Circuits and Magnetism Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 4

Soru 61Soru →

A technician is troubleshooting a simple series circuit powered by a 12-volt DC battery and containing a single incandescent light bulb. When the switch is closed, the bulb does not light up. Measuring with a digital voltmeter directly across the bulb's socket terminals shows a reading of 12 volts. What fault condition does this measurement indicate?

A direct short circuit across the light bulb socket

An open circuit in the light bulb filament

A blown fuse at the battery positive terminal

An incorrect multimeter setting configured for resistance measurement

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Cevap: An open circuit in the light bulb filament

Cevap

An open circuit in the light bulb filament

When a filament burns out, it creates an open circuit. Because the circuit is open, current cannot flow (

I = 0\text{ A}

), meaning no voltage is dropped across the closed switch or connecting wires. Consequently, connecting a voltmeter across the open filament places the meter across the full potential of the 12-volt battery.

Adım Adım Çözüm

1

Analyze the circuit state and the voltmeter reading.

The voltmeter measures the full supply voltage of

12\text{ V}

directly across the bulb terminals, yet no light is produced.

In a series circuit with an open fault, current flow ceases entirely (

I = 0\text{ A}

). With no current flowing, there are no voltage drops across any normal conductors or closed switches.

2

Apply voltage laws to identify the location of the open fault.

The full source potential difference (

12\text{ V}

) appears across the open gap in the filament.

According to Kirchhoff's Voltage Law, the sum of voltage drops must equal the source voltage. Because the open filament has infinite resistance compared to the wiring, the entire

12\text{ V}

drops across the break.

Anahtar Kavram

An open component in a series circuit stops all current flow and drops the full source voltage across its terminals.

Tahmini Süre:45s

Soru 62Soru →

An electrical technician is testing an electromagnetic induction system where a straight copper conductor moves through a uniform, stationary magnetic field. According to Faraday's law of induction, which of the following actions will increase the magnitude of the electromotive force (EMF) induced across the conductor?

Moving the conductor parallel to the magnetic flux lines rather than perpendicular to them

Increasing the velocity at which the conductor cuts across the magnetic flux lines

Applying a steady direct current (DC) voltage across the conductor to maintain a static electric field

Replacing the copper conductor with a dielectric insulator to store electrostatic charge

Cevabı ve açıklamayı göster

Cevap: Increasing the velocity at which the conductor cuts across the magnetic flux lines

Cevap

Increasing the velocity at which the conductor cuts across the magnetic flux lines

According to Faraday's law of electromagnetic induction, the magnitude of induced voltage (

e

) across a conductor is determined by the formula

e = B \cdot l \cdot v \cdot \sin\theta

. Higher velocity (

v

) increases the rate of cutting magnetic flux lines (

\frac{\Delta \Phi}{\Delta t}

), thereby directly increasing the induced electromotive force.

Adım Adım Çözüm

1

Identify the governing formula for induced electromotive force in a moving conductor.

e = B \cdot l \cdot v \cdot \sin\theta

, where

e

is induced EMF,

B

is magnetic flux density,

l

is conductor length,

v

is velocity, and

\theta

is the angle between velocity and the magnetic field.

Faraday's law dictates that induced voltage depends directly on the rate at which magnetic flux lines are cut.

2

Analyze how each parameter affects the magnitude of the induced EMF.

Increasing conductor velocity (

v

) directly increases

\frac{\Delta \Phi}{\Delta t}

(the rate of magnetic flux cutting), which proportionally increases the induced voltage (

e

).

Faster relative motion between the conductor and magnetic field lines generates greater electrical potential across the conductor.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction

Tahmini Süre:1m 0s

Soru 63Soru →

A technician is testing a direct-current (DC) control circuit connected to a $20\text{ V}$ power supply. An ammeter in the loop indicates a steady current of $0.5\text{ A}$ . The circuit contains three resistors wired in series: two have fixed values of $12\ \Omega$ and $16\ \Omega$ , while the third is an unlabelled component. What is the resistance of the unlabelled resistor?

28\ \Omega

12\ \Omega

10\ \Omega

40\ \Omega

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Cevap: $12\ \Omega$

Cevap

12\ \Omega

The correct answer of

12\ \Omega

is found by first calculating the total circuit resistance using Ohm's Law (

R_T = \frac{20\text{ V}}{0.5\text{ A}} = 40\ \Omega

). In a series circuit, total resistance is the sum of all individual components (

R_T = R_1 + R_2 + R_3

). Subtracting the two known values (

12\ \Omega + 16\ \Omega = 28\ \Omega

) from

40\ \Omega

leaves exactly

12\ \Omega

for the unlabelled resistor.

Adım Adım Çözüm

1

Calculate the total resistance (

R_T

) of the series circuit using Ohm's Law.

