Circuits and Magnetism

163 soru

Soru 41Soru →

An electrical technician connects a 12 V12\text{ V} DC battery across the primary winding of an iron-core transformer through a switch. After the switch is closed and a steady current is established in the primary winding, what voltage behavior is observed across the secondary winding?

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Cevap: A brief voltage pulse occurs when the switch closes, followed by zero output while current remains steady.

Cevap

A brief voltage pulse occurs when the switch closes, followed by zero output while current remains steady.
According to Faraday's Law of Electromagnetic Induction, an electromotive force (voltage) is induced in a secondary winding only when there is a changing magnetic flux (e=−NΔΦΔte = -N \frac{\Delta \Phi}{\Delta t}). When the switch is initially closed, the current rises from zero to its maximum level, generating an expanding magnetic field that induces a brief voltage pulse. Once the DC current reaches a steady state, the magnetic field stops changing (ΔΦΔt=0\frac{\Delta \Phi}{\Delta t} = 0), causing the induced secondary voltage to drop immediately to zero.

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1
Analyze the conditions required for electromagnetic induction according to Faraday's Law.
Induced voltage is directly proportional to the rate of change of magnetic flux (e=−NΔΦΔte = -N \frac{\Delta \Phi}{\Delta t}).
A magnetic field must be expanding, collapsing, or moving relative to a conductor to induce an electromotive force (EMF).
2
Determine the magnetic flux behavior during switch closure versus steady DC state.
During switch closure, current rises from 0 A0\text{ A} to steady state, creating a rapidly expanding magnetic field (brief voltage pulse). Once steady state is reached, current is constant, making ΔΦΔt=0\frac{\Delta \Phi}{\Delta t} = 0, resulting in zero induced secondary voltage.
Steady direct current produces a stationary, unchanging magnetic field that cannot induce voltage in a stationary secondary coil.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction and Transformer Operation
Tahmini Süre:1m 0s
Soru 42Soru →

A technician sets up an experiment with two electrically isolated wire coils wrapped around a shared soft-iron core. Coil 1 is connected in series with a 12 V12\text{ V} DC power supply and a mechanical switch. Coil 2 is connected directly to a sensitive center-zero galvanometer. The switch is closed and kept closed until current stabilizes, after which the switch is abruptly opened. What behavior will the galvanometer display during the period of steady current and at the exact instant the switch is opened?

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Cevap: The meter reads zero while the current is steady, then momentarily deflects in response to the rapidly collapsing magnetic flux when the switch is opened.

Cevap

The meter reads zero while current is steady, but momentarily deflects when the switch is opened due to the rapid collapse of magnetic flux.
Electromagnetic induction strictly requires a changing magnetic flux cutting through a conductor (Vinduced=−NΔΦΔtV_{\text{induced}} = -N \frac{\Delta \Phi}{\Delta t}). While steady DC current flows in the primary coil, the magnetic field is static, producing no voltage in the secondary coil (meter reads zero). When the switch is opened, the magnetic field collapses rapidly to zero, generating a steep rate of flux change that induces a momentary voltage spike, causing the galvanometer to deflect briefly.

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1
Analyze the electromagnetic state during steady-state DC current flow.
Magnetic flux is constant (ΔΦΔt=0\frac{\Delta \Phi}{\Delta t} = 0), resulting in 0 V0\text{ V} induced in the secondary coil.
According to Faraday's Law of Electromagnetic Induction (Vinduced=−NΔΦΔtV_{\text{induced}} = -N\frac{\Delta \Phi}{\Delta t}), an electromotive force is generated only when magnetic flux changes over time.
2
Analyze the electromagnetic transition at the instant the primary switch is opened.
Current drops abruptly from steady-state value to zero, causing a rapid collapse of magnetic flux (ΔΦΔt≠0\frac{\Delta \Phi}{\Delta t} \neq 0).
The sudden collapse of magnetic lines of force cuts across the secondary windings, inducing a transient voltage pulse and causing a momentary galvanometer deflection.
3
Synthesize the two operational states to evaluate secondary meter response.
The meter must read zero during steady current and show a sharp, temporary deflection when the circuit is broken.
Combines Faraday's requirement for dynamic flux with the transient behavior of inductive circuits.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction and transient flux changes
Tahmini Süre:1m 30s
Soru 43Soru →

A DC power supply is connected across a series circuit containing two resistive components. Fill in the blanks with the correct numerical values to complete the circuit analysis.

Aşağıdaki boşlukları doldurun

A 50 V50\text{ V} DC source is connected in series with a 15 Ω15\ \Omega resistor and a second resistor R2R_2. If the circuit draws a current of 2 A2\text{ A}, the total circuit resistance is Ω\Omega, and the resistance of R2R_2 is Ω\Omega.
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Cevap

The total circuit resistance is 25 Ω25\ \Omega, and the resistance of R2R_2 is 10 Ω10\ \Omega.
Applying Ohm's Law to the total circuit yields a total resistance of RT=50 V2 A=25 ΩR_T = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega. In a series circuit, total resistance equals the sum of the individual component resistances (RT=R1+R2R_T = R_1 + R_2). Subtracting the known resistor value (15 Ω15\ \Omega) from the total resistance (25 Ω25\ \Omega) results in R2=10 ΩR_2 = 10\ \Omega.

