Circuits and Magnetism Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB) — Sayfa 5

Soru 81Soru →

An electronics technician is testing an aircraft panel indicator circuit. When connected to a $24\text{ V}$ DC power supply, the circuit draws a steady current of $4\text{ A}$ . What is the resistance of the circuit, and what current will flow if the supply voltage is reduced to $12\text{ V}$ ?

Aşağıdaki boşlukları doldurun

According to Ohm's Law, the resistance of the circuit is

\Omega

, and when the voltage is decreased to

12\text{ V}

, the new current drawn by the circuit is

\text{A}

.

Cevabı ve açıklamayı göster

Cevap

The resistance of the circuit is

6\ \Omega

and the current under the reduced voltage is

2\text{ A}

.

Applying Ohm's Law, the resistance is found by dividing the initial voltage by the initial current (

R = \frac{24\text{ V}}{4\text{ A}} = 6\ \Omega

). Keeping this resistance constant, halving the voltage to

12\text{ V}

results in half the current (

I = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}

).

Adım Adım Çözüm

1

Calculate the circuit resistance using Ohm's Law with the initial parameters (

V_1 = 24\text{ V}

,

I_1 = 4\text{ A}

).

R = \frac{V_1}{I_1} = \frac{24\text{ V}}{4\text{ A}} = 6\ \Omega

Ohm's Law states that resistance equals voltage divided by current (

R = \frac{V}{I}

).

2

Determine the new current when the voltage is adjusted to

V_2 = 12\text{ V}

across the constant

6\ \Omega

resistance.

I_2 = \frac{V_2}{R} = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}

Current is directly proportional to voltage for a fixed resistance (

I = \frac{V}{R}

).

Anahtar Kavram

Ohm's Law expresses the fundamental relationship among voltage, current, and resistance:

V = I \times R

,

I = \frac{V}{R}

, and

R = \frac{V}{I}

.

Tahmini Süre:45s

Soru 82Soru →

A technician performs electrical diagnostic checks on various direct-current (DC) circuit components. Match each multimeter measurement or symptom to the correct circuit condition or fault.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Match the open coil to infinite resistance, intact fuse to 0 V drop, shorted load to 0 V with high current surge / blown fuse, and high-resistance contact fault to elevated closed-switch resistance.

Each diagnostic measurement corresponds to a fundamental circuit state: infinite coil resistance reflects an open circuit; zero volts across an intact fuse demonstrates normal low-resistance conduction; zero volts with blown protection indicates a short circuit; and elevated resistance across closed switch contacts indicates a high-resistance contact fault.

Adım Adım Çözüm

1

Analyze the isolated coil resistance test reading

An infinite resistance (O.L.) measurement signifies an internal discontinuity or open circuit.

Current cannot pass through a severed conductor, resulting in infinite measured resistance.

2

Evaluate the in-circuit voltage drop across the closed fuse

A 0 V drop across an intact conducting fuse indicates normal low-resistance operation.

According to Ohm's law (

V = I \times R

), a conductor with negligible resistance produces virtually no voltage drop.

3

Identify the fault causing 0 V across a load with blown overcurrent protection

A short circuit parallel to or across the load bypasses normal resistance and draws excessive current.

Zero branch resistance eliminates voltage drop across the parallel terminals while causing current to surge past the fuse rating.

4

Interpret the elevated resistance reading across closed switch contacts

A reading of

180\,\Omega

represents an abnormal high-resistance fault.

Clean, closed mechanical contacts should read less than

1\,\Omega

; oxidation and pitting insert unwanted resistance into the path.

Anahtar Kavram

Circuit Fault Diagnosis and Multimeter Interpretation

Tahmini Süre:1m 15s

Soru 83Soru →

A $12\text{ V}$ direct-current (DC) circuit consists of a power source, a single-pole toggle switch, and an indicator lamp wired in series. When the switch is flipped to the ON (closed) position, the lamp fails to illuminate. A technician connects a digital multimeter across the two terminals of the closed switch and reads $12\text{ V}$ . Which of the following faults is indicated by this measurement?

The indicator lamp has a short circuit across its filament.

The circuit is operating normally and drawing maximum rated current.

The switch contacts have failed open internally.

The indicator lamp filament is burned open.

Cevabı ve açıklamayı göster

Cevap: The switch contacts have failed open internally.

Cevap

The switch contacts have failed open internally.

In a series circuit, a closed switch should exhibit continuity (

0\ \Omega

) and produce a

0\text{ V}

drop. When a digital multimeter measures the entire source voltage (

12\text{ V}

) across a switch in the ON position, it proves that the switch contacts have failed to close electrically, acting as an open circuit that drops the entire available potential.

Adım Adım Çözüm

1

Analyze the expected behavior of a properly closed switch in a series circuit.