R_T = \frac{V_{\text{total}}}{I} = \frac{20\text{ V}}{0.5\text{ A}} = 40\ \Omega

The total resistance of a DC circuit is equal to the total applied voltage divided by the total circuit current.

2

Sum the resistances of the two known series resistors.

R_{\text{known}} = 12\ \Omega + 16\ \Omega = 28\ \Omega

In a series circuit, individual resistances add directly to form the total equivalent resistance.

3

Subtract the sum of the known resistors from the total circuit resistance to find the unknown resistance (

R_3

).

R_3 = R_T - R_{\text{known}} = 40\ \Omega - 28\ \Omega = 12\ \Omega

Because

R_T = R_1 + R_2 + R_3

, the unlabelled resistor's value is the remaining resistance needed to reach

40\ \Omega

.

Anahtar Kavram

In a DC series circuit, total resistance is the sum of all individual resistances (

R_T = R_1 + R_2 + \dots + R_n

), and the total current is determined by Ohm's Law (

I = \frac{V_T}{R_T}

).

Tahmini Süre:1m 0s

Soru 64Soru →

A technician is troubleshooting a direct-current parallel lighting system. Match each circuit modification on the left with its direct electrical effect on the system on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

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Cevap

The circuit modifications match their electrical effects as follows: adding an additional branch decreases total resistance and increases total supply current; disconnecting one branch increases total resistance while allowing remaining branches to operate normally; increasing resistance of a single branch reduces current only through that branch; creating a short circuit across one branch bypasses all branches and causes a dangerous surge in total current.

Each pair reflects core parallel circuit behavior: adding branches lowers equivalent resistance and raises total current; disconnecting a branch raises equivalent resistance without interrupting other branches; changing one branch resistance alters only that branch's current; and shorting across one branch shorts the entire common bus.

Adım Adım Çözüm

1

Determine the impact of adding a parallel branch

Total equivalent resistance drops and total supply current rises (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

).

Adding a parallel branch creates an extra pathway for current flow.

2

Determine the impact of opening or disconnecting an active branch

Total equivalent resistance increases, but unaffected branches stay powered.

Removing a branch eliminates a current path while full source voltage remains across remaining branches.

3

Determine the impact of increasing a single branch's resistance

Current through that specific branch decreases (

I = V/R

) without changing the other branches.

Branch voltages in a parallel network are independent and equal to source voltage.

4

Determine the impact of short-circuiting one branch

The entire power source is shorted, bypassing all parallel loads.

All parallel branches connect directly to the same common positive and negative supply nodes.

Anahtar Kavram

Parallel Circuit Branch Behavior and Independence

Tahmini Süre:45s

Soru 65Soru →

A technician is troubleshooting a series circuit powered by a $12\text{ V}$ DC source that normally operates with three identical $10\,\Omega$ resistive loads ( $R_1$ , $R_2$ , and $R_3$ ). Match each specific multimeter reading or diagnostic symptom on the left to its corresponding circuit fault condition on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The multimeter readings correctly pair with the specific circuit faults as follows: measuring full supply voltage across a single component with zero current corresponds to an open circuit in that component; measuring zero volts across a component while current and remaining voltage drops increase corresponds to a shorted component; reduced voltage across all loads with an unexpected voltage drop at a junction corresponds to a high-resistance connection; and measuring zero ohms to chassis before loads corresponds to a direct short-to-ground fault.

Each fault produces a distinct electrical signature in a series circuit: an open component exhibits infinite resistance and drops the full source voltage with zero current; a shorted component exhibits zero resistance and zero voltage drop while increasing circuit current; a high-resistance connection adds unwanted series resistance that drops voltage and reduces current to all loads; and a short-to-ground creates a direct zero-ohm path to ground before the load network.

Adım Adım Çözüm

1

Analyze the voltage and current behavior for the open circuit symptom

Full source voltage (

12\text{ V}

) across

R_2

with

0\text{ A}

current indicates an open circuit at

R_2

.

An open circuit creates infinite resistance, stopping all current (

I = 0

) so no voltage drops occur across intact series components, placing total source EMF across the open terminals.

2

Analyze the voltage and current behavior for the shorted component symptom

Zero volts across

R_2

with increased current from

0.4\text{ A}

to

0.6\text{ A}

indicates a short circuit across

R_2

.

A shorted component has

0\,\Omega

, dropping

0\text{ V}

while decreasing total circuit resistance (

R_{total} = 20\,\Omega

) and increasing current (

I = 12\text{ V} / 20\,\Omega = 0.6\text{ A}

).

3

Analyze the effect of an unwanted series resistance

Lower voltage drops across all loads combined with an unexpected drop across a terminal indicates a high-resistance connection.

Corrosion or a loose terminal adds resistance to the series string, lowering total current and taking a portion of the source voltage away from the loads.

4

Evaluate the de-energized ohmmeter reading to chassis

A reading of

0\,\Omega

from the power lead to chassis ground prior to the loads confirms a short-to-ground fault.