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1
Calculate the total circuit resistance (RTR_T) using Ohm's Law with source voltage and total current.
RT=VTI=50 V2 A=25 ΩR_T = \frac{V_T}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega
In any complete circuit, the total equivalent resistance equals total source voltage divided by the total circuit current.
2
Calculate the resistance of R2R_2 using the series circuit resistance addition formula (RT=R1+R2R_T = R_1 + R_2).
R2=RT−R1=25 Ω−15 Ω=10 ΩR_2 = R_T - R_1 = 25\ \Omega - 15\ \Omega = 10\ \Omega
In a series circuit, total resistance is the direct sum of individual component resistances.

Anahtar Kavram

In a series circuit, total resistance is the direct sum of individual resistances (RT=R1+R2+…R_T = R_1 + R_2 + \dots), and the circuit current is constant through all components as governed by Ohm's Law (RT=VT/IR_T = V_T / I).
Tahmini Süre:1m 0s
Soru 44Soru →

A transformer can continuously induce a voltage in its secondary winding when powered by a steady, unchanging direct current (DC) source.

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Cevap: False

Cevap

False
The statement is false because continuous electromagnetic induction requires a time-varying magnetic flux. A steady DC current produces a stationary, unchanging magnetic field, resulting in zero continuous induced voltage across the secondary winding.

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1
Analyze the electrical power source and component operation described in the statement.
The statement specifies applying steady, constant direct current (DC) to the primary winding of a transformer to produce continuous secondary voltage.
Transformers rely on mutual electromagnetic induction between primary and secondary windings.
2
Evaluate the condition required for continuous electromagnetic induction.
Faraday's law of induction dictates that induced voltage is directly proportional to the rate of change of magnetic flux (ΔΦ/Δt\Delta \Phi / \Delta t). A steady DC source produces a constant magnetic field (ΔΦ/Δt=0\Delta \Phi / \Delta t = 0), meaning no continuous voltage is induced.
Continuous induction requires a continuously changing current and magnetic flux, such as alternating current (AC).

Anahtar Kavram

Electromagnetic induction in transformers requires a time-varying magnetic flux, meaning a steady direct current (DC) cannot maintain continuous secondary induction.
Tahmini Süre:45s
Soru 45Soru →

Inserting a soft iron core into the center of an energized coil of wire increases the overall strength of the magnetic field compared to an air-core coil with the same current and number of turns.

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Cevap: True

Cevap

True
The statement is true because soft iron is a ferromagnetic material with high magnetic permeability, which greatly concentrates magnetic flux lines and amplifies the electromagnetic field strength for a given current.

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1
Identify the factors that determine magnetic flux density in a solenoid
Magnetic flux density is governed by the relation B=μNLIB = \mu \frac{N}{L} I, where μ\mu is core permeability, NN is number of turns, LL is coil length, and II is current.
Core material permeability directly determines the magnetic flux density for a given amount of magnetomotive force.
2
Compare the magnetic properties of soft iron to air
Soft iron is a ferromagnetic substance with relative magnetic permeability hundreds to thousands of times greater than air.
Higher permeability provides a low-reluctance path that concentrates lines of magnetic force inside and around the core.
3
Determine the effect of adding an iron core on overall field strength
Adding the soft iron core significantly amplifies the magnetic field produced by the coil, confirming the statement is true.
This physical principle is the foundation for creating practical electromagnets, relays, and inductors.

Anahtar Kavram

Effect of ferromagnetic core permeability on electromagnetic field strength
Tahmini Süre:45s
Soru 46Soru →

A direct-current power distribution box supplies power to two parallel branches. The first branch has a resistance of 20 Ω20\ \Omega, and the second branch has a resistance of 5 Ω5\ \Omega. What is the total equivalent resistance of the network?

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Cevap: 4 Ω4\ \Omega

Cevap

4 Ω4\ \Omega
In a parallel circuit with two branches, the equivalent resistance is calculated using the product-over-sum formula Req=R1×R2R1+R2R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}. Substituting 20 Ω20\ \Omega and 5 Ω5\ \Omega gives 20×520+5=10025=4 Ω\frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega. Total resistance in parallel is always less than the lowest individual resistor value.

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1
Identify the formula for the equivalent resistance of two parallel resistors.
Req=R1×R2R1+R2R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} or 1Req=1R1+1R2\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}
Parallel branches provide multiple paths for current, resulting in a total resistance that is less than any individual branch resistance.
2
Substitute the given values (R1=20 ΩR_1 = 20\ \Omega and R2=5 ΩR_2 = 5\ \Omega) into the product-over-sum formula.
Req=20×520+5=10025R_{\text{eq}} = \frac{20 \times 5}{20 + 5} = \frac{100}{25}
Simplifying the numerator and denominator allows direct calculation of equivalent resistance.
3
Divide the product by the sum.
Req=4 ΩR_{\text{eq}} = 4\ \Omega
Computing 10025\frac{100}{25} yields the final equivalent resistance.

Anahtar Kavram

Total Equivalent Resistance in Parallel Circuits
Tahmini Süre:45s
Soru 47Soru →

When the direct current flowing through an inductor decreases, the self-induced electromotive force (counter-EMF) acts in the same direction as the original current to oppose the reduction in current flow.

