Basic Electrical Concepts Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 4

Soru 61Soru →

An off-grid solar power system stores electrical energy in a $24\text{ V}$ direct current (DC) battery bank. To operate standard household appliances that require $120\text{ V}$ alternating current (AC), which electronic device must be used to perform the power conversion?

A bridge rectifier

A power inverter

A step-up transformer

A forward-biased diode

Cevabı ve açıklamayı göster

Cevap: A power inverter

Cevap

A power inverter is required to convert direct current (DC) from a battery into alternating current (AC) for household appliances.

A power inverter is designed specifically to take a steady, unidirectional direct current (DC) input from a source such as a battery and electronically convert it into an oscillating alternating current (AC) output suitable for standard appliances.

Adım Adım Çözüm

1

Identify the electrical characteristics of the input source and the required output load

The battery provides steady unidirectional direct current (DC), whereas household appliances require periodically reversing alternating current (AC).

Determining the input and output current types is necessary to select the correct conversion device.

2

Evaluate the functional role of conversion and transformation devices

An inverter switches and shapes DC into AC voltage, whereas a rectifier converts AC to DC, and a transformer changes AC voltage levels without converting DC to AC.

Direct current requires active electronic switching to produce the oscillating waveform of alternating current.

Anahtar Kavram

Direct Current (DC) vs. Alternating Current (AC) conversion principles

Soru 62Soru →

Match each voltage-related electrical term on the left with its correct technical definition or operational characteristic on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Electromotive Force corresponds to the total energy per unit charge supplied by an active source; Voltage Drop corresponds to the loss of potential across a circuit component; Volt corresponds to one joule per coulomb; and Voltmeter corresponds to the high-internal-resistance instrument connected in parallel.

Each term accurately matches its foundational principle: EMF describes the electrical force provided by a source, voltage drop denotes the potential consumed across a load, the volt is the SI unit defining joules per coulomb, and a voltmeter is the high-impedance instrument used in parallel across components.

Adım Adım Çözüm

1

Define Electromotive Force (EMF)

EMF represents the source voltage generated by converting chemical, mechanical, or thermal energy into electrical energy per unit charge.

An active energy source establishes the potential difference that pushes electrons through the external circuit.

2

Distinguish Voltage Drop across a load

Voltage drop reflects the potential energy consumed when charges traverse a resistive element.

Ohm's Law governs potential loss (

V = IR

) as electric energy is converted into heat, light, or work.

3

Identify the standard physical definition of the Volt

One volt (

1\text{ V}

) is mathematically equivalent to one joule per coulomb (

1\text{ J/C}

).

Voltage quantifies potential energy difference per unit electric charge.

4

Determine the operational method of a Voltmeter

A voltmeter measures potential difference between two nodes and requires a parallel connection with high internal resistance.

High impedance ensures the meter does not draw significant current from the branch under test.

Anahtar Kavram

Voltage represents electrical pressure and potential energy per unit charge, manifesting as source EMF, component voltage drops, and measured in volts using parallel-connected voltmeters.

Tahmini Süre:1m 0s

Soru 63Soru →

Match each electrical quantity from Ohm's law on the left with its correct algebraic formula on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Voltage (

V

) matches with

I \times R

, Current (

I

) matches with

\frac{V}{R}

, and Resistance (

R

) matches with

\frac{V}{I}

.

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

V = IR

). Solving for current isolates

I

by dividing voltage by resistance (

I = \frac{V}{R}

), and solving for resistance isolates

R

by dividing voltage by current (

R = \frac{V}{I}

).

Adım Adım Çözüm

1

Identify the base equation of Ohm's law that calculates potential difference.

V = I \times R

Ohm's law defines voltage as the product of current and resistance.

2

Rearrange the base formula to solve for current by isolating

I

.

I = \frac{V}{R}

Dividing both sides of the base formula by resistance isolates current.

3

Rearrange the base formula to solve for resistance by isolating

R

.

R = \frac{V}{I}

Dividing both sides of the base formula by current isolates resistance.

Anahtar Kavram

Ohm's Law algebraic variations (

V = IR

,

I = \frac{V}{R}

,

R = \frac{V}{I}

)

Tahmini Süre:45s

Soru 64Soru →

In a direct-current (DC) circuit powered by a battery, the physical drift velocity of individual free electrons through a solid copper conductor approaches the speed of light, which directly accounts for the nearly instantaneous delivery of electrical energy to a connected load.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False

The statement is false. Individual free electrons in a copper wire drift at an extremely slow pace—typically less than

1\text{ mm/s}

—because of continuous collisions with lattice ions. The virtually instantaneous illumination of a lamp or activation of a load occurs because the electric field travels through the wire at nearly the speed of light, setting all free electrons throughout the circuit into motion almost simultaneously.