An ohmmeter reading continuity (

0\,\Omega

) to chassis ground before the loads shows an unintended low-resistance return path that bypasses the load network entirely.

Anahtar Kavram

Circuit Fault Diagnosis and Multimeter Reading Interpretation in DC Series Circuits

Tahmini Süre:1m 30s

Soru 66Soru →

A technician is troubleshooting a direct-current circuit that suddenly stopped operating. In what order should the technician perform the standard diagnostic steps to locate and resolve the fault safely?

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct troubleshooting sequence begins with a visual inspection, followed by verifying supply voltage, then de-energizing the circuit, and finally measuring isolated component resistance.

Systematic troubleshooting follows a logical hierarchy: first perform a visual check for obvious physical damage; second, verify that input voltage is present; third, de-energize the circuit for safety; and fourth, perform resistance checks on isolated components to identify the specific faulty device.

Adım Adım Çözüm

1

Inspect the physical condition of the circuit.

Identifies obvious signs of overheating, loose terminals, or mechanical damage without disturbing connections.

Visual checks are fast, safe, and can resolve obvious problems immediately.

2

Measure input voltage while the circuit is connected to power.

Confirms whether proper operational voltage reaches the circuit input terminals.

If supply voltage is missing, troubleshooting must shift to the upstream power source rather than the circuit itself.

3

Shut off and isolate the power supply.

Removes all live electrical energy from the circuit.

Safety protocol requires de-energizing the circuit before testing resistance or removing parts.

4

Use an ohmmeter to measure continuity and resistance across individual components.

Pinpoints open coils, shorted semiconductor junctions, or out-of-tolerance resistors.

Ohmmeters supply their own internal battery voltage and will be damaged if used on an active, energized circuit.

Anahtar Kavram

Standard systematic troubleshooting progresses from non-invasive visual inspection to live voltage verification, followed by complete power isolation before conducting ohmmeter resistance measurements.

Tahmini Süre:45s

Soru 67Soru →

In an alternating current (AC) generator, increasing the rotational speed of the armature coil within a constant magnetic field increases the output frequency but has no effect on the magnitude of the peak induced electromotive force (EMF).

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

False

The statement is false because induced electromotive force depends directly on how fast the magnetic field lines are cut. Spinning the armature coil faster increases both the frequency of the alternating current and the peak induced voltage.

Adım Adım Çözüm

1

Identify the relationship between rotational speed and AC frequency in a generator.

Higher rotational speed increases the number of electrical cycles completed per second, directly increasing frequency (

f

).

Each mechanical revolution corresponds to a fixed number of electrical cycles based on the generator's pole pairs.

2

Apply Faraday's Law of Electromagnetic Induction to determine how rotational speed affects induced voltage amplitude.

Faster rotation increases the angular velocity (

\omega

), causing magnetic flux lines to be cut more rapidly over time (

\frac{\Delta \Phi}{\Delta t}

).

Faraday's Law states that induced electromotive force is directly proportional to the rate of change of magnetic flux (

E_{\text{peak}} = N B A \omega

).

3

Evaluate the statement based on the combined physical effects.

Since rotational speed increases both output frequency and peak induced EMF, the claim that peak EMF remains unaffected is incorrect.

A generator produces higher peak voltage when spun faster because conductor velocity through the magnetic field increases.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction in AC Generation

Tahmini Süre:1m 0s

Soru 68Soru →

A $60\text{ V}$ direct-current power supply feeds a circuit containing three parallel branches. The total current drawn from the supply is $10\text{ A}$ . If the first two branches have resistances of $20\ \Omega$ and $30\ \Omega$ , what is the resistance of the third branch?

6\ \Omega

12\ \Omega

18\ \Omega

50\ \Omega

Cevabı ve açıklamayı göster

Cevap: $12\ \Omega$

Cevap

The resistance of the third branch is

12\ \Omega

.

Because the circuit is wired in parallel, the full supply voltage of

60\text{ V}

is applied across each branch. Using Ohm's Law (

I = V / R

), the first branch draws

60\text{ V} / 20\ \Omega = 3\text{ A}

and the second branch draws

60\text{ V} / 30\ \Omega = 2\text{ A}

. The sum of these two branch currents is

5\text{ A}

. Since the total source current is

10\text{ A}

, the third branch must carry the remaining

10\text{ A} - 5\text{ A} = 5\text{ A}

. Applying Ohm's Law to the third branch yields

R_3 = 60\text{ V} / 5\text{ A} = 12\ \Omega

.

Adım Adım Çözüm

1

Determine the voltage across each parallel branch.

Each branch receives the full source voltage of

60\text{ V}

.

In a parallel circuit, the voltage across every parallel branch is identical to the source voltage.

2

Calculate the individual currents flowing through the first two known branches using Ohm's Law (

I = V / R

).

