrons drifting through the atomic lattice under an applied electric field.

Adım Adım Çözüm

1

Identify the charge carriers present in solid metallic conductors.

Metals have an atomic structure characterized by a lattice of fixed positive ions surrounded by loosely held, mobile valence electrons.

Determining which atomic particles are free to move establishes the physical mechanism of conduction.

2

Evaluate the particle movement that constitutes electrical current.

When a voltage source is applied, these free valence electrons drift collectively through the conductor, which defines the physical flow of electric current.

Atomic nuclei remain locked in the crystal lattice while only the free electrons travel through the wire.

Anahtar Kavram

Physical electric current in solid metallic conductors is the collective drift of free valence electrons.

Tahmini Süre:30s

Soru 96Soru →

A technician connects a primary coil wound around a soft iron core to a constant $12\text{ V}$ direct current (DC) supply, establishing a steady magnetic field through an adjacent secondary coil. Which of the following modifications will produce a continuous, sustained output voltage across the secondary coil?

Increasing the steady DC supply voltage from

12\text{ V}

to

24\text{ V}

while holding the current constant

Replacing the DC supply with an alternating current (AC) source to continuously vary the magnetic flux

Reversing the DC terminal polarity so that steady electron flow moves in the opposite direction

Doubling the number of turns in the secondary winding while maintaining the steady DC input

Cevabı ve açıklamayı göster

Cevap: Replacing the DC supply with an alternating current (AC) source to continuously vary the magnetic flux

Cevap

Replacing the DC supply with an alternating current (AC) source to continuously vary the magnetic flux

According to Faraday's law of induction, an electromotive force (EMF) is only induced when there is relative motion or a changing magnetic flux linking the conductor. An alternating current (AC) continuously cycles in amplitude and direction, generating an expanding and collapsing magnetic field that sustains a continuous output voltage in the secondary coil.

Adım Adım Çözüm

1

Apply Faraday's Law of Electromagnetic Induction

The induced electromotive force is given by

\mathcal{E} = -N \frac{d\Phi}{dt}

, where

\frac{d\Phi}{dt}

represents the rate of change of magnetic flux.

Induction strictly depends on a dynamic, changing magnetic field rather than a static magnetic field.

2

Evaluate the magnetic field produced by steady DC

A steady direct current creates a constant magnetic flux (

\frac{d\Phi}{dt} = 0

), producing

0\text{ V}

across the secondary winding in steady state.

Static direct currents cannot sustain continuous transformer action or mutual induction.

3

Determine the condition for continuous, sustained induction

Supplying alternating current (AC) causes the magnetic flux to expand and collapse continuously, providing a perpetual non-zero

\frac{d\Phi}{dt}

.

Continuous induction requires a continuously oscillating or varying current source.

Anahtar Kavram

Electromagnetic Induction and Time-Varying Magnetic Flux

Tahmini Süre:1m 0s

Soru 97Soru →

A direct current (DC) solenoid valve in an industrial pneumatic system has an internal coil resistance of $15\text{ }\Omega$ . When connected to a $30\text{ V}$ DC power supply, what is the current flowing through the solenoid coil in amperes ( $\text{A}$ )?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The current flowing through the solenoid coil is

2\text{ A}

.

According to Ohm's law, electrical current (

I

) equals voltage (

V

) divided by resistance (

R

). Substituting the given values yields

I = \frac{30\text{ V}}{15\text{ }\Omega} = 2\text{ A}

.

Adım Adım Çözüm

1

Identify the given circuit parameters

Voltage

V = 30\text{ V}

, Resistance

R = 15\text{ }\Omega

Extract the given values to identify which Ohm's law formula variation is required.

2

State Ohm's law formula for current

I = \frac{V}{R}

Current is directly proportional to applied voltage and inversely proportional to total circuit resistance.

3

Substitute the values and evaluate

I = \frac{30\text{ V}}{15\text{ }\Omega} = 2\text{ A}

Dividing

30

by

15

yields the circuit current in amperes.

Anahtar Kavram

Ohm's Law (

I = \frac{V}{R}

)

Tahmini Süre:45s

Soru 98Soru →

In electronics and solid-state physics, materials are fundamentally classified by the number of valence electrons in their outermost atomic shell and their conductivity characteristics. An element contains exactly four valence electrons in its outer shell, forming covalent bonds in a crystalline lattice that exhibits moderate resistivity at room temperature and increased conductivity as temperature rises. Which of the following substances possesses these specific atomic and electrical properties?

Copper

Silicon

Mica

Aluminum

Cevabı ve açıklamayı göster

Cevap: Silicon

Cevap

Silicon, because it is a semiconductor with four valence electrons in its outer shell and a negative temperature coefficient of resistance.