Adım Adım Çözüm

1

Differentiate between charge carrier drift velocity and electromagnetic field propagation in a conductor.

Drift velocity describes the slow, net physical displacement of electrons through the lattice, whereas field propagation is the speed at which the electromotive force and electric field establish across the circuit.

Understanding electrical current requires distinguishing the physical transport of matter and charge from the propagation of the energy-carrying electromagnetic wave.

2

Analyze the typical magnitude of electron drift velocity (

v_d

) in metallic conductors.

Due to frequent collisions with vibrating copper atoms in the lattice, the net drift velocity of electrons under direct current is on the order of millimeters per second or less (

10^{-4}\text{ m/s}

to

10^{-3}\text{ m/s}

).

Thermal motion is fast and random, but the directional drift caused by the applied electric field is very slow.

3

Identify the mechanism responsible for near-instantaneous energy transfer when closing a circuit switch.

When the circuit closes, an electric field propagates through the conductor at a significant fraction of the speed of light (

c

), simultaneously exerting a force on all free electrons already present in the wire.

Energy is transferred through the field rather than by individual electrons racing from the source to the load.

4

Evaluate the truth value of the overall technical statement.

Because physical electron drift velocity is microscopic rather than approaching the speed of light, the assertion is false.

The statement erroneously attributes energy delivery speed to physical carrier velocity instead of field propagation.

Anahtar Kavram

Electron drift velocity versus electric field propagation in current flow

Tahmini Süre:45s

Soru 65Soru →

Rechargeable batteries store electrical energy as direct current (DC) and cannot be charged directly from an alternating current (AC) source without first converting the current to DC.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True

Chemical energy storage in batteries relies on unidirectional electron flow (direct current) to sustain the oxidation-reduction reactions necessary for charging. Because alternating current reverses direction at regular intervals, it cannot build up a net chemical charge without first being converted to direct current by a rectifier.

Adım Adım Çözüm

1

Identify the operating principle of battery charging and electrochemical energy storage

Chemical batteries operate via fixed-polarity reduction-oxidation reactions, requiring unidirectional direct current (DC) to drive electrons into the negative electrode.

Electrochemical storage depends on electrons flowing consistently in a single direction to reverse cell discharge.

2

Analyze the behavior of alternating current (AC) applied to battery terminals

AC periodically reverses direction (e.g., 60 times per second in a

60\text{ Hz}

system), attempting to charge during one half-cycle and discharge during the next.

Without converting AC to DC via a rectifier, there is zero net accumulation of chemical charge.

3

Evaluate the validity of the statement

The statement is True.

Batteries store energy as DC and require AC-to-DC conversion (rectification) before charging from an AC source.

Anahtar Kavram

Direct current (DC) maintains unidirectional electron flow required for electrochemical storage, whereas alternating current (AC) periodically reverses direction and cannot charge chemical batteries without rectification.

Tahmini Süre:45s

Soru 66Soru →

An electronic circuit contains a $20\text{ k}\Omega$ precision resistor that dissipates $180\text{ mW}$ of electrical power. What is the DC voltage drop across this resistor?

9\text{ V}

60\text{ V}

6\text{ V}

3\text{ V}

Cevabı ve açıklamayı göster

Cevap: $60\text{ V}$

Cevap

The DC voltage drop across the resistor is

60\text{ V}

.

According to Watt's Law, electrical power dissipated in a resistive circuit can be expressed as

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

. Solving for voltage yields

V = \sqrt{P \times R}

. Converting the prefix values gives

180\text{ mW} = 0.18\text{ W}

and

20\text{ k}\Omega = 20{}000\ \Omega

. Multiplying these values gives

V^2 = 0.18 \times 20{}000 = 3600\text{ V}^2

, and taking the square root gives

\sqrt{3600} = 60\text{ V}

.

Adım Adım Çözüm

1

Convert given values to standard base units (Watts and Ohms)

P = 180\text{ mW} = 180 \times 10^{-3}\text{ W} = 0.18\text{ W}

;

R = 20\text{ k}\Omega = 20 \times 10^{3}\ \Omega = 20{}000\ \Omega

Standard electrical formulas require base SI units to prevent magnitude errors.