An ideal closed switch provides continuity with approximately

0\ \Omega

resistance and a voltage drop of nearly

0\text{ V}

.

According to Ohm's Law (

V = I \times R

), current flowing through near-zero resistance produces negligible voltage drop.

2

Evaluate the meter reading across the switch terminals.

The meter measures

12\text{ V}

, which equals the full source voltage.

In a series circuit, an open point has infinite resistance and drops the entire supply voltage because no current can flow through the rest of the circuit.

3

Determine the exact fault location based on the voltage drop.

Because the full

12\text{ V}

appears across the switch itself, the break in continuity is inside the switch.

If the lamp were open instead, the voltmeter across the intact closed switch would read

0\text{ V}

.

Anahtar Kavram

Voltage Drop Across Open vs. Closed Series Components

Tahmini Süre:1m 0s

Soru 84Soru →

In direct-current (DC) series circuit analysis, distinct electrical principles govern how current, voltage, resistance, and faults behave across a single closed pathway. Match each series circuit parameter or condition on the left with its correct governing rule on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Total equivalent resistance matches the direct sum of all individual resistive values; Circuit current matches remaining constant at every point; Total source voltage matches the sum of the individual voltage drops across all loads; An open-circuit fault matches immediately halting all current flow.

Series circuits are defined by a single pathway for current. Consequently, current (

I

) is identical at all points, total resistance (

R_T

) is the arithmetic sum of the component resistances, total voltage (

V_T

) equals the sum of the individual component voltage drops, and an interruption anywhere breaks the complete circuit, dropping total current to zero.

Adım Adım Çözüm

1

Identify the resistance relationship for series circuits.

Total resistance increases with each added resistor and is found by adding resistances directly (

R_T = R_1 + R_2 + \dots

).

Electrons must pass sequentially through every resistive element along the single pathway.

2

Determine the current distribution rule in a single-loop circuit.

Current magnitude is uniform everywhere (

I_T = I_1 = I_2

).

Charge cannot accumulate or branch off in an unbranched series loop.

3

Apply Kirchhoff's Voltage Law to the series loop.

Total applied voltage equals the sum of the series voltage drops (

V_T = V_1 + V_2 + \dots

).

The electrical potential energy supplied by the source must be completely dissipated across the series loads.

4

Evaluate the effect of an open-circuit fault.

An open condition creates infinite resistance, stopping all loop current (

I = 0\text{ A}

).

A continuous physical path is required for DC electron flow.

Anahtar Kavram

Fundamental series circuit characteristics: additive resistance, constant current, additive voltage drops (Kirchhoff's Voltage Law), and single-point failure dependency.

Tahmini Süre:1m 0s

Soru 85Soru →

In electrical circuits, the power consumed or dissipated can be calculated using different formulas depending on which circuit parameters are known. Match each set of known circuit parameters on the left with the correct electrical power formula on the right. How should each parameter combination be paired with its corresponding power formula?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Voltage and Current pairs with

P = V \times I

; Current and Resistance pairs with

P = I^2 R

; Voltage and Resistance pairs with

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

; Electrical Energy and Time pairs with

P = \frac{E}{t}

.

Each parameter combination matches its standard mathematical expression: Voltage and Current relate directly through Watt's Law (

P = V \times I

), Current and Resistance relate through

P = I^2 R

, Voltage and Resistance relate through

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

, and Energy and Time match the foundational definition of power (

P = \frac{E}{t}

).

Adım Adım Çözüm

1

Identify the base relationship of electrical power from fundamental physical definitions

Power is defined as energy divided by time:

P = \frac{E}{t}

.

This establishes the core definition of power in Joules per second (Watts).

2

Identify Watt's Law for circuits when potential difference and flow rate are known

Power is given by

P = V \times I

.

Multiplying potential difference (Volts) by current (Amperes) yields electrical power in Watts.

3

Use Ohm's Law substitutions to find power formulas for resistive circuits

When current and resistance are known:

P = I^2 R

. When voltage and resistance are known:

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

.

Substituting

V = IR

or

I = \frac{V}{R}

allows power to be determined when one electrical variable is omitted.

Anahtar Kavram

Electrical Power and Watt's Law Formulations

Tahmini Süre:45s

Soru 86Soru →

A benchtop DC power supply delivers $45\text{ V}$ across a fixed resistor in a test circuit. A series ammeter measures a steady current flow of $5\text{ A}$ through the component. What is the electrical resistance of the resistor?

225\ \Omega

9\ \Omega

40\ \Omega

90\ \Omega

Cevabı ve açıklamayı göster

Cevap: $9\ \Omega$

Cevap

9\ \Omega

According to Ohm's Law, electrical resistance is equal to the voltage across a component divided by the current passing through it (

R = V / I

). Substituting

45\text{ V}

and

5\text{ A}

yields

R = 45 / 5 = 9\ \Omega

.