Silicon is an elemental semiconductor belonging to Group IV of the periodic table, meaning each atom has four valence electrons. In a solid crystal lattice, these four electrons form covalent bonds with adjacent atoms. At room temperature, thermal excitation frees a small fraction of these electrons across the narrow energy gap into the conduction band, giving silicon an intermediate level of conductivity that increases as thermal energy rises.

Adım Adım Çözüm

1

Identify the material category based on the outer shell valence electron count.

An atom with exactly four valence electrons is classified as a semiconductor (Group IV element).

Good conductors typically have 1 to 3 valence electrons, semiconductors have exactly 4, and insulators generally have 5 to 8 tightly held valence electrons.

2

Evaluate the electrical conduction behavior described in the scenario.

The material exhibits moderate resistivity and a negative temperature coefficient of resistance (conductivity increases with temperature).

In intrinsic semiconductors, added thermal energy breaks covalent bonds, releasing electron-hole pairs into the conduction band to facilitate charge flow.

3

Match the specified atomic and electrical characteristics to the given material choices.

Silicon is a Group IV semiconductor, whereas copper and aluminum are conductors and mica is an insulator.

Silicon has an atomic number of 14 with an electron configuration of 2-8-4, giving it precisely four valence electrons.

Anahtar Kavram

Semiconductor atomic structure and valence electron properties

Tahmini Süre:1m 15s

Soru 99Soru →

A direct current (DC) emergency lighting circuit operates from a $28\text{ V}$ power supply. An in-line ammeter measures a steady current flow of $350\text{ mA}$ through the circuit. What is the total electrical resistance of the lighting circuit?

0.08\text{ }\Omega

9.8\text{ }\Omega

12.5\text{ }\Omega

80\text{ }\Omega

Cevabı ve açıklamayı göster

Cevap: $80\text{ }\Omega$

Cevap

The total electrical resistance of the lighting circuit is

80\text{ }\Omega

.

According to Ohm's law, resistance is found by dividing voltage by current (

R = V / I

). Converting

350\text{ mA}

into amperes yields

0.35\text{ A}

. Dividing

28\text{ V}

by

0.35\text{ A}

gives the correct value of

80\text{ }\Omega

.

Adım Adım Çözüm

1

Convert the current measurement from milliamperes to amperes.

I = 350\text{ mA} = \frac{350}{1000}\text{ A} = 0.35\text{ A}

Standard Ohm's law calculations require current to be in base SI units (amperes) to produce resistance in ohms.

2

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

).

R = \frac{V}{I}

Resistance equals the electromotive force (voltage) across the circuit divided by the resulting current flow.

3

Substitute the known voltage and converted current values into the rearranged formula.

R = \frac{28\text{ V}}{0.35\text{ A}} = \frac{2800}{35}\text{ }\Omega = 80\text{ }\Omega

Dividing

28\text{ V}

by

0.35\text{ A}

determines the load resistance in ohms.

Anahtar Kavram

Ohm's Law (

V = I \times R

) defines the mathematical relationship between voltage, current, and resistance in DC circuits, requiring proper metric unit conversion before calculation.

Tahmini Süre:1m 0s

Soru 100Soru →

A remote weather station is powered by a battery bank containing two identical parallel branches. Each branch is composed of three $12\text{ V}$ batteries connected in series aiding. What total electromotive force (EMF) is delivered across the main output terminals of the bank?

18\text{ V}

72\text{ V}

36\text{ V}

12\text{ V}

Cevabı ve açıklamayı göster

Cevap: $36\text{ V}$

Cevap

36\text{ V}

In a battery bank containing series strings wired in parallel, each series branch sums the voltages of its three

12\text{ V}

batteries to produce

36\text{ V}

(

12\text{ V} + 12\text{ V} + 12\text{ V} = 36\text{ V}

). Connecting two identical

36\text{ V}

strings in parallel keeps the potential difference across the output terminals at

36\text{ V}

while doubling current capacity.

Adım Adım Çözüm

1

Calculate the total electromotive force of a single series-connected branch.

12\text{ V} + 12\text{ V} + 12\text{ V} = 36\text{ V}

When DC voltage sources are wired in series aiding, their individual voltages add directly.

2

Determine the combined electromotive force delivered by the parallel configuration.

36\text{ V}

Connecting identical voltage branches in parallel provides the same terminal voltage as a single branch while expanding overall current capacity.

Anahtar Kavram

Series-parallel voltage principles dictate that series-connected sources sum their electromotive forces, whereas connecting identical strings in parallel maintains that branch voltage across the load terminals.

Basic Electrical Concepts

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