2

Select and rearrange the appropriate Watt's Law formula relating power, voltage, and resistance

P = \frac{V^2}{R} \implies V^2 = P \times R \implies V = \sqrt{P \times R}

The known variables are power (

P

) and resistance (

R

), and the unknown variable is voltage (

V

).

3

Calculate the product of power and resistance

V^2 = (180 \times 10^{-3}\text{ W}) \times (20 \times 10^{3}\ \Omega) = 180 \times 20 = 3600\text{ V}^2

The metric prefix multipliers

10^{-3}

and

10^{3}

cancel each other out (

10^{-3} \times 10^{3} = 1

).

4

Take the square root to determine the voltage

V = \sqrt{3600\text{ V}^2} = 60\text{ V}

Taking the square root of

V^2

yields the electromotive force across the component.

Anahtar Kavram

Watt's Law power calculation using

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

and rearranging to solve for voltage with metric prefix conversion.

Alternatif Yöntem

First, find the current using

P = I^2 \times R \implies I = \sqrt{\frac{P}{R}} = \sqrt{\frac{0.18\text{ W}}{20{}000\ \Omega}} = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3}\text{ A} = 3\text{ mA}

. Then apply Ohm's Law to find voltage:

V = I \times R = (3 \times 10^{-3}\text{ A}) \times (20 \times 10^{3}\ \Omega) = 60\text{ V}

.

Tahmini Süre:1m 30s

Soru 67Soru →

In a direct-current (DC) circuit connected to a battery, what creates electrical current through the metallic conductor?

The accumulation of stationary electric potential without the movement of physical charges

The flow of free electrons moving from the negative terminal toward the positive terminal

The continuous movement of electrons directed from the positive terminal toward the negative terminal

The periodic back-and-forth oscillation of electrons without net forward drift

Cevabı ve açıklamayı göster

Cevap: The flow of free electrons moving from the negative terminal toward the positive terminal

Cevap

The flow of free electrons moving from the negative terminal toward the positive terminal

Electric current in metallic conductors consists of the movement of free electrons. Because electrons carry a negative electrical charge, the electromotive force from a DC source repels them from the negative terminal and attracts them toward the positive terminal, creating a continuous unidirectional drift.

Adım Adım Çözüm

1

Identify the mobile charge carriers in a metallic conductor.

In metals, loosely bound valence electrons (free electrons) move between atoms.

Atomic nuclei are fixed in the metallic lattice, leaving free electrons as the mobile carriers of electric charge.

2

Determine the direction of electron drift under a DC voltage source.

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

According to electrostatic principles, like charges repel and opposite charges attract.

Anahtar Kavram

Electric current in a metallic DC circuit is the physical flow of free electrons drifting from the negative terminal toward the positive terminal.

Tahmini Süre:35s

Soru 68Soru →

An automotive technician tests a $12\text{ V}$ direct current (DC) circuit powering an auxiliary cooling fan. If the multimeter measures a current of $3\text{ A}$ flowing through the fan motor, what is the electrical resistance of the motor?

36\text{ }\Omega

0.25\text{ }\Omega

4\text{ }\Omega

9\text{ }\Omega

Cevabı ve açıklamayı göster

Cevap: $4\text{ }\Omega$

Cevap

4\text{ }\Omega

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

R = \frac{V}{I}

). Dividing

12\text{ V}

by

3\text{ A}

yields

4\text{ }\Omega

.

Adım Adım Çözüm

1

Identify the known values and target quantity from Ohm's law.

Voltage (

V

) =

12\text{ V}

, Current (

I

) =

3\text{ A}

, Resistance (

R

) = unknown.

Setting up the given parameters ensures the proper mathematical formula is selected.

2

Rearrange Ohm's law (

V = I \times R

) to solve for resistance (

R

).

R = \frac{V}{I}

Dividing both sides by current isolates resistance as the subject.

3

Substitute the measured values into the formula and calculate.

R = \frac{12\text{ V}}{3\text{ A}} = 4\text{ }\Omega

Carrying out the division gives the total electrical resistance of the fan motor.

Anahtar Kavram

Ohm's Law:

R = \frac{V}{I}

Tahmini Süre:45s

Soru 69Soru →

An avionics technician reads a specification sheet indicating that an auxiliary battery provides an electromotive force (EMF) of $28\text{ V}$ . What does this $28\text{ V}$ EMF rating fundamentally represent?

The battery supplies

28\text{ joules}

of electrical energy for every

1\text{ coulomb}

of charge moved.

The battery maintains a continuous current flow of

28\text{ coulombs}

of charge per second.

The battery exerts an internal opposition to electron flow equal to

28\text{ ohms}

.