Adım Adım Çözüm

1

Identify the given circuit parameters and standard Ohm's Law formula

Voltage

V = 45\text{ V}

, Current

I = 5\text{ A}

, Ohm's Law:

V = I \times R

Ohm's Law defines the relationship between potential difference, current flow, and resistance in a DC circuit.

2

Rearrange Ohm's Law to solve for resistance (

R

)

R = \frac{V}{I}

Dividing both sides of

V = I \times R

by current isolates the unknown resistance.

3

Substitute the given values into the formula and evaluate

R = \frac{45\text{ V}}{5\text{ A}} = 9\ \Omega

Dividing 45 by 5 yields the exact resistance value in ohms.

Anahtar Kavram

Ohm's Law Resistance Calculation (

R = V / I

)

Tahmini Süre:45s

Soru 87Soru →

An industrial control panel utilizes a direct-current bus to power three parallel branches. Branch 1 contains a $15\ \Omega$ resistor, Branch 2 contains a $30\ \Omega$ resistor, and Branch 3 contains a $10\ \Omega$ resistor. If an ammeter connected in Branch 2 measures a current of $3\text{ A}$ , what is the total current delivered by the power supply to the entire circuit?

4.5\text{ A}

6.0\text{ A}

9.0\text{ A}

18.0\text{ A}

Cevabı ve açıklamayı göster

Cevap: $18.0\text{ A}$

Cevap

18.0\text{ A}

The supply voltage across the parallel circuit is constant. Using Branch 2,

V = 3\text{ A} \times 30\ \Omega = 90\text{ V}

. The remaining branch currents are

I_1 = 90\text{ V} / 15\ \Omega = 6\text{ A}

and

I_3 = 90\text{ V} / 10\ \Omega = 9\text{ A}

. Summing all three branch currents yields

I_{\text{total}} = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

.

Adım Adım Çözüm

1

Calculate the common supply voltage across the parallel circuit using Ohm's Law on Branch 2.

V = I_2 \times R_2 = 3\text{ A} \times 30\ \Omega = 90\text{ V}

In a parallel circuit, the voltage drop across every individual branch is equal to the total supply voltage.

2

Determine the current flowing through Branch 1 and Branch 3 using the common voltage of

90\text{ V}

.

I_1 = \frac{90\text{ V}}{15\ \Omega} = 6\text{ A}

and

I_3 = \frac{90\text{ V}}{10\ \Omega} = 9\text{ A}

Current in each individual branch is determined by dividing the constant branch voltage by that branch's resistance.

3

Apply Kirchhoff's Current Law to sum the currents of all parallel branches.

I_{\text{total}} = I_1 + I_2 + I_3 = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

The total current leaving the power supply equals the sum of the individual branch currents in a parallel circuit.

Anahtar Kavram

In a parallel circuit, voltage remains constant across all branches (

V_{\text{total}} = V_1 = V_2 = V_3

), and total circuit current is the sum of all individual branch currents (

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

).

Alternatif Yöntem

Calculate total equivalent resistance first:

\frac{1}{R_{\text{eq}}} = \frac{1}{15\ \Omega} + \frac{1}{30\ \Omega} + \frac{1}{10\ \Omega} = \frac{2 + 1 + 3}{30} = \frac{6}{30} = \frac{1}{5\ \Omega}

, giving

R_{\text{eq}} = 5\ \Omega

. With

V = 90\text{ V}

, total current is

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{90\text{ V}}{5\ \Omega} = 18\text{ A}

.

Tahmini Süre:1m 30s

Soru 88Soru →

While troubleshooting a direct-current ( $12\text{ V}$ ) automotive auxiliary circuit, a technician observes that the branch fuse blows immediately every time the power switch is closed. Which of the following electrical faults is the most probable cause of this issue?

An open circuit in the wiring leading to the load

A short circuit directly to ground before the load

High contact resistance due to a corroded connector terminal

Reversed polarity connection at the power source

Cevabı ve açıklamayı göster

Cevap: A short circuit directly to ground before the load

Cevap

A short circuit directly to ground before the load is the most probable cause because it creates a near-zero resistance path that draws excessive current, exceeding the fuse rating.

A short circuit directly to ground creates a zero- or near-zero-resistance bypass around the intended load. According to Ohm's law (

I = \frac{V}{R}

), as resistance approaches zero, the circuit current rapidly exceeds the protective fuse's current rating, instantly melting the fuse link when the switch is closed.

Adım Adım Çözüm

1

Analyze the observed symptom and the operational role of a fuse.

A fuse is an overcurrent protection device designed to melt and open when current exceeds its specified ampere rating.

Understanding why fuses blow establishes that the root cause must be excessive current draw.

2

Apply Ohm's law (

I = \frac{V}{R}

) to evaluate how each potential fault affects circuit current.