I_1 = 60\text{ V} / 20\ \Omega = 3\text{ A}

and

I_2 = 60\text{ V} / 30\ \Omega = 2\text{ A}

.

Branch currents must be determined to find how much of the total current is accounted for by the known branches.

3

Find the current flowing through the third branch by subtracting the known branch currents from the total supply current (

I_3 = I_{\text{total}} - I_1 - I_2

).

I_3 = 10\text{ A} - (3\text{ A} + 2\text{ A}) = 5\text{ A}

.

Total current in a parallel circuit equals the sum of the individual branch currents.

4

Calculate the resistance of the third branch using Ohm's Law (

R_3 = V / I_3

).

R_3 = 60\text{ V} / 5\text{ A} = 12\ \Omega

.

Applying Ohm's Law to the third branch gives its required resistance value.

Anahtar Kavram

In a parallel circuit, voltage remains constant across all branches, the total current equals the sum of all individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

), and each branch resistance satisfies Ohm's Law (

R_n = V / I_n

).

Tahmini Süre:1m 30s

Soru 69Soru →

Analyze the series circuit parameters under normal operation and after a short-circuit fault by filling in the correct numerical values (use decimal format without units, e.g., '0.5' or '10').

Aşağıdaki boşlukları doldurun

An aviation warning beacon is powered by a

120\text{ V}

DC supply connected in series with three resistive components:

R_1 = 100\ \Omega

,

R_2 = 150\ \Omega

, and

R_3 = 50\ \Omega

. Under normal operation, the total current flowing through the circuit is A and the voltage drop across

R_1

is V. If component

R_2

experiences a catastrophic failure and completely shorts out (its internal resistance drops to

0\ \Omega

), the equivalent resistance of the entire circuit drops to

\Omega

, causing the voltage drop across

R_3

to become V.

Cevabı ve açıklamayı göster

Cevap

Under normal operation, the current is 0.4 A, and the voltage drop across R1 is 40 V. After R2 shorts out, the total resistance drops to 150 ohms, and the increased current causes the voltage drop across R3 to become 40 V.

In a series circuit, total resistance is the linear sum of all resistive components. Current remains uniform throughout the entire loop and is determined by the total voltage divided by the total resistance. When a component fails via a short circuit, it effectively acts as a direct wire (zero ohms). This absence of resistance lowers the total circuit resistance, which consequently forces a higher current to flow from the constant voltage source. The increased current then results in proportionately higher voltage drops across the surviving resistors.

Adım Adım Çözüm

1

Calculate the total equivalent resistance under normal operating conditions.

R_{total} = R_1 + R_2 + R_3 = 100 + 150 + 50 = 300\ \Omega

In a series circuit, the total resistance is the sum of all individual resistances.

2

Use Ohm's Law to find the total current flowing through the normal circuit.

I_{total} = \frac{V_{total}}{R_{total}} = \frac{120\text{ V}}{300\ \Omega} = 0.4\text{ A}

(This is blank 1).

Total current is equal to the source voltage divided by the total equivalent resistance.

3

Calculate the voltage drop across

R_1

under normal conditions.

V_{R1} = I_{total} \times R_1 = 0.4\text{ A} \times 100\ \Omega = 40\text{ V}

(This is blank 2).

The voltage drop across any single component in a series circuit is the total current multiplied by that component's resistance.

4

Determine the new total resistance after

R_2

shorts out.

R_{new\_total} = 100 + 0 + 50 = 150\ \Omega

(This is blank 3).

A shorted component has

0\ \Omega

resistance. The circuit remains closed, but the total resistance decreases.

5

Calculate the new total current and the resulting new voltage drop across

R_3

.

I_{new} = \frac{120\text{ V}}{150\ \Omega} = 0.8\text{ A}

. Then,

V_{R3} = 0.8\text{ A} \times 50\ \Omega = 40\text{ V}

(This is blank 4).

The lowered total resistance causes a higher total current to flow from the same voltage source, which proportionally increases the voltage drop across the remaining functional resistors.

Anahtar Kavram

Series Circuit Analysis and Component Fault (Short Circuit) Behavior

Soru 70Soru →

A DC cooling fan installed in an equipment enclosure has an internal resistance of $16\ \Omega$ and draws a steady current of $1.5\text{ A}$ during normal operation. According to Ohm's law, what is the voltage drop across the fan in volts ( $\text{V}$ )?

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Cevap: 24

Cevap

24 V

According to Ohm's law, voltage (

V

) is found by multiplying current (

I

) by resistance (

R

). With a current of

1.5\text{ A}

and a resistance of

16\ \Omega

, the voltage is

1.5 \times 16 = 24\text{ V}

.

Adım Adım Çözüm

1

Identify the known electrical quantities from the problem statement.

I = 1.5\text{ A}

and

R = 16\ \Omega

Ohm's law requires current and resistance values to calculate voltage.