The battery produces alternating polarity reversals at a rate of

28\text{ cycles}

per second.

Cevabı ve açıklamayı göster

Cevap: The battery supplies $28\text{ joules}$ of electrical energy for every $1\text{ coulomb}$ of charge moved.

Cevap

The battery supplies

28\text{ joules}

of electrical energy for every

1\text{ coulomb}

of charge moved.

Electromotive force (EMF), measured in volts, represents the electrical potential energy imparted to charge carriers by a source. By definition, one volt equals one joule of energy per one coulomb of charge (

1\text{ V} = 1\text{ J/C}

). Therefore, a

28\text{ V}

rating indicates that the battery supplies

28\text{ joules}

of energy for each coulomb of charge transferred.

Adım Adım Çözüm

1

Recall the fundamental physical definition of electromotive force (EMF) and electric potential difference.

Voltage (

V

) is defined as electrical work or energy (

W

) per unit charge (

Q

), expressed by the formula

V = \frac{W}{Q}

.

Electromotive force represents the electric pressure or energy supplied to charge carriers by a power source.

2

Apply the standard units of measurement to the voltage formula.

1\text{ volt} = 1\text{ joule per coulomb}

(

1\text{ V} = 1\text{ J/C}

).

Joules measure energy or work done, while coulombs measure quantity of electric charge.

3

Evaluate a potential difference of

28\text{ V}

using these units.

A rating of

28\text{ V}

means

28\text{ joules}

of electrical energy are supplied to each coulomb of charge.

Direct multiplication:

28\text{ V} = 28\text{ J/C}

.

Anahtar Kavram

Voltage and Electromotive Force (EMF) as electrical potential energy per unit charge (

1\text{ V} = 1\text{ J/C}

)

Tahmini Süre:1m 0s

Soru 70Soru →

In a closed direct-current (DC) circuit operating under steady-state conditions, continuous electrical current requires a complete conducting loop through both the external load and the power source. Which of the following correctly describes the direction of physical electron movement throughout the complete circuit?

Electrons flow from the positive terminal to the negative terminal through the external load, and from the negative terminal to the positive terminal inside the power source.

Electrons flow from the negative terminal to the positive terminal through the external load, and are driven from the positive terminal to the negative terminal inside the power source.

Electrons circulate in a single uninterrupted direction from the negative terminal to the positive terminal through both the external conductor and inside the power source.

Electrons and conventional current both travel from the positive terminal to the negative terminal through the external load, driven directly by electromotive force.

Cevabı ve açıklamayı göster

Cevap: Electrons flow from the negative terminal to the positive terminal through the external load, and are driven from the positive terminal to the negative terminal inside the power source.

Cevap

Electrons flow from the negative terminal to the positive terminal through the external load, and are driven from the positive terminal to the negative terminal inside the power source.

In any direct-current circuit, free electrons in the external conductors drift from the negative terminal toward the positive terminal due to electrostatic forces. To sustain continuous current, the power source utilizes electromotive force to pump electrons internally from the positive terminal back to the negative terminal, completing the closed cycle.

Adım Adım Çözüm

1

Analyze external electron movement

External drift is from the negative terminal to the positive terminal.

Free electrons in metallic conductors carry a negative electric charge and are repelled by the negative battery terminal and attracted toward the positive terminal.

2

Analyze closed-loop continuity inside the source

Internal drift is from the positive terminal to the negative terminal.

To maintain a continuous current without charge buildup, the source's electromotive force (chemical or mechanical energy) does work against the internal electric field to transport electrons from the positive terminal back to the negative terminal.

3

Synthesize full loop current dynamics

External flow is negative-to-positive; internal transport is positive-to-negative.

A complete electrical circuit forms a closed loop, ensuring conservation of electric charge across all nodes.

Anahtar Kavram

Closed-Loop Electron Drift and Charge Conservation in DC Circuits

Tahmini Süre:45s

Soru 71Soru →

Electric utility companies distribute power across regional power grids using high-voltage alternating current (AC) rather than direct current (DC). Which of the following best explains why alternating current is preferred for long-distance power distribution?

Alternating current maintains a continuous, unidirectional electron flow that completely eliminates heat dissipation in transmission cables.

Alternating current allows power transmission using voltage alone without requiring any electrical current flow through the conductors.

Alternating current voltage levels can be efficiently stepped up and down using transformers to minimize transmission power losses.

Alternating current can be directly stored in standard chemical battery banks at substations without needing rectification.