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Cevap: True

Cevap

True
The statement is true because Lenz's law dictates that self-induced voltage opposes the change in current rather than the absolute direction of current flow. When circuit current decreases, the collapsing magnetic field induces a voltage of such polarity that it acts in the same direction as the original current to sustain the flow.

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1
Analyze the fundamental principle of electromagnetic induction and Lenz's law for inductors.
Lenz's law states that the polarity of an induced electromotive force (EMF) always opposes the specific change in magnetic flux or current that creates it (e=−LΔIΔte = -L \frac{\Delta I}{\Delta t}).
Inductors store energy within their magnetic field and oppose instantaneous changes in current magnitude.
2
Evaluate the behavior of the magnetic field when current decreases.
As current decreases, the magnetic field around the inductor collapses across the coil windings, inducing an electromotive force.
A changing magnetic flux through a conductor induces an electric potential.
3
Determine the direction and polarity of the induced EMF relative to the original current.
To oppose the reduction in current (ΔI/Δt<0\Delta I / \Delta t < 0), the induced EMF acts as an aiding voltage source in the same direction as the original current flow.
Opposing a decay in current requires supplying voltage to keep current moving in the original direction.
4
Conclude the truth value of the statement.
The statement correctly describes the directional behavior of self-induced counter-EMF during current decay, confirming the statement is True.
Counter-EMF opposes the direction of current only when current is increasing; during current decrease, it reinforces the original current direction.

Anahtar Kavram

Lenz's Law and Self-Induction in Inductors
Tahmini Süre:45s
Soru 48Soru →

A direct current (DC) circuit consists of a 48 V48\text{ V} power supply connected in series with three resistors: R1=12 ΩR_1 = 12\ \Omega, R2=24 ΩR_2 = 24\ \Omega, and R3=24 ΩR_3 = 24\ \Omega. In its baseline operating state, the total circuit resistance is RT=60 ΩR_T = 60\ \Omega and the circuit loop current is I=0.8 AI = 0.8\text{ A}.

Match each circuit alteration on the left to its corresponding electrical outcome on the right.

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Öğeler

A zero-resistance jumper wire (short circuit) is placed directly across resistor R3R_3.
Resistor R2R_2 burns out internally, creating an open circuit condition.
A fourth resistor with a value of R4=20 ΩR_4 = 20\ \Omega is connected in series with the existing chain.
The power supply voltage is increased from 48 V48\text{ V} to 72 V72\text{ V} while keeping all components unchanged.

Eşleşmeler

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Cevap

The correct matches pair: (1) shorting resistor R3R_3 with total resistance dropping to 36 Ω36\ \Omega, current rising to 1.33 A1.33\text{ A}, and 0 V0\text{ V} across R3R_3; (2) an open at R2R_2 with loop current dropping to 0 A0\text{ A} and 48 V48\text{ V} appearing across R2R_2; (3) adding a 20 Ω20\ \Omega series resistor with total resistance rising to 80 Ω80\ \Omega, current dropping to 0.6 A0.6\text{ A}, and 7.2 V7.2\text{ V} across R1R_1; and (4) raising the source to 72 V72\text{ V} with total resistance remaining 60 Ω60\ \Omega, current rising to 1.2 A1.2\text{ A}, and 28.8 V28.8\text{ V} across R2R_2.
Each modification directly follows the governing principles of series DC circuits. In series circuits, total resistance equals the direct sum of all individual resistances (RT=R1+R2+⋯+RnR_T = R_1 + R_2 + \dots + R_n), and current is constant throughout the single loop (IT=VTRTI_T = \frac{V_T}{R_T}). Short circuits eliminate individual component resistance and drop component voltage to zero while raising loop current. Open circuits eliminate current entirely and place full source voltage across the open terminals. Adding series resistance reduces current throughout the loop, and increasing source voltage increases current and all individual voltage drops proportionally.

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1
Analyze the short circuit condition across R3R_3
RT=12+24=36 ΩR_T = 12 + 24 = 36\ \Omega; I=4836≈1.33 AI = \frac{48}{36} \approx 1.33\text{ A}; V3=0 VV_3 = 0\text{ V}
A short circuit bypasses the component with zero resistance, eliminating its resistance from the total series sum and producing zero potential difference across itself.
2
Analyze the open circuit condition at R2R_2
I=0 AI = 0\text{ A}; V1=0 VV_1 = 0\text{ V}; V3=0 VV_3 = 0\text{ V}; V2=48 VV_2 = 48\text{ V}
An open in a series circuit interrupts the sole conductive loop, causing current to cease everywhere; infinite resistance across the open component causes the full source potential to appear across the gap.
3
Analyze the addition of a fourth series resistor (R4=20 ΩR_4 = 20\ \Omega)
RT=60+20=80 ΩR_T = 60 + 20 = 80\ \Omega; I=4880=0.6 AI = \frac{48}{80} = 0.6\text{ A}; V1=0.6×12=7.2 VV_1 = 0.6 \times 12 = 7.2\text{ V}
In series circuits, equivalent resistance is additive (RT=∑RR_T = \sum R). Increasing total resistance reduces total current according to Ohm's Law (I=VRTI = \frac{V}{R_T}), which proportionally reduces individual component voltage drops.
4
Analyze the voltage source increase to 72 V72\text{ V}
RT=60 ΩR_T = 60\ \Omega; I=7260=1.2 AI = \frac{72}{60} = 1.2\text{ A}; V2=1.2×24=28.8 VV_2 = 1.2 \times 24 = 28.8\text{ V}
Resistor values are independent of source voltage. Applying Ohm's Law with higher voltage across constant resistance yields directly proportional increases in loop current and individual voltage drops.