2

State Ohm's law formula arranged to solve for voltage (

V

).

V = I \times R

The potential difference across a resistive element is directly proportional to the product of current and resistance.

3

Multiply the current by the resistance to find the voltage.

V = 1.5\text{ A} \times 16\ \Omega = 24\text{ V}

Performing the multiplication yields the potential difference in volts.

Anahtar Kavram

Ohm's Law (

V = I \times R

)

Tahmini Süre:30s

Soru 71Soru →

A diesel generator preheat circuit is powered by a $24\text{ V}$ DC source connected in series with a ballast resistor $R_1 = 1.8\ \Omega$ and two identical glow plugs ( $R_2$ and $R_3$ ). If an ammeter in the circuit measures a steady current of $4.0\text{ A}$ , what is the resistance of a single glow plug in ohms ( $\Omega$ )?

Cevabı ve açıklamayı göster

Cevap: 2.1

Cevap

The resistance of a single glow plug is

2.1\ \Omega

.

Applying Ohm's Law gives a total series loop resistance of

R_{\text{total}} = 24\text{ V} / 4.0\text{ A} = 6.0\ \Omega

. In a series circuit, total resistance is the sum of all individual resistances (

R_{\text{total}} = R_1 + R_2 + R_3

). Subtracting the ballast resistor (

1.8\ \Omega

) leaves

4.2\ \Omega

across the two identical glow plugs. Dividing

4.2\ \Omega

equally between them gives

2.1\ \Omega

per glow plug.

Adım Adım Çözüm

1

Calculate the total circuit resistance using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{24\text{ V}}{4.0\text{ A}} = 6.0\ \Omega

In a closed DC loop, total resistance equals total source voltage divided by total current.

2

Subtract the known ballast resistance from the total resistance to determine the combined resistance of both glow plugs.

R_2 + R_3 = R_{\text{total}} - R_1 = 6.0\ \Omega - 1.8\ \Omega = 4.2\ \Omega

Resistances in a series circuit are directly additive (

R_{\text{total}} = R_1 + R_2 + R_3

).

3

Divide the combined glow plug resistance equally between the two identical units.

R_{\text{plug}} = \frac{4.2\ \Omega}{2} = 2.1\ \Omega

Because the two glow plugs are identical, each provides exactly half of the remaining series resistance.

Anahtar Kavram

Series Total Resistance and Ohm's Law

Tahmini Süre:1m 0s

Soru 72Soru →

An electrician tests a de-energized heating circuit using a digital multimeter set to measure resistance. When the test probes are placed across the disconnected heating element, the meter displays an infinite resistance reading (OL). What fault condition does this measurement indicate?

A short circuit across the heating element

An open circuit within the heating element

Excessive electrical current flowing through the circuit

A properly functioning element operating at its rated resistance

Cevabı ve açıklamayı göster

Cevap: An open circuit within the heating element

Cevap

An open circuit within the heating element

An ohmmeter displaying infinite resistance (indicated as 'OL' for open loop or out of limits) across a component means there is a physical break in the internal conductor, which defines an open circuit.

Adım Adım Çözüm

1

Analyze the multimeter mode and circuit condition

The meter is configured as an ohmmeter measuring a de-energized, isolated load.

Resistance must be tested with power removed to prevent meter damage and obtain true component resistance.

2

Interpret the display reading of infinite resistance (OL)

Infinite resistance signifies that there is no continuous electrical path between the meter leads.

An intact heating element has a continuous resistive wire that provides a measurable finite resistance.

3

Identify the corresponding circuit fault

A break or gap in the internal filament represents an open circuit.

Open circuits present infinite resistance and completely prevent current flow under normal operating voltage.

Anahtar Kavram

Identifying open circuits using ohmmeter resistance measurements

Tahmini Süre:45s

Soru 73Soru →

A $48\text{ V}$ direct-current telecommunications power bus supplies electricity to three parallel branches. The first branch draws a current of $2\text{ A}$ , the second branch contains a load with a resistance of $24\ \Omega$ , and the total current supplied by the power bus is $7\text{ A}$ . What is the electrical resistance of the third branch, in ohms?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The resistance of the third branch is

16\ \Omega

.

Because the circuit is parallel, the full supply voltage of

48\text{ V}

is present across every branch. The second branch with

24\ \Omega

draws

I_2 = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

. Subtracting the known currents of Branch 1 (

2\text{ A}

) and Branch 2 (

2\text{ A}

) from the total current of

7\text{ A}

leaves

3\text{ A}

flowing through Branch 3. Using Ohm's Law,

R_3 = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

.

Adım Adım Çözüm

1

Calculate the current flowing through Branch 2

I_2 = \frac{V}{R_2} = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

In a parallel circuit, the voltage across each parallel branch is identical to the supply voltage (

V_1 = V_2 = V_3 = 48\text{ V}

).