Cevabı ve açıklamayı göster

Cevap: Alternating current voltage levels can be efficiently stepped up and down using transformers to minimize transmission power losses.

Cevap

Alternating current voltage levels can be efficiently stepped up and down using transformers to minimize transmission power losses.

Alternating current produces a continuously alternating magnetic field that allows transformers to efficiently step up voltage to very high levels for long-distance transmission, which lowers current and minimizes resistive heat losses (

I^2R

). Transformers can then step the voltage down to standard consumer levels.

Adım Adım Çözüm

1

Identify the primary electrical requirement for long-distance power distribution.

Long-distance lines suffer power loss proportional to the square of current (

P_{\text{loss}} = I^2R

). Reducing current by raising voltage reduces energy loss.

Transmission cables have inherent resistance over miles of wire, creating significant power dissipation if current is high.

2

Evaluate the operational principle of transformers with alternating current versus direct current.

Transformers rely on mutual electromagnetic induction, which requires a constantly changing magnetic field generated by alternating current.

Steady direct current does not produce a continuous rate of change in magnetic flux, preventing transformers from stepping direct current voltages up or down.

3

Connect transformer functionality to the preference for alternating current in power distribution grids.

Alternating current can be stepped up to hundreds of kilovolts for efficient transmission and stepped down to safe voltages for end users.

This ease of voltage transformation makes alternating current the standard choice for long-distance power grids.

Anahtar Kavram

Transformer compatibility and transmission efficiency differences between alternating current and direct current.

Tahmini Süre:45s

Soru 72Soru →

In electrical circuits, an electric current of one ampere ( $1\text{ A}$ ) is defined as one coulomb of electrical charge passing through a given point in one second.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True

Electric current is the measure of how much electrical charge flows past a specific point per unit of time. Under the International System of Units (SI), a current of one ampere is exactly one coulomb of charge traveling past a reference plane per second.

Adım Adım Çözüm

1

Identify the standard formula relating electric current, electrical charge, and time.

The formula is

I = \frac{Q}{t}

, where

I

is current,

Q

is charge in coulombs, and

t

is time in seconds.

Current quantifies the rate of electrical charge movement across a conductor cross-section.

2

Substitute standard SI base units into the rate equation.

1\text{ Ampere} = \frac{1\text{ Coulomb}}{1\text{ Second}}

(

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

).

This establishes that one ampere directly equals one coulomb per second.

Anahtar Kavram

Definition of electric current as the rate of electrical charge flow in coulombs per second (

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

).

Tahmini Süre:30s

Soru 73Soru →

A DC circuit consists of a $20\text{ V}$ power source connected in series with a $4\text{ }\Omega$ resistor ( $R_1$ ). After passing through $R_1$ , the circuit splits into two parallel branches containing an $8\text{ }\Omega$ resistor ( $R_2$ ) and a $24\text{ }\Omega$ resistor ( $R_3$ ). The branches then recombine and connect back to the negative terminal of the power source. What is the electrical power dissipated by the $8\text{ }\Omega$ resistor ( $R_2$ )?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

18 Watts

To find the power dissipated by a single component in a compound circuit, you must isolate the parameters for that specific component. By finding the total resistance (

10\text{ }\Omega

) and total current (

2\text{ A}

), we can determine that the voltage dropping specifically across the parallel section is

12\text{ V}

. Because components in parallel share the same voltage, the

8\text{ }\Omega

resistor receives the full

12\text{ V}

. Using Watt's Law,

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

, we find

\frac{12^2}{8} = 18\text{ W}

.

Adım Adım Çözüm

1

Calculate the equivalent resistance of the parallel section (

R_p

).

R_p = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\text{ }\Omega

Before analyzing the whole circuit, parallel branches must be simplified into a single equivalent resistance.

2

Calculate the total resistance of the circuit (

R_t

).

R_t = 4\text{ }\Omega + 6\text{ }\Omega = 10\text{ }\Omega

The equivalent parallel resistance is in series with the

4\text{ }\Omega

resistor.

3

Calculate the total current (

I_t

) leaving the power source.

I_t = \frac{20\text{ V}}{10\text{ }\Omega} = 2\text{ A}

Ohm's law (

I = \frac{V}{R}

) relates the total source voltage to the total circuit resistance.

4

Calculate the voltage drop across the parallel section (

V_p

).

V_p = 2\text{ A} \times 6\text{ }\Omega = 12\text{ V}

The total current flows through the equivalent parallel resistance. This gives the voltage available specifically to the parallel branches.

5

Calculate the power dissipated by the

8\text{ }\Omega

resistor (

R_2

).