Anahtar Kavram

Series Circuit Governing Rules: Additive Resistance, Current Continuity, Ohm's Law, and Fault Dynamics
Tahmini Süre:2m 0s
Soru 49Soru →

A direct-current marine radio circuit contains two resistors connected in parallel: a 10 Ω10\ \Omega resistor and a 40 Ω40\ \Omega resistor. What is the total equivalent resistance of this parallel circuit, in ohms?

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Cevap: 8

Cevap

8 Ω
For two resistors in parallel with values of 10 Ω10\ \Omega and 40 Ω40\ \Omega, the total equivalent resistance is calculated using the product-over-sum formula: Req=R1×R2R1+R2=10×4010+40=40050=8 ΩR_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 40}{10 + 40} = \frac{400}{50} = 8\ \Omega. In any parallel circuit, the total equivalent resistance is always smaller than the smallest branch resistor.

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1
Select the appropriate formula for calculating the equivalent resistance of two parallel resistors.
R_eq = (R_1 * R_2) / (R_1 + R_2) or 1/R_eq = 1/R_1 + 1/R_2
In parallel circuits, total resistance is found by taking the reciprocal of the sum of the reciprocals, which simplifies to the product-over-sum rule for two branches.
2
Substitute the given resistor values (10 Ω and 40 Ω) into the product-over-sum formula.
R_eq = (10 * 40) / (10 + 40) = 400 / 50
Multiplying the branch values gives the numerator, and adding them gives the denominator.
3
Divide the numerator by the denominator to find the equivalent resistance.
R_eq = 8 Ω
400 divided by 50 equals 8, which is less than the smallest branch resistance (10 Ω), confirming the validity of the parallel calculation.

Anahtar Kavram

Parallel Circuit Equivalent Resistance
Tahmini Süre:45s
Soru 50Soru →

Which of the following occurs when a direct current flows through an insulated wire coiled around a soft-iron core?

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Cevap: The iron core becomes temporarily magnetized and creates an external magnetic field

Cevap

The iron core becomes temporarily magnetized and creates an external magnetic field.
When direct current passes through a coil wrapped around a soft-iron core, the magnetic field generated by the current aligns the magnetic domains in the iron. This creates a temporary magnet (an electromagnet) that produces an external magnetic field as long as current flows.

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1
Identify the physical effect of electric current passing through a coiled conductor
Current moving through the loops produces a concentrated magnetic field along the axis of the coil
All electric currents generate an associated magnetic field according to the principles of electromagnetism
2
Analyze how placing a soft-iron core inside the coil affects this field
The ferromagnetic iron core aligns its magnetic domains with the coil field, greatly strengthening it and acting as a temporary magnet
Soft iron has high magnetic permeability, allowing it to easily magnetize when exposed to a magnetic field and demagnetize when the current stops

Anahtar Kavram

Electromagnetism in Coils with Ferromagnetic Cores
Tahmini Süre:45s
Soru 51Soru →

When the direct current flowing through an energized inductor coil is suddenly interrupted by opening a switch, a high-voltage spark often arcs across the switch contacts. Which electromagnetic effect causes this sudden voltage surge?

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Cevap: Counter-electromotive force (CEMF) induced by the rapidly collapsing magnetic field

Cevap

Counter-electromotive force (CEMF) induced by the rapidly collapsing magnetic field causes the high-voltage spike.
According to Faraday's law of induction and Lenz's law, when current through an inductor is suddenly cut, the magnetic field around the coil rapidly collapses. Because the rate of change of magnetic flux is extremely high, the coil generates a massive counter-electromotive force (CEMF) or inductive kickback voltage that arcs across the opening switch contacts.

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1
Analyze the state of the circuit before and after the switch opens
Steady current established a magnetic field around the inductor coil; opening the switch abruptly drops current to zero.
Inductors store energy in magnetic fields proportional to the current flowing through them.
2
Apply Faraday's and Lenz's laws of electromagnetic induction
The rapid collapse of magnetic flux induces a large counter-electromotive force (V=−LΔIΔtV = -L \frac{\Delta I}{\Delta t}) across the coil terminals.
A faster change in flux generates a higher induced voltage spike that attempts to keep current flowing.
3
Identify the cause of the arc across the contacts
The induced voltage spike exceeds the dielectric breakdown strength of the small air gap between the opening switch contacts.
This inductive kickback produces a momentary high-voltage spark.

Anahtar Kavram

Self-Induction and Counter-Electromotive Force (CEMF)
Tahmini Süre:1m 0s
Soru 52Soru →

A naval vessel alarm panel is powered by a 72 V72\text{ V} DC supply connected in series with three resistive loads: an alert strobe (R1=8 ΩR_1 = 8\ \Omega), an auxiliary buzzer (R2=10 ΩR_2 = 10\ \Omega), and a line-balancing resistor (R3=6 ΩR_3 = 6\ \Omega). What is the voltage drop across the auxiliary buzzer (R2R_2)?