To produce massive current (

I

) from a constant

12\text{ V}

source, the total resistance (

R

) must drop close to

0\,\Omega

.

Ohm's law dictates an inverse relationship between resistance and current.

3

Identify the circuit condition that causes near-zero resistance.

A short circuit bypasses the normal load resistance directly to ground, causing an instantaneous current spike that blows the fuse.

Open circuits and high-resistance connections reduce or eliminate current flow rather than increasing it.

Anahtar Kavram

A short circuit provides an unintended low-resistance bypass path that causes current to surge above safe levels, blowing protective fuses.

Tahmini Süre:45s

Soru 89Soru →

A portable electric heater is connected to a $120\text{ V}$ direct-current power supply. If the heating element has an internal resistance of $24\ \Omega$ , what is the current flowing through the circuit?

0.2\text{ A}

5\text{ A}

50\text{ A}

2{}880\text{ A}

Cevabı ve açıklamayı göster

Cevap: $5\text{ A}$

Cevap

5\text{ A}

Using Ohm's Law (

I = \frac{V}{R}

), dividing the applied voltage of

120\text{ V}

by the resistance of

24\ \Omega

yields exactly

5\text{ A}

.

Adım Adım Çözüm

1

Identify the known values and state Ohm's Law formula for current.

Voltage

V = 120\text{ V}

, Resistance

R = 24\ \Omega

, Formula:

I = \frac{V}{R}

Ohm's Law states that current is directly proportional to voltage and inversely proportional to resistance.

2

Substitute the circuit values into the formula and solve for current (

I

).

I = \frac{120\text{ V}}{24\ \Omega} = 5\text{ A}

Dividing

120

by

24

computes the magnitude of current flow in amperes.

Anahtar Kavram

Ohm's Law expresses the fundamental relationship among voltage, current, and resistance (

I = V / R

).

Tahmini Süre:45s

Soru 90Soru →

A multi-component series circuit has failed and is drawing zero current due to an open fault. Arrange the following troubleshooting steps in the correct chronological sequence to safely and logically locate the open component using a digital multimeter (DMM) while the circuit is energized.

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct sequence is to first verify the source voltage, anchor the negative reference lead, systematically probe downstream with the positive lead, observe the full source voltage across intact paths, and pinpoint the fault where the voltage abruptly drops to zero.

The voltage drop method is the most effective and professional way to locate an open fault in an energized circuit. In an open circuit, current cannot flow. According to Ohm's Law (

V = I \times R

), if current (

I

) is zero, the voltage drop across any intact resistor is zero. Therefore, measuring anywhere between the positive terminal and the open fault will yield the full source voltage. Measuring anywhere after the open fault will yield zero volts because that section is physically isolated from the positive terminal but electrically connected to the negative reference. The steps outlined provide the logical, chronological, and safe procedure to trace this voltage path and isolate the break.

Adım Adım Çözüm

1

Understand the electrical behavior of an open circuit.

Current is 0 Amps, meaning intact resistors will have a voltage drop of 0 Volts (

V = 0 \times R = 0

).

This principle dictates that the full source voltage will be measurable all the way up to the break.

2

Verify the baseline power supply.

The source voltage is confirmed to be present at the main terminals.

Troubleshooting downstream is useless if the main power source is dead or the battery is disconnected.

3

Establish a consistent measurement reference.

The black lead is fixed to ground, ensuring all readings are relative to 0 Volts.

Moving both leads simultaneously can cause confusion; anchoring one lead provides a steady reference point.

4

Trace the voltage potential through the circuit.

The meter reads full source voltage at each test point prior to the open fault.

The circuit wiring is intact up to the fault, extending the positive potential forward.

5

Identify the point of failure.

The fault is located exactly between the last point reading full voltage and the first point reading zero volts.

The open break physically separates the positive potential from the remainder of the circuit, causing the downstream voltage reading to drop to zero relative to ground.

Anahtar Kavram

Voltage Drop Troubleshooting Method

Tahmini Süre:1m 30s

Soru 91Soru →

An aircraft avionics system is powered by a $28\text{ V}$ DC supply connected in series with three components: a $40\ \Omega$ indicator lamp, a $70\ \Omega$ current-limiting resistor, and a $30\ \Omega$ monitoring resistor. What is the voltage drop across the $70\ \Omega$ current-limiting resistor?

28\text{ V}

14\text{ V}

8\text{ V}

5.6\text{ V}

Cevabı ve açıklamayı göster

Cevap: $14\text{ V}$

Cevap

14\text{ V}

The correct answer is

14\text{ V}

. In a series circuit, total resistance equals the sum of the individual resistors:

40\ \Omega + 70\ \Omega + 30\ \Omega = 140\ \Omega

. The circuit current is

I = 28\text{ V} / 140\ \Omega = 0.2\text{ A}

. Multiplying this current by the

70\ \Omega

resistance gives a voltage drop of

0.2\text{ A} \times 70\ \Omega = 14\text{ V}

.