P = \frac{12^2}{8} = \frac{144}{8} = 18\text{ W}

Using Watt's Law (

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

), we apply the voltage specific to the parallel branch (

12\text{ V}

) and the individual resistance (

8\text{ }\Omega

).

Anahtar Kavram

Calculating power dissipation in a specific component of a series-parallel circuit requires finding the exact voltage or current acting on that specific component, not the entire circuit.

Soru 74Soru →

A field communications transceiver is powered by a battery string consisting of 7 series-connected cells. Six of the cells each supply an electromotive force (EMF) of $2.4\text{ V}$ in a series-aiding configuration, while one $2.4\text{ V}$ cell is connected in series opposing. What is the net electromotive force, in volts ( $\text{V}$ ), delivered by the battery string?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The net electromotive force delivered by the battery string is

12\text{ V}

.

In a series circuit, aiding EMFs add together while opposing EMFs subtract:

V_{\text{net}} = (6 \times 2.4\text{ V}) - (1 \times 2.4\text{ V}) = 14.4\text{ V} - 2.4\text{ V} = 12.0\text{ V}

.

Adım Adım Çözüm

1

Calculate the total electromotive force contributed by the series-aiding cells.

6 \times 2.4\text{ V} = 14.4\text{ V}

When cells are connected in series aiding (the positive terminal of one connected to the negative terminal of the next), their voltages add together directly.

2

Subtract the electromotive force of the series-opposing cell from the aiding total.

14.4\text{ V} - 2.4\text{ V} = 12.0\text{ V}

A cell connected in reverse polarity opposes the circuit electromotive force, subtracting its potential difference from the total.

Anahtar Kavram

Series electromotive force addition and subtraction: aiding cells add potential (

+V

), whereas opposing cells subtract potential (

-V

).

Alternatif Yöntem

Find the effective number of aiding cells:

6 - 1 = 5

net aiding cells. Then calculate net voltage:

5 \times 2.4\text{ V} = 12.0\text{ V}

.

Tahmini Süre:45s

Soru 75Soru →

A technician connects a direct current (DC) indicator lamp across a $10\text{ V}$ power supply. A digital multimeter connected in series measures a current of $0.5\text{ A}$ flowing through the circuit. What is the electrical resistance of the indicator lamp in ohms ( $\Omega$ )?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The resistance of the indicator lamp is

20\text{ }\Omega

.

According to Ohm's Law, resistance is determined by dividing voltage by current (

R = \frac{V}{I}

). Substituting

10\text{ V}

and

0.5\text{ A}

gives

R = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega

.

Adım Adım Çözüm

1

Identify the given circuit parameters from the problem statement.

Voltage

V = 10\text{ V}

, Current

I = 0.5\text{ A}

Applying Ohm's law requires identifying the known electromotive force and circuit current.

2

Select and arrange the Ohm's law equation to solve for resistance (

R

).

R = \frac{V}{I}

Resistance is equal to the voltage divided by the current flowing through the component.

3

Substitute the measured values into the formula and solve.

R = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega

Dividing

10\text{ V}

by

0.5\text{ A}

yields the total resistance of

20\text{ }\Omega

.

Anahtar Kavram

Ohm's Law defines the relationship between voltage, current, and resistance in an electrical circuit:

V = I \times R

, which rearranges to

R = \frac{V}{I}

.

Tahmini Süre:45s

Soru 76Soru →

Match each circuit modification on the left with its corresponding effect on power dissipation relative to the baseline power ( $P_0$ ) on the right.

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Cevabı ve açıklamayı göster

Cevap

Doubling voltage with constant resistance matches

4P_0

; Halving current with constant resistance matches

\frac{1}{4}P_0

; Doubling resistance with constant voltage matches

\frac{1}{2}P_0

; Tripling current with one-third resistance matches

3P_0

.

Each modification is correctly determined by applying Watt's Law formulas (

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

and

P = I^2 R

). Doubling voltage quadruples power (

2^2 = 4

), halving current reduces power to one-fourth (

(\frac{1}{2})^2 = \frac{1}{4}

), doubling resistance at constant voltage reduces power by half (

\frac{1}{2}

), and tripling current while reducing resistance to one-third increases power threefold (

3^2 \times \frac{1}{3} = 3

).

Adım Adım Çözüm

1

Select the appropriate form of Watt's Law for each scenario depending on which parameters are held constant and which are modified.

Use

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

when voltage and resistance are given, and

P = I^2 R

when current and resistance are given.

Using the formula directly containing the relevant independent variables simplifies scaling calculations.