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Cevap: 30 V30\text{ V}

Cevap

The voltage drop across the auxiliary buzzer (R2R_2) is 30 V30\text{ V}.
The correct answer is 30 V30\text{ V}. First, sum the resistances in series to obtain RT=8 Ω+10 Ω+6 Ω=24 ΩR_T = 8\ \Omega + 10\ \Omega + 6\ \Omega = 24\ \Omega. Next, determine the loop current using Ohm's Law: I=72 V24 Ω=3 AI = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}. Finally, multiply the current by the resistance of the auxiliary buzzer (10 Ω10\ \Omega) to get V2=3 A×10 Ω=30 VV_2 = 3\text{ A} \times 10\ \Omega = 30\text{ V}.

Adım Adım Çözüm

1
Calculate the total equivalent resistance (RTR_T) of the series circuit.
RT=R1+R2+R3=8 Ω+10 Ω+6 Ω=24 ΩR_T = R_1 + R_2 + R_3 = 8\ \Omega + 10\ \Omega + 6\ \Omega = 24\ \Omega
In a series circuit, total resistance is the algebraic sum of all individual resistances.
2
Determine the common loop current (II) flowing through the circuit.
I=VTRT=72 V24 Ω=3 AI = \frac{V_T}{R_T} = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}
According to Ohm's Law, current equals total voltage divided by total resistance, and current remains constant throughout a series path.
3
Calculate the voltage drop across the target component (R2R_2).
V2=I×R2=3 A×10 Ω=30 VV_2 = I \times R_2 = 3\text{ A} \times 10\ \Omega = 30\text{ V}
The voltage drop across a specific resistor in series is the product of the circuit current and that resistor's value.

Anahtar Kavram

In a series DC circuit, total resistance is the sum of all individual resistances, current is identical through every component, and the voltage drop across any individual resistor is proportional to its resistance (V=I×RV = I \times R).
Tahmini Süre:1m 0s
Soru 53Soru →

An industrial security monitoring loop is powered by a 36 V36\text{ V} DC power supply connected in series with a current-limiting resistor R1=150 ΩR_1 = 150\ \Omega, an optical sensor module R2R_2, and a line termination resistor R3=250 ΩR_3 = 250\ \Omega. When the circuit is active, a voltmeter measures a 12 V12\text{ V} drop across the optical sensor module R2R_2. What is the electrical resistance of the optical sensor module R2R_2?

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Cevap: 200 Ω200\ \Omega

Cevap

The electrical resistance of the optical sensor module R2R_2 is 200 Ω200\ \Omega.
The correct answer is determined by first finding the remaining voltage across the known resistors (36 V−12 V=24 V36\text{ V} - 12\text{ V} = 24\text{ V}). Dividing this by their combined resistance (150 Ω+250 Ω=400 Ω150\ \Omega + 250\ \Omega = 400\ \Omega) yields a constant loop current of 0.06 A0.06\text{ A}. Dividing the sensor's 12 V12\text{ V} drop by 0.06 A0.06\text{ A} gives 200 Ω200\ \Omega.

Adım Adım Çözüm

1
Calculate the combined voltage drop across the known series resistors R1R_1 and R3R_3.
V1,3=Vtotal−V2=36 V−12 V=24 VV_{1,3} = V_{\text{total}} - V_2 = 36\text{ V} - 12\text{ V} = 24\text{ V}
In a series circuit, Kirchhoff's Voltage Law dictates that the sum of all individual voltage drops equals the total supply voltage.
2
Find the sum of the known series resistances.
R1,3=R1+R3=150 Ω+250 Ω=400 ΩR_{1,3} = R_1 + R_3 = 150\ \Omega + 250\ \Omega = 400\ \Omega
Resistances in a series path are directly additive.
3
Determine the common loop current flowing through the circuit.
I=V1,3R1,3=24 V400 Ω=0.06 AI = \frac{V_{1,3}}{R_{1,3}} = \frac{24\text{ V}}{400\ \Omega} = 0.06\text{ A}
Current is constant at all points in a series circuit according to Ohm's Law.
4
Calculate the resistance of R2R_2 using its measured voltage drop and the loop current.
R2=V2I=12 V0.06 A=200 ΩR_2 = \frac{V_2}{I} = \frac{12\text{ V}}{0.06\text{ A}} = 200\ \Omega
Applying Ohm's Law (R=VIR = \frac{V}{I}) to the individual component provides its exact resistance.

Anahtar Kavram

Series circuit current constancy and voltage division using Ohm's Law and Kirchhoff's Voltage Law

Alternatif Yöntem

Using the voltage divider formula: V2Vtotal=R2R1+R2+R3  ⟹  1236=R2400+R2  ⟹  13=R2400+R2  ⟹  400+R2=3R2  ⟹  2R2=400  ⟹  R2=200 Ω\frac{V_2}{V_{\text{total}}} = \frac{R_2}{R_1 + R_2 + R_3} \implies \frac{12}{36} = \frac{R_2}{400 + R_2} \implies \frac{1}{3} = \frac{R_2}{400 + R_2} \implies 400 + R_2 = 3R_2 \implies 2R_2 = 400 \implies R_2 = 200\ \Omega.
Tahmini Süre:1m 30s
Soru 54Soru →

An aircraft instrument panel contains three identical 12 Ω12\ \Omega indicator lights connected in parallel across a direct-current power bus. What is the total equivalent resistance of this parallel circuit?