2

Calculate the current flowing through Branch 3

I_3 = I_{\text{total}} - I_1 - I_2 = 7\text{ A} - 2\text{ A} - 2\text{ A} = 3\text{ A}

The total current entering a parallel network equals the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + I_3

).

3

Calculate the resistance of Branch 3

R_3 = \frac{V}{I_3} = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) using the common

48\text{ V}

supply voltage and the calculated

3\text{ A}

branch current yields the branch resistance.

Anahtar Kavram

In a parallel DC circuit, voltage remains constant across all branches, total current is the algebraic sum of individual branch currents, and branch resistance is determined by Ohm's Law (

R = \frac{V}{I}

).

Soru 74Soru →

A $24\text{ V}$ direct current (DC) power supply delivers a steady current of $3\text{ A}$ to a resistive heating element. What is the total electrical power consumed by the heating element in watts ( $\text{W}$ )?

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Cevap: 72

Cevap

The total electrical power consumed is

72\text{ W}

.

Electrical power in a direct current circuit is calculated using the formula

P = V \times I

. Substituting the given voltage of

24\text{ V}

and current of

3\text{ A}

yields

P = 24 \times 3 = 72\text{ W}

.

Adım Adım Çözüm

1

Identify the given values from the circuit problem

Voltage

V = 24\text{ V}

and current

I = 3\text{ A}

Electrical power calculations require knowing at least two circuit parameters.

2

Select the appropriate electrical power formula

P = V \times I

Watt's Law defines electrical power as the product of voltage across a component and the current flowing through it.

3

Substitute the values and calculate the final result

P = 24\text{ V} \times 3\text{ A} = 72\text{ W}

Multiplying volts by amperes gives power directly in watts.

Anahtar Kavram

Electrical power is the rate at which electrical energy is transferred in a circuit, calculated using Watt's Law (

P = V \times I

).

Tahmini Süre:45s

Soru 75Soru →

Match each standard electrical circuit fault with its correct definition and behavior.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

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Cevap

Open circuit pairs with a complete break in the path that stops current flow; Short circuit pairs with an unintended zero-resistance bypass around a load that causes excessive current; Short to ground pairs with an unintended low-resistance connection to the chassis or frame; High-resistance connection pairs with an unwanted increase in opposition creating an unintended voltage drop.

Each electrical fault produces distinct physical and electrical symptoms: open circuits prevent current flow via continuity loss; short circuits draw maximum current by bypassing load resistance; ground faults divert current directly into the chassis/ground; and high-resistance joints cause partial voltage loss without completely breaking continuity.

Adım Adım Çözüm

1

Analyze the definition of an open circuit.

Identify that an open creates infinite resistance and interrupts all current, matching the break in the conductive path.

An open breaks continuity.

2

Analyze the definition of a short circuit.

Identify that a short bypasses the load with near-zero resistance, matching excessive current flow and tripped protective devices.

Ohm's law dictates that near-zero resistance at supply voltage results in high current.

3

Analyze the definition of a short to ground.

Identify that contacting the equipment frame or chassis defines a ground fault.

Chassis/ground serves as the common return or protective reference.

4

Analyze the definition of a high-resistance connection.

Identify that dirty, loose, or corroded terminals act like unwanted series resistors that drop voltage.

Contact resistance reduces available voltage across the operating load.

Anahtar Kavram

Basic Circuit Fault Characteristics

Alternatif Yöntem

Use the fundamental relationship

I = \frac{V}{R}

: when

R = \infty

(open),

I = 0

; when

R \approx 0

(short),

I

is extremely large.

Tahmini Süre:45s

Soru 76Soru →

An emergency backup lighting circuit is powered by a $48\text{ V}$ DC supply connected in series with a $30\ \Omega$ current-limiting resistor and two identical LED lamp modules. If a steady current of $0.4\text{ A}$ flows through the circuit during operation, what is the voltage drop across each individual LED lamp module?

12\text{ V}

36\text{ V}

18\text{ V}

24\text{ V}

Cevabı ve açıklamayı göster

Cevap: $18\text{ V}$

Cevap

18\text{ V}

In a series circuit, the total voltage of

48\text{ V}

is distributed across all components according to Ohm's Law. With

0.4\text{ A}

flowing through the

30\ \Omega

resistor, that resistor drops

12\text{ V}

. Subtracting

12\text{ V}

from the total

48\text{ V}

source leaves

36\text{ V}

dropped across the two identical LED modules. Because the modules are identical and in series, they share this remaining voltage equally, resulting in

18\text{ V}

across each module.

Adım Adım Çözüm

1

Calculate the voltage drop across the

30\ \Omega

series resistor using Ohm's Law (

V = I \times R

).

V_{R} = 0.4\text{ A} \times 30\ \Omega = 12\text{ V}

Current is identical through every component in a series circuit.