Adım Adım Çözüm

1

Calculate the total equivalent resistance (

R_{\text{total}}

) of the series circuit.

R_{\text{total}} = 40\ \Omega + 70\ \Omega + 30\ \Omega = 140\ \Omega

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

2

Determine the common circuit current (

I

) using Ohm's Law.

I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{28\text{ V}}{140\ \Omega} = 0.2\text{ A}

Current remains constant through every component along a single series path.

3

Calculate the voltage drop across the

70\ \Omega

resistor (

V_{70\Omega}

).

V_{70\Omega} = I \times R = 0.2\text{ A} \times 70\ \Omega = 14\text{ V}

Applying Ohm's Law (

V = I \times R

) to the specific component yields its proportional voltage drop.

Anahtar Kavram

Series Circuit Resistance and Voltage Division

Alternatif Yöntem

Use the voltage divider formula directly:

V_{70\Omega} = 28\text{ V} \times \left(\frac{70\ \Omega}{40\ \Omega + 70\ \Omega + 30\ \Omega}\right) = 28\text{ V} \times \frac{70}{140} = 28\text{ V} \times 0.5 = 14\text{ V}

.

Tahmini Süre:1m 0s

Soru 92Soru →

An industrial control panel is powered by a $36\text{ V}$ DC source. The main circuit splits into three parallel branches. Branch one powers a cooling unit that dissipates $108\text{ W}$ of power. Branch two contains an $18\ \Omega$ limiting resistor. Branch three contains a sensor module with an unknown resistance. If a technician measures the total equivalent resistance of this entire parallel network to be exactly $4\ \Omega$ , what is the resistance of the sensor module?

4\ \Omega

9\ \Omega

12\ \Omega

18\ \Omega

Cevabı ve açıklamayı göster

Cevap: $9\ \Omega$

Cevap

The resistance of the sensor module in the third branch is

9\ \Omega

.

To find the resistance of a single parallel branch, you need the voltage across it and the current through it. The voltage is the source voltage (

36\text{ V}

). The total current is

9\text{ A}

(

36\text{ V} / 4\ \Omega

). Subtracting the current from the first branch (

108\text{ W} / 36\text{ V} = 3\text{ A}

) and the second branch (

36\text{ V} / 18\ \Omega = 2\text{ A}

) leaves

4\text{ A}

flowing through the third branch. Using Ohm's Law (

R = V/I

),

36\text{ V} / 4\text{ A}

equals

9\ \Omega

.

Adım Adım Çözüm

1

Calculate the total current (

I_{total}

) drawn by the entire circuit.

I_{total} = \frac{V}{R_{total}} = \frac{36\text{ V}}{4\ \Omega} = 9\text{ A}

Ohm's law applies to the circuit as a whole just as it does to individual components.

2

Calculate the current flowing through Branch 1 (

I_1

).

I_1 = \frac{P_1}{V} = \frac{108\text{ W}}{36\text{ V}} = 3\text{ A}

In a parallel circuit, the voltage across each branch is equal to the source voltage. We use the power formula (

P = V \times I

) rearranged to solve for current.

3

Calculate the current flowing through Branch 2 (

I_2

).

I_2 = \frac{V}{R_2} = \frac{36\text{ V}}{18\ \Omega} = 2\text{ A}

Using Ohm's law (

I = V/R

) with the known voltage and resistance for this specific branch.

4

Determine the current flowing through the unknown Branch 3 (

I_3

).

I_3 = I_{total} - (I_1 + I_2) = 9\text{ A} - (3\text{ A} + 2\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current entering a parallel network must equal the sum of the currents in individual branches.

5

Calculate the resistance of Branch 3 (

R_3

).

R_3 = \frac{V}{I_3} = \frac{36\text{ V}}{4\text{ A}} = 9\ \Omega

Applying Ohm's law once more to the specific values of the third branch.

Anahtar Kavram

Parallel Circuit Current and Power Laws

Soru 93Soru →

A heating resistor with a resistance of $20\text{ \Omega}$ has a steady current of $3\text{ A}$ flowing through it. What is the total electrical power dissipated by the resistor?

60\text{ W}

120\text{ W}

180\text{ W}

18\text{ W}

Cevabı ve açıklamayı göster

Cevap: $180\text{ W}$

Cevap

180\text{ W}

Electrical power dissipated by a resistor when current and resistance are known is determined by the formula

P = I^2 \times R

. Substituting the given values gives

P = (3\text{ A})^2 \times 20\text{ \Omega} = 9 \times 20 = 180\text{ W}

.

Adım Adım Çözüm

1

Identify the known circuit parameters and the appropriate power formula.