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Cevap: 4 Ω4\ \Omega

Cevap

4 Ω4\ \Omega
For parallel circuits containing equal resistors, the total equivalent resistance is found by dividing the individual resistance value by the number of parallel branches (Rtotal=R/NR_{\text{total}} = R / N). Here, 12 Ω/3=4 Ω12\ \Omega / 3 = 4\ \Omega.

Adım Adım Çözüm

1
Identify the number of parallel branches and branch resistance
Branch resistance R=12 ΩR = 12\ \Omega, number of branches N=3N = 3
When all resistors in parallel have identical resistance values, the equal-resistor shortcut formula can be directly applied.
2
Calculate total equivalent resistance using Rtotal=RNR_{\text{total}} = \frac{R}{N}
Rtotal=12 Ω3=4 ΩR_{\text{total}} = \frac{12\ \Omega}{3} = 4\ \Omega
Connecting resistors in parallel provides multiple parallel paths for current, reducing the total equivalent resistance below any individual branch.

Anahtar Kavram

Equal-Resistor Parallel Shortcut Rule (Rtotal=R/NR_{\text{total}} = R / N)
Tahmini Süre:45s
Soru 55Soru →

During the calibration of a naval radar control system, engineers measure a continuous current of 1.5 A1.5\text{ A} flowing through a series safety loop. The loop is energized by a 60 V60\text{ V} direct current source and contains three elements: a 15 Ω15\ \Omega current-limiting resistor (R1R_1), a calibration rheostat (R2R_2), and a primary sensor coil (R3R_3). If the primary sensor coil dissipates exactly 36 W36\text{ W} of power, what is the required resistance of the calibration rheostat (R2R_2)?

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Cevap: 9 Ω9\ \Omega

Cevap

The required resistance of the calibration rheostat is 9 Ω9\ \Omega.
In this series circuit, the total resistance must equal the source voltage divided by the total current (60 V/1.5 A=40 Ω60\text{ V} / 1.5\text{ A} = 40\ \Omega). Since components are in series, they share the same 1.5 A1.5\text{ A} current. The sensor coil dissipates 36 W36\text{ W}, meaning its resistance is P/I2P / I^2 (36/2.25=16 Ω36 / 2.25 = 16\ \Omega). Subtracting the 15 Ω15\ \Omega resistor and the 16 Ω16\ \Omega coil from the 40 Ω40\ \Omega total leaves 9 Ω9\ \Omega for the calibration rheostat.

Adım Adım Çözüm

1
Calculate the total equivalent resistance of the series circuit using Ohm's Law.
RT=VTIT=60 V1.5 A=40 ΩR_T = \frac{V_T}{I_T} = \frac{60\text{ V}}{1.5\text{ A}} = 40\ \Omega
Total resistance is needed to find the missing component value in a series string.
2
Determine the resistance of the primary sensor coil (R3R_3) using the power formula P=I2RP = I^2R.
R3=PI2=36 W(1.5 A)2=362.25=16 ΩR_3 = \frac{P}{I^2} = \frac{36\text{ W}}{(1.5\text{ A})^2} = \frac{36}{2.25} = 16\ \Omega
Since the circuit is in series, the full 1.5 A1.5\text{ A} current flows through all components, allowing the coil's resistance to be calculated from its power dissipation.
3
Subtract the known series resistances from the total resistance to find the rheostat's value (R2R_2).
R2=RT−R1−R3=40 Ω−15 Ω−16 Ω=9 ΩR_2 = R_T - R_1 - R_3 = 40\ \Omega - 15\ \Omega - 16\ \Omega = 9\ \Omega
In a series circuit, total resistance is the sum of all individual resistances (RT=R1+R2+R3R_T = R_1 + R_2 + R_3).

Anahtar Kavram

Series Circuit Resistance and Power Dissipation
Soru 56Soru →

Match each electromagnetic device on the left with its correct operating principle or primary functional description on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Electromechanical Relay
Linear Solenoid
Step-Up Transformer
Inductor Coil

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Electromechanical Relay matches with using an energized coil to pull an armature to open or close circuit contacts; Linear Solenoid matches with converting an electromagnetic field into linear mechanical motion to actuate a plunger; Step-Up Transformer matches with using mutual induction to increase AC voltage while decreasing AC current; Inductor Coil matches with opposing changes in electrical current via self-induced counter-EMF.
Each electromagnetic device corresponds to a distinct electromagnetic mechanism: the relay magnetically toggles switch contacts; the solenoid converts magnetic fields into linear mechanical stroke; the step-up transformer uses mutual induction to increase AC voltage; and the inductor coil generates self-induced counter-EMF to oppose changes in current.