2

Determine the remaining voltage available for the two LED modules by applying Kirchhoff's Voltage Law (

V_{\text{total}} - V_R

).

V_{\text{remaining}} = 48\text{ V} - 12\text{ V} = 36\text{ V}

The sum of all voltage drops in a series loop must equal the total supply voltage.

3

Divide the remaining voltage equally between the two identical LED modules.

V_{\text{module}} = \frac{36\text{ V}}{2} = 18\text{ V}

Identical series components carry the same current and therefore drop equal portions of the remaining voltage.

Anahtar Kavram

In a series circuit, current is constant throughout the single loop, and the total supply voltage equals the sum of the individual component voltage drops.

Tahmini Süre:1m 0s

Soru 77Soru →

An industrial direct-current (DC) sensor loop is powered by a $50\text{ V}$ power source. The loop consists of two series-connected components: a fixed resistor $R_1 = 10\ \Omega$ and an unknown resistor $R_2$ . An ammeter placed in the circuit measures a continuous current of $2\text{ A}$ . What is the resistance of $R_2$ , and what is the voltage drop across $R_2$ ?

Aşağıdaki boşlukları doldurun

The resistance of

R_2

is

\Omega

, and the voltage drop across

R_2

is

\text{V}

.

Cevabı ve açıklamayı göster

Cevap

The resistance of

R_2

is

15\ \Omega

, and the voltage drop across

R_2

is

30\text{ V}

.

In a series circuit, the total resistance is calculated as

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

. Since

R_{\text{total}} = R_1 + R_2

, the unknown resistance is

R_2 = 25\ \Omega - 10\ \Omega = 15\ \Omega

. The voltage drop across

R_2

is found by multiplying the common series current by

R_2

, yielding

V_2 = 2\text{ A} \times 15\ \Omega = 30\text{ V}

(which also satisfies Kirchhoff's Voltage Law:

50\text{ V} - 20\text{ V} = 30\text{ V}

).

Adım Adım Çözüm

1

Calculate the total equivalent resistance (

R_{\text{total}}

) of the series circuit using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

In a closed DC loop, total resistance equals total source voltage divided by total current.

2

Determine the value of the unknown resistor

R_2

using the series resistance rule.

R_2 = R_{\text{total}} - R_1 = 25\ \Omega - 10\ \Omega = 15\ \Omega

Total resistance in a series circuit is the sum of individual resistances (

R_{\text{total}} = R_1 + R_2

).

3

Calculate the individual voltage drop across

R_2

(

V_2

).

V_2 = I \times R_2 = 2\text{ A} \times 15\ \Omega = 30\text{ V}

The same current flows through all series components; voltage drop across any component equals current multiplied by its resistance (or

V_{\text{total}} - V_1 = 50\text{ V} - 20\text{ V} = 30\text{ V}

).

Anahtar Kavram

Series Circuit Resistance and Voltage Division

Tahmini Süre:1m 0s

Soru 78Soru →

A $24\text{ V}$ DC bridge circuit consists of two parallel branches connected between a constant $+24\text{ V}$ DC supply rail and ground:
- Left branch: Resistor $R_1 = 1.0\,\text{k}\Omega$ connected in series with resistor $R_2 = 1.0\,\text{k}\Omega$ to ground, with Test Point A ( $V_A$ ) located between them.
- Right branch: Resistor $R_3 = 1.0\,\text{k}\Omega$ connected in series with resistor $R_4 = 1.0\,\text{k}\Omega$ to ground, with Test Point B ( $V_B$ ) located between them.

Under normal operation, the total current drawn from the supply is $24\text{ mA}$ , and $V_A = V_B = 12.0\text{ V}$ . During diagnostic testing of a faulty unit, a technician measures the following values:
- $V_A = 12.0\text{ V}$ (measured with respect to ground)
- $V_B = 24.0\text{ V}$ (measured with respect to ground)
- Total current drawn from the $24\text{ V}$ source = $36\text{ mA}$

Which of the following circuit faults accounts for these diagnostic readings?

An open circuit fault in resistor

R_4

A short circuit fault across resistor

R_4

A short circuit fault across resistor

R_3

An open circuit fault in resistor

R_3

Cevabı ve açıklamayı göster

Cevap: A short circuit fault across resistor $R_3$

Cevap

A short circuit fault across resistor

R_3

A short circuit across resistor

R_3

bypasses the upper resistor, placing Test Point B directly at the

+24\text{ V}

supply rail. This leaves only resistor

R_4

(

1.0\,\text{k}\Omega

) in the right branch, increasing the right branch current to

24\text{ V} / 1.0\,\text{k}\Omega = 24\text{ mA}

. Combining this with the normal

12\text{ mA}

from the left branch produces the exact measured total current of

36\text{ mA}

.

Adım Adım Çözüm

1

Analyze the operational status and current of the left branch.

Left branch current

I_{\text{left}} = 12\text{ mA}

.