Known values:

I = 3\text{ A}

,

R = 20\text{ \Omega}

. Applicable formula:

P = I^2 \times R

.

When current and resistance are known, Watt's law in the form

P = I^2 R

gives electrical power directly.

2

Square the current value.

I^2 = (3\text{ A})^2 = 9\text{ A}^2

.

Power dissipation is proportional to the square of the current flowing through a resistor.

3

Multiply the squared current by the resistance.

P = 9 \times 20 = 180\text{ W}

.

Multiplying squared current by resistance yields total dissipated power in watts.

Anahtar Kavram

Electrical power dissipation in a resistive component using Joule's/Watt's law (

P = I^2 R

).

Tahmini Süre:45s

Soru 94Soru →

A $36\text{ V}$ DC control circuit supplies power to a single heating element. An inline ammeter measures a steady current flow of $3\text{ A}$ through the circuit. What is the electrical resistance of the heating element in ohms?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

12 ohms

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

R = \frac{V}{I}

). Dividing the source voltage of

36\text{ V}

by the current of

3\text{ A}

yields

12\ \Omega

.

Adım Adım Çözüm

1

Identify the known electrical values provided in the problem

Voltage (

V

) =

36\text{ V}

, Current (

I

) =

3\text{ A}

Ohm's Law relates potential difference, current flow, and circuit resistance.

2

Rearrange Ohm's Law to express resistance in terms of voltage and current

R = \frac{V}{I}

Resistance is defined as the ratio of voltage across a component to the current flowing through it.

3

Substitute the numerical values into the equation and solve

R = \frac{36\text{ V}}{3\text{ A}} = 12\ \Omega

Dividing 36 by 3 gives the resistance directly in ohms.

Anahtar Kavram

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance (

V = I \times R

, which rearranges to

R = \frac{V}{I}

).

Tahmini Süre:30s

Soru 95Soru →

An electronics technician is troubleshooting a $24\text{ V}$ direct-current (DC) control circuit that repeatedly trips its circuit breaker due to a suspected short circuit to ground. Arrange the following diagnostic procedures in the correct chronological order to safely locate and resolve the fault.

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct troubleshooting procedure starts with de-energizing the circuit and verifying zero voltage, followed by visual inspection, disconnecting branch loads for isolation, measuring resistance to ground using an ohmmeter, and finally repairing the shorted component and verifying resistance before restoring power.

Effective and safe troubleshooting follows a structured logical progression: electrical safety isolation must occur first, followed by visual inspection, systematic branch isolation, ohmmeter resistance testing to locate the shorted path, and pre-power verification following repair.

Adım Adım Çözüm

1

De-energize and verify zero voltage on the circuit.

Protects the technician from electrical shock and prevents damaging the ohmmeter.

Ohmmeters supply internal sensing voltage and will be damaged if connected to an energized circuit.

2

Visually inspect conductors, terminals, and components.

Identifies obvious burned insulation, pinched wires, or loose metallic debris.

Visual inspection is the fastest and least invasive diagnostic method before mechanical disassembly.

3

Disconnect branch loads to isolate parallel circuit sections.

Separates parallel current paths from one another.

In parallel networks, a single short circuit to ground will reflect across all interconnected branches unless they are isolated.

4

Measure resistance from each isolated conductor to ground using an ohmmeter.

Locates the exact branch exhibiting abnormal near-zero resistance (

0\ \Omega

).

A dead short creates an unintended low-resistance connection directly to chassis ground.

5

Repair the defective component and verify resistance prior to restoring power.

Ensures normal circuit resistance is restored and prevents immediate breaker tripping upon restart.

Verifying high resistance to ground on the repaired branch confirms the short has been eliminated before power is re-applied.

Anahtar Kavram

Systematic Isolation and Troubleshooting of a Short Circuit to Ground

Tahmini Süre:1m 15s

Soru 96Soru →

A workshop bench grinder is connected to a $120\text{ V}$ electrical supply line and draws a steady current of $5\text{ A}$ . What is the total electrical power consumed by the grinder in watts ( $\text{W}$ )?

Cevabı ve açıklamayı göster

Cevap: 600

Cevap

600 W

Electrical power is calculated using the formula

P = V \times I

, where

P

is power in watts,

V

is electromotive force in volts, and

I

is current in amperes. Substituting the given values gives

P = 120\text{ V} \times 5\text{ A} = 600\text{ W}

.

Adım Adım Çözüm

1

Identify the given circuit parameters

Voltage

V = 120\text{ V}

and Current

I = 5\text{ A}

Determining the known values is necessary to select the correct power formula.

2

Apply Watt's Law formula for power

P = V \times I

Watt's Law defines electrical power in watts as the product of potential difference in volts and current in amperes.