Adım Adım Çözüm

1
Identify the operating mechanism of an electromechanical relay
Matched to using an energized coil to pull a ferromagnetic armature and toggle electrical contacts.
Relays function as electrically operated switches, using electromagnetism to control contacts in a secondary circuit without direct electrical connection.
2
Identify the operating mechanism of a linear solenoid
Matched to converting magnetic flux into linear mechanical motion of an internal plunger.
Solenoids are actuators that translate electromagnetic coil energization into physical linear pushing or pulling work.
3
Identify the operating mechanism of a step-up transformer
Matched to using mutual induction to increase AC voltage while proportionally decreasing current.
Transformers rely on dynamic magnetic flux linkage between primary and secondary coils to scale alternating voltage levels according to turns ratios.
4
Identify the operating mechanism of an inductor coil
Matched to opposing current variations through self-induced counter-electromotive force.
According to Lenz's law and Faraday's law of self-induction, an inductor generates a counter-voltage that directly resists changes in the current passing through it.

Anahtar Kavram

Operating principles and applications of core electromagnetic components: relays, solenoids, transformers, and inductors
Tahmini Süre:1m 0s
Soru 57Soru →

An electronics technician is analyzing the operational principles of a DC series circuit. Match each series circuit characteristic on the left with its correct electrical rule or behavioral description on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

Total circuit resistance (RTR_T)
Current flow (II)
Individual component voltage drop (VnV_n)
Open-circuit fault

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

Total circuit resistance matches with the sum of all individual load values; Current flow matches with remaining constant and identical at every point; Individual component voltage drop matches with being directly proportional to resistance (I×RnI \times R_n); Open-circuit fault matches with reducing total loop current to zero and disabling all components.
In a series circuit, all components are connected sequentially in a single loop. Therefore, total resistance is the direct sum of all resistor values, the current is uniform throughout, the individual voltage drops are proportional to component resistance via Ohm's Law, and an open circuit completely eliminates current flow across all components.

Adım Adım Çözüm

1
Identify the relationship for total resistance in a series string
RT=R1+R2+⋯+RnR_T = R_1 + R_2 + \dots + R_n
Series components add their oppositions consecutively to the flow of current.
2
Determine the path and behavior of electrical current in a series loop
IT=I1=I2=⋯=InI_T = I_1 = I_2 = \dots = I_n
A single conducting path allows no branching, ensuring the rate of charge flow is uniform everywhere.
3
Apply Ohm's Law to find the voltage drop across an individual resistor
Vn=I×RnV_n = I \times R_n
With current constant, the voltage drop across each component depends directly on its resistance value.
4
Analyze the impact of a discontinuity (open circuit)
I=0 AI = 0\text{ A} throughout the entire loop
Breaking any point in a single closed loop stops all current from circulating.

Anahtar Kavram

Core governing laws of DC series circuits: additive resistance, uniform loop current, proportional voltage division, and single-path continuity dependency.
Tahmini Süre:1m 0s
Soru 58Soru →

A 24 V24\text{ V} DC power supply energizes a circuit consisting of a 100 Ω100\,\Omega series resistor (R1R_1) connected in series with a parallel network composed of two 200 Ω200\,\Omega branch loads. Under normal operating conditions, the circuit draws 120 mA120\text{ mA}, dropping 12 V12\text{ V} across R1R_1 and 12 V12\text{ V} across the parallel network. During a troubleshooting check, a technician records a total circuit current of 80 mA80\text{ mA}, a voltage drop of 8 V8\text{ V} across R1R_1, and a voltage drop of 16 V16\text{ V} across the parallel network. Which fault condition in the circuit explains these diagnostic readings?

Cevabı ve açıklamayı göster

Cevap: An open circuit occurred in one of the two parallel branch loads.

Cevap

An open circuit occurred in one of the two parallel branch loads.
When one parallel branch develops an open circuit, current can only flow through the remaining 200 Ω200\,\Omega branch. This increases the equivalent parallel resistance from 100 Ω100\,\Omega to 200 Ω200\,\Omega, bringing total circuit resistance to 300 Ω300\,\Omega. By Ohm's law, total current drops to I=24 V300 Ω=80 mAI = \frac{24\text{ V}}{300\,\Omega} = 80\text{ mA}. Consequently, the voltage across R1R_1 drops to 8 V8\text{ V} (80 mA×100 Ω80\text{ mA} \times 100\,\Omega) while the parallel network drops the remaining 16 V16\text{ V} (80 mA×200 Ω80\text{ mA} \times 200\,\Omega).