Since

V_A = 12.0\text{ V}

across

R_2 = 1.0\,\text{k}\Omega

, the left branch is normal and draws

I_{\text{left}} = 24\text{ V} / (1.0\,\text{k}\Omega + 1.0\,\text{k}\Omega) = 12\text{ mA}

.

2

Determine the current flowing through the right branch under fault conditions.

Right branch current

I_{\text{right}} = 24\text{ mA}

.

Subtract the normal left branch current from the measured total current:

I_{\text{right}} = I_{\text{total}} - I_{\text{left}} = 36\text{ mA} - 12\text{ mA} = 24\text{ mA}

.

3

Calculate the effective resistance of the right branch and evaluate voltage drop across components.

R_{\text{right}} = 1.0\,\text{k}\Omega

, with

0\text{ V}

dropped across

R_3

.

R_{\text{right}} = V_{\text{source}} / I_{\text{right}} = 24\text{ V} / 24\text{ mA} = 1.0\,\text{k}\Omega

. Since

R_4

is

1.0\,\text{k}\Omega

and

V_B = 24.0\text{ V}

, the full

24\text{ V}

drops across

R_4

, meaning

R_3

has

0\text{ V}

drop and zero resistance (a direct short circuit).

Anahtar Kavram

Distinguishing between open and short circuit faults using branch voltage and total current analysis

Tahmini Süre:1m 30s

Soru 79Soru →

A technician connects an alternating current (AC) power source to the primary winding of an iron-core transformer to power a low-voltage circuit. Which of the following statements correctly explains how electrical energy is transferred from the primary winding to the secondary winding?

The primary coil stores electrical energy in an electrostatic field between the windings, which discharges periodically into the secondary circuit.

The alternating current in the primary winding creates a dynamic magnetic field that cuts across the secondary winding, inducing an alternating voltage through mutual induction.

A steady direct current magnetizes the iron core, enabling electrons to physically conduct through the core material into the secondary winding.

High primary voltage forces electrons to ionize the insulation and jump across the air gap directly into the secondary conductors.

Cevabı ve açıklamayı göster

Cevap: The alternating current in the primary winding creates a dynamic magnetic field that cuts across the secondary winding, inducing an alternating voltage through mutual induction.

Cevap

The alternating current in the primary winding creates a dynamic magnetic field that cuts across the secondary winding, inducing an alternating voltage through mutual induction.

Energy transfer in a transformer occurs via mutual induction. When alternating current flows through the primary winding, it produces a continuously changing magnetic flux. This flux travels through the ferromagnetic core and links with the secondary winding, inducing an alternating voltage across the secondary terminals according to Faraday's Law of Induction without requiring any direct electrical connection.

Adım Adım Çözüm

1

Identify the electrical input supplied to the primary winding

Alternating current (AC) flows through the primary coil, providing a continuously varying current over time.

A time-varying current is necessary to generate a continuously changing magnetic field.

2

Analyze the behavior of the magnetic field generated within the core

The changing primary current creates an expanding and collapsing magnetic flux that circulates through the ferromagnetic core.

According to Ampere's Law and the principles of electromagnetism, electric current passing through a wound coil establishes a concentrated magnetic field proportional to the current.

3

Determine the mechanism producing electrical energy in the secondary circuit

The moving magnetic flux lines cut through the secondary coil windings, generating an induced electromotive force (voltage) across its terminals.

Faraday's Law of Electromagnetic Induction dictates that a changing magnetic flux passing through a closed loop induces a voltage without any direct physical electrical connection between the coils.

Anahtar Kavram

Mutual Induction in Transformers

Tahmini Süre:1m 0s

Soru 80Soru →

An aircraft landing light is connected to a $28\text{ V}$ DC bus and draws a current of $4\text{ A}$ . What is the total electrical power consumed by the light?

7\text{ W}

32\text{ W}

112\text{ W}

1120\text{ W}

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Cevap: $112\text{ W}$

Cevap

112\text{ W}

According to Watt's Law, electrical power (

P

) is equal to the product of voltage (

V

) and current (

I

). Multiplying

28\text{ V}

by

4\text{ A}

yields

112\text{ W}

.

Adım Adım Çözüm

1

Identify the given electrical values from the circuit

Voltage

V = 28\text{ V}

, Current

I = 4\text{ A}

These are the two known parameters required to calculate power directly.

2

Select Watt's Law power formula that relates voltage, current, and power

P = V \times I

Electrical power in watts is equal to electromotive force in volts multiplied by current in amperes.

3

Calculate the total power dissipation

P = 28\text{ V} \times 4\text{ A} = 112\text{ W}

Multiplying

28

by

4

gives the power consumption in watts.

Anahtar Kavram

Electrical Power Calculation using Watt's Law (

P = V \times I

)

Tahmini Süre:45s

Circuits and Magnetism

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