3

Calculate the total power

P = 120\text{ V} \times 5\text{ A} = 600\text{ W}

Multiplying voltage by current yields the total power consumed by the electrical device.

Anahtar Kavram

Electrical power (

P

) in a circuit is the rate at which electrical energy is converted into another form of energy, calculated using Watt's Law as

P = V \times I

.

Tahmini Süre:45s

Soru 97Soru →

Magnetic lines of flux form continuous closed loops that define the magnetic field of a permanent bar magnet. In which direction do these lines of force travel outside of the magnet?

From the south pole to the north pole

From the north pole to the south pole

From the positive terminal to the negative terminal

From the center of the magnet outward in all directions

Cevabı ve açıklamayı göster

Cevap: From the north pole to the south pole

Cevap

From the north pole to the south pole

External to a permanent magnet, magnetic field lines (lines of flux) always leave the north pole and curve through space to enter the south pole.

Adım Adım Çözüm

1

Identify the fundamental physical convention for magnetic field lines around a magnet.

Magnetic flux lines exit the north pole and curve through the surrounding space to enter the south pole.

Outside a magnet, lines of force always travel from the north magnetic pole toward the south magnetic pole to complete an external path.

Anahtar Kavram

Direction of external magnetic flux lines

Tahmini Süre:30s

Soru 98Soru →

A marine communication console operates on a $36\text{ V}$ DC power bus. The circuit consists of three components connected in series: a status indicator resistor $R_1 = 18\ \Omega$ , an internal filter resistor $R_2 = 24\ \Omega$ , and an unknown current-limiting resistor $R_3$ . An inline ammeter measures the total circuit current as $0.5\text{ A}$ . What is the resistance of resistor $R_3$ ?

18\ \Omega

30\ \Omega

24\ \Omega

72\ \Omega

Cevabı ve açıklamayı göster

Cevap: $30\ \Omega$

Cevap

The resistance of resistor

R_3

is

30\ \Omega

.

Using Ohm's Law, the total series resistance is

R_T = \frac{V_T}{I} = \frac{36\text{ V}}{0.5\text{ A}} = 72\ \Omega

. Since resistances in a series circuit are additive (

R_T = R_1 + R_2 + R_3

), the unknown resistance is found by subtracting the sum of the known resistors (

18\ \Omega + 24\ \Omega = 42\ \Omega

) from the total:

R_3 = 72\ \Omega - 42\ \Omega = 30\ \Omega

.

Adım Adım Çözüm

1

Calculate the total series circuit resistance (

R_T

) using Ohm's Law.

R_T = \frac{V_T}{I_T} = \frac{36\text{ V}}{0.5\text{ A}} = 72\ \Omega

In a series circuit, total resistance equals total applied voltage divided by the common loop current.

2

Sum the resistances of the known series components.

R_{1+2} = R_1 + R_2 = 18\ \Omega + 24\ \Omega = 42\ \Omega

Resistances add directly in a series configuration.

3

Subtract the sum of the known resistances from the total circuit resistance to isolate

R_3

.

R_3 = R_T - (R_1 + R_2) = 72\ \Omega - 42\ \Omega = 30\ \Omega

The total resistance is the sum of all individual resistors:

R_T = R_1 + R_2 + R_3

.

Anahtar Kavram

In a series circuit, current remains constant throughout the entire loop, and total resistance equals the sum of all individual resistances (

R_T = R_1 + R_2 + R_3

). Ohm's Law (

R = \frac{V}{I}

) relates total voltage to total resistance and current.

Tahmini Süre:1m 0s

Soru 99Soru →

An aircraft auxiliary power bus supplies a constant $24\text{ V}$ DC across three parallel branch circuits. Branch 1 draws a current of $3\text{ A}$ , Branch 2 contains an unknown resistor $R_2$ , and Branch 3 contains a $12\ \Omega$ resistor. If the total electrical power consumed by all three branches combined is $192\text{ W}$ , what is the resistance of resistor $R_2$ ?

4\ \Omega

8\ \Omega

6\ \Omega

3\ \Omega

Cevabı ve açıklamayı göster

Cevap: $8\ \Omega$

Cevap

The resistance of resistor

R_2

is

8\ \Omega

.

The correct resistance is

8\ \Omega

. First, finding total current from the total power gives

I_{\text{total}} = \frac{192\text{ W}}{24\text{ V}} = 8\text{ A}

. Since the voltage across each parallel branch is

24\text{ V}

, Branch 3 draws

I_3 = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}

. Subtracting the known branch currents (

3\text{ A}

and

2\text{ A}

) from the total current (

8\text{ A}

) leaves

3\text{ A}

flowing through Branch 2. Applying Ohm's Law yields

R_2 = \frac{24\text{ V}}{3\text{ A}} = 8\ \Omega

.

Adım Adım Çözüm

1

Calculate the total circuit current using total power and supply voltage.