2

Calculate the scaling factor for doubling voltage with constant resistance.

P' = \frac{(2V)^2}{R} = 4\frac{V^2}{R} = 4P_0

.

Power is directly proportional to the square of the voltage across a fixed load.

3

Calculate the scaling factor for halving current with constant resistance.

P' = \left(\frac{1}{2}I\right)^2 R = \frac{1}{4}I^2 R = \frac{1}{4}P_0

.

Power is directly proportional to the square of current through a fixed load.

4

Calculate the scaling factor for doubling resistance at constant voltage.

P' = \frac{V^2}{2R} = \frac{1}{2}\frac{V^2}{R} = \frac{1}{2}P_0

.

At constant voltage, power is inversely proportional to resistance.

5

Calculate the scaling factor for tripling current while reducing resistance to one-third.

P' = (3I)^2 \cdot \left(\frac{1}{3}R\right) = 9I^2 \cdot \frac{1}{3}R = 3I^2 R = 3P_0

.

The squared current factor of 9 multiplied by the fractional resistance factor of 1/3 results in an overall factor of 3.

Anahtar Kavram

Watt's Law Power Scaling Relationships:

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

Tahmini Süre:1m 30s

Soru 77Soru →

In a direct current (DC) circuit, an ammeter measures the rate of charge flow past a specific point. If exactly $1.248 \times 10^{19}$ electrons flow past this point in $4\text{ seconds}$ , and $1\text{ Coulomb}$ is defined as approximately $6.24 \times 10^{18}$ electrons, which of the following correctly identifies the current magnitude and describes the standard convention for its flow direction relative to the actual electrons?

0.5\text{ A}

; conventional current flows from positive to negative, opposite to the actual electron flow.

0.05\text{ A}

; conventional current flows from positive to negative, opposite to the actual electron flow.

0.5\text{ A}

; conventional current flows from negative to positive, identical to the actual electron flow.

5.0\text{ A}

; conventional current flows from negative to positive, identical to the actual electron flow.

Cevabı ve açıklamayı göster

Cevap: $0.5\text{ A}$ ; conventional current flows from positive to negative, opposite to the actual electron flow.

Cevap

The current is

0.5\text{ A}

, and conventional current is analyzed as flowing from positive to negative, which is opposite to the actual electron flow.

The total charge is found by dividing the number of electrons by the electrons per Coulomb:

1.248 \times 10^{19} / 6.24 \times 10^{18} = 2\text{ Coulombs}

. The current is the charge divided by time:

2\text{ C} / 4\text{ s} = 0.5\text{ A}

. Standard electrical convention dictates that current flows from positive to negative, which is physically opposite to the actual movement of negatively charged electrons.

Adım Adım Çözüm

1

Convert the given number of electrons into Coulombs of charge.

1.248 \times 10^{19} \text{ electrons} / 6.24 \times 10^{18} \text{ electrons/C} = 2\text{ Coulombs}

.

Current is defined as the rate of charge flow in Coulombs per second, not electrons per second.

2

Calculate the current magnitude in Amperes.

I = Q / t = 2\text{ C} / 4\text{ s} = 0.5\text{ A}

.

One Ampere is equivalent to one Coulomb of charge passing a point in one second.

3

Determine the relationship between conventional current and actual electron flow.

Conventional current is directed positive to negative, opposite to actual electron flow (negative to positive).

Standard circuit analysis uses conventional current (positive charge flow), while solid conductors actually move negatively charged electrons.

Anahtar Kavram

Electrical Current Calculations and Flow Conventions

Soru 78Soru →

A technician is testing a direct current (DC) relay coil connected to a $48\text{ V}$ power supply. An ammeter placed in series with the circuit measures a steady current of $150\text{ mA}$ . Based on Ohm's law, what is the internal electrical resistance of the relay coil in ohms ( $\Omega$ )?

Cevabı ve açıklamayı göster

Cevap: 320

Cevap

The internal electrical resistance of the relay coil is

320\text{ }\Omega

.

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

R = \frac{V}{I}

). Converting

150\text{ mA}

to

0.15\text{ A}

and dividing the applied voltage of

48\text{ V}

by

0.15\text{ A}

yields exactly

320\text{ }\Omega

.

Adım Adım Çözüm

1

Convert the current measurement from milliamperes to base amperes

I = 150\text{ mA} = 150 \times 10^{-3}\text{ A} = 0.15\text{ A}

Standard electrical calculations using Ohm's law require current in amperes.