I_{\text{total}} = \frac{P_{\text{total}}}{V} = \frac{192\text{ W}}{24\text{ V}} = 8\text{ A}

In any DC circuit, total power is the product of total voltage and total current (

P = V \times I

).

2

Calculate the current drawn by Branch 3 using Ohm's Law.

I_3 = \frac{V}{R_3} = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}

In a parallel circuit, full supply voltage (

24\text{ V}

) appears across every parallel branch.

3

Determine the current flowing through Branch 2 using Kirchhoff's Current Law for parallel circuits.

I_2 = I_{\text{total}} - I_1 - I_3 = 8\text{ A} - 3\text{ A} - 2\text{ A} = 3\text{ A}

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

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

).

4

Calculate the resistance of

R_2

using Ohm's Law.

R_2 = \frac{V}{I_2} = \frac{24\text{ V}}{3\text{ A}} = 8\ \Omega

Branch resistance is found by dividing the branch voltage by the branch current (

R = \frac{V}{I}

).

Anahtar Kavram

Parallel circuit rules: constant branch voltage (

V_{\text{total}} = V_1 = V_2 = V_3

), additive branch currents (

I_{\text{total}} = \sum I_n

), and power relationships (

P = V \times I

).

Alternatif Yöntem

Calculate individual branch powers directly: Branch 1 consumes

P_1 = V \times I_1 = 24\text{ V} \times 3\text{ A} = 72\text{ W}

. Branch 3 consumes

P_3 = \frac{V^2}{R_3} = \frac{576}{12} = 48\text{ W}

. The power dissipated in Branch 2 is

P_2 = P_{\text{total}} - P_1 - P_3 = 192\text{ W} - 72\text{ W} - 48\text{ W} = 72\text{ W}

. Using the power-resistance formula,

R_2 = \frac{V^2}{P_2} = \frac{576}{72} = 8\ \Omega

.

Tahmini Süre:1m 30s

Soru 100Soru →

A series circuit consists of a $48\text{ V}$ direct-current (DC) power source and three identical $16\,\Omega$ resistors labeled $R_1$ , $R_2$ , and $R_3$ . Under normal operation, each resistor drops $16\text{ V}$ . During troubleshooting, a technician measures $24\text{ V}$ across $R_1$ , $0\text{ V}$ across $R_2$ , and $24\text{ V}$ across $R_3$ , while current continues to flow through the circuit. Which of the following conditions explains this circuit fault?

Resistor

R_2

has developed an open circuit.

Resistor

R_2

has developed a short circuit.

Resistor

R_2

has doubled its internal resistance value.

The power supply voltage has decreased by half.

Cevabı ve açıklamayı göster

Cevap: Resistor $R_2$ has developed a short circuit.

Cevap

Resistor

R_2

has developed a short circuit.

When a component becomes shorted in a series circuit, its effective resistance drops to zero, resulting in a measured voltage drop of

0\text{ V}

across its terminals. The total circuit resistance decreases from

48\,\Omega

to

32\,\Omega

, increasing the circuit current to

1.5\text{ A}

. This higher current causes the remaining two

16\,\Omega

resistors to each drop

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

, exactly matching the technician's meter readings.

Adım Adım Çözüm

1

Analyze the voltage drop across the faulted component (

R_2

).

The voltage across

R_2

is measured at

0\text{ V}

.

A component with zero resistance (a short circuit) drops zero volts according to Ohm's Law (

V = I \times R = I \times 0 = 0\text{ V}

).

2

Evaluate the remaining circuit voltages across

R_1

and

R_3

.

The sum of voltages is

24\text{ V} + 24\text{ V} = 48\text{ V}

, matching the source voltage.

Kirchhoff's Voltage Law confirms the full source voltage is distributed across the remaining active components in the loop.

3

Calculate the total circuit resistance and current under shorted conditions.

R_{\text{total}} = 16\,\Omega + 0\,\Omega + 16\,\Omega = 32\,\Omega

, producing a current of

I = \frac{48\text{ V}}{32\,\Omega} = 1.5\text{ A}

.

A shorted component decreases total circuit resistance, which increases total current flow through the remaining healthy components.

Anahtar Kavram

Effect of a short circuit in a series circuit

Alternatif Yöntem

Use Kirchhoff's Voltage Law (

V_{\text{source}} = V_1 + V_2 + V_3

). Substituting the measured values gives

48\text{ V} = 24\text{ V} + 0\text{ V} + 24\text{ V}

. Because

V_2 = 0\text{ V}

while current is actively flowing,

R_2

must have zero resistance (

R_2 = \frac{0\text{ V}}{I} = 0\,\Omega

), which is the definition of a short circuit.

Tahmini Süre:1m 0s

Circuits and Magnetism

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Alternatif Yöntem

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Alternatif Yöntem