2

Rearrange Ohm's law formula (

V = I \times R

) to isolate resistance (

R

)

R = \frac{V}{I}

The circuit voltage and current are known, and resistance is the required unknown quantity.

3

Substitute the given voltage and converted current into the formula and solve

R = \frac{48\text{ V}}{0.15\text{ A}} = 320\text{ }\Omega

Dividing

48

by

0.15

(or multiplying

48 \times \frac{20}{3}

) computes the exact resistance value.

Anahtar Kavram

Ohm's law states that electrical resistance (

R

) is the ratio of voltage (

V

) across a component to the current (

I

) flowing through it:

R = \frac{V}{I}

.

Tahmini Süre:1m 0s

Soru 79Soru →

Consider the physical principles governing electrical resistance ( $R = \rho \frac{L}{A}$ ) and material resistivity. Match each specific physical modification of a cylindrical wire conductor with its resulting net effect on the wire's total electrical resistance. Which pairings accurately reflect these proportional and thermal relationships?

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Cevap

Stretching a wire to double length at constant volume pairs with increasing resistance by a factor of 4; doubling wire diameter pairs with decreasing resistance to one-fourth; doubling both length and radius pairs with decreasing resistance to one-half; and heating a copper conductor pairs with increasing resistance due to higher resistivity.

Each match strictly obeys the physical governing equation for resistance

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

. For cylindrical geometry,

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

. Uniform stretching alters length and cross-sectional area inversely (

2L \times 0.5A \to 4R

). Diameter changes scale quadratically (

2d \to 4A \to 0.25R

). Simultaneous doubling of length and radius yields a ratio of

2/4 = 0.5R

. Thermal elevation in conductors increases resistivity (

\rho

) due to enhanced electron-phonon collisions.

Adım Adım Çözüm

1

Analyze volume conservation under mechanical stretching

Stretching to double length (

L' = 2L

) causes cross-sectional area to halve (

A' = A/2

), giving

R' = \rho \frac{2L}{A/2} = 4R

.

Since volume

V = A \cdot L

remains constant, stretching alters both length and cross-sectional area simultaneously.

2

Calculate the geometric effect of doubling diameter

Doubling diameter (

d' = 2d

) quadruples area (

A' = 4A

), resulting in

R' = \rho \frac{L}{4A} = 0.25R

.

Cross-sectional area varies quadratically with diameter (

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

).

3

Calculate simultaneous scaling of length and radius

Doubling length (

2\times

) and radius (

4\times

area) produces

R' = \frac{2}{4}R = 0.5R

.

The length factor of 2 in the numerator is divided by the area factor of 4 in the denominator.

4

Evaluate the thermal effect on metallic conductors

Increasing temperature increases copper's resistivity

\rho

, raising resistance without altering geometry.

Metals possess a positive temperature coefficient of resistance where lattice vibrations hinder electron drift.

Anahtar Kavram

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

R = \rho \frac{L}{\pi r^2}

), with metallic resistivity increasing as temperature rises.

Tahmini Süre:1m 30s

Soru 80Soru →

An electronics technician tests a direct current (DC) alarm buzzer with a measured resistance of $40\text{ }\Omega$ . If an ammeter connected in the circuit measures a current flow of $0.5\text{ A}$ , what is the voltage applied across the buzzer?

80\text{ V}

20\text{ V}

0.0125\text{ V}

20\text{ mV}

Cevabı ve açıklamayı göster

Cevap: $20\text{ V}$

Cevap

20\text{ V}

According to Ohm's law, voltage across a component equals the current flowing through it multiplied by its resistance (

V = I \times R

). Multiplying

0.5\text{ A}

by

40\text{ }\Omega

gives exactly

20\text{ V}

.

Adım Adım Çözüm

1

Identify the given values and unknown variable from the problem statement.

Current (

I

) =

0.5\text{ A}

, Resistance (

R

) =

40\text{ }\Omega

, Voltage (

V

) = ?

Establishing the known quantities determines which algebraic arrangement of Ohm's law is required.

2

Select and apply the standard Ohm's law formula for voltage.

V = I \times R

Voltage is equal to the product of current in amperes and resistance in ohms.

3

Substitute the given numerical values into the formula and calculate.

V = 0.5\text{ A} \times 40\text{ }\Omega = 20\text{ V}

Multiplying the decimal current by the integer resistance yields the total potential difference in volts.

Anahtar Kavram

Ohm's Law expresses the fundamental relationship between voltage, current, and resistance in a DC circuit (

V = I \times R

).

Tahmini Süre:45s

Basic Electrical Concepts

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