Adım Adım Çözüm

1
Analyze normal baseline circuit parameters.
Normal equivalent parallel resistance is Rparallel=200 Ω×200 Ω200 Ω+200 Ω=100 ΩR_{\text{parallel}} = \frac{200\,\Omega \times 200\,\Omega}{200\,\Omega + 200\,\Omega} = 100\,\Omega. Total normal resistance is Rtotal=100 Ω+100 Ω=200 ΩR_{\text{total}} = 100\,\Omega + 100\,\Omega = 200\,\Omega. Normal current is I=24 V200 Ω=120 mAI = \frac{24\text{ V}}{200\,\Omega} = 120\text{ mA}, producing equal 12 V12\text{ V} drops across R1R_1 and the parallel bank.
Establishing normal values is required to determine whether total circuit resistance increased or decreased under fault conditions.
2
Calculate the faulted circuit's total resistance and component values using the measured readings.
With Imeasured=80 mAI_{\text{measured}} = 80\text{ mA} and Vsource=24 VV_{\text{source}} = 24\text{ V}, total circuit resistance is Rtotal, fault=24 V0.08 A=300 ΩR_{\text{total, fault}} = \frac{24\text{ V}}{0.08\text{ A}} = 300\,\Omega. Since R1R_1 drops 8 V8\text{ V} at 80 mA80\text{ mA} (R1=8 V0.08 A=100 ΩR_1 = \frac{8\text{ V}}{0.08\text{ A}} = 100\,\Omega, unchanged), the parallel bank now exhibits Rparallel, fault=16 V0.08 A=200 ΩR_{\text{parallel, fault}} = \frac{16\text{ V}}{0.08\text{ A}} = 200\,\Omega.
Ohm's law confirms where the additional resistance is located.
3
Determine the physical fault that produces a 200 Ω200\,\Omega parallel bank resistance.
Because each individual branch load is rated at 200 Ω200\,\Omega, an equivalent resistance of 200 Ω200\,\Omega across the bank means only one 200 Ω200\,\Omega branch is conducting, indicating the other branch has an open circuit.
Removing one branch of a two-branch identical parallel network doubles the network's equivalent resistance to the value of the remaining single branch.

Anahtar Kavram

In a series-parallel circuit, an open fault in a parallel branch increases total circuit resistance, reduces total line current, and causes a higher proportion of the source voltage to drop across the parallel section.
Tahmini Süre:1m 30s
Soru 59Soru →

A specialized weather station telemetry circuit is powered by a 96 V96\text{ V} direct current (DC) power source connected in series with four resistive components: a surge-limiting resistor R1=18 ΩR_1 = 18\ \Omega, a status indicator R2=27 ΩR_2 = 27\ \Omega, a variable humidity sensor R3R_3, and a reference resistor R4=15 ΩR_4 = 15\ \Omega. A technician measures the voltage drop across the status indicator (R2R_2) to be 21.6 V21.6\text{ V}. What is the resistance of the humidity sensor R3R_3 in ohms (Ω\Omega)?

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Cevap: 60

Cevap

The resistance of the humidity sensor R3R_3 is 60 Ω60\ \Omega.
In a series circuit, current is uniform throughout the entire loop. Using the measured voltage drop across R2R_2 (21.6 V21.6\text{ V}) and its known resistance (27 Ω27\ \Omega), the loop current is I=V2R2=21.6 V27 Ω=0.8 AI = \frac{V_2}{R_2} = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}. The total equivalent resistance connected across the 96 V96\text{ V} source is RT=VsourceI=96 V0.8 A=120 ΩR_T = \frac{V_{\text{source}}}{I} = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega. Because series resistances are directly additive (RT=R1+R2+R3+R4R_T = R_1 + R_2 + R_3 + R_4), the unknown resistance is determined by subtracting the known values from the total: R3=120 Ω−(18 Ω+27 Ω+15 Ω)=120 Ω−60 Ω=60 ΩR_3 = 120\ \Omega - (18\ \Omega + 27\ \Omega + 15\ \Omega) = 120\ \Omega - 60\ \Omega = 60\ \Omega.

Adım Adım Çözüm

1
Calculate loop current from the known component voltage drop and resistance.
I=21.6 V27 Ω=0.8 AI = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}
In a series circuit, current is identical through all components.
2
Calculate total circuit resistance using Ohm's Law and the source voltage.
RT=96 V0.8 A=120 ΩR_T = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega
Total resistance equals source voltage divided by total circuit current.
3
Calculate the sum of all known resistances.
Rknown=18 Ω+27 Ω+15 Ω=60 ΩR_{\text{known}} = 18\ \Omega + 27\ \Omega + 15\ \Omega = 60\ \Omega
Resistances in a series circuit are directly additive.
4
Subtract the sum of known resistors from total circuit resistance to isolate R3R_3.
R3=120 Ω−60 Ω=60 ΩR_3 = 120\ \Omega - 60\ \Omega = 60\ \Omega
RT=R1+R2+R3+R4R_T = R_1 + R_2 + R_3 + R_4, so R3=RT−(R1+R2+R4)R_3 = R_T - (R_1 + R_2 + R_4).

Anahtar Kavram

Series Circuit Loop Current and Total Resistance Analysis

Alternatif Yöntem

Calculate individual voltage drops using loop current I=0.8 AI = 0.8\text{ A}: V1=0.8 A×18 Ω=14.4 VV_1 = 0.8\text{ A} \times 18\ \Omega = 14.4\text{ V} and V4=0.8 A×15 Ω=12.0 VV_4 = 0.8\text{ A} \times 15\ \Omega = 12.0\text{ V}. By Kirchhoff's Voltage Law, the voltage across R3R_3 is V3=96 V−(14.4 V+21.6 V+12.0 V)=48.0 VV_3 = 96\text{ V} - (14.4\text{ V} + 21.6\text{ V} + 12.0\text{ V}) = 48.0\text{ V}. Applying Ohm's Law directly to R3R_3 yields R3=V3I=48.0 V0.8 A=60 ΩR_3 = \frac{V_3}{I} = \frac{48.0\text{ V}}{0.8\text{ A}} = 60\ \Omega.
Tahmini Süre:1m 30s
Soru 60Soru →

An electronics technician is analyzing a circuit where a heavy-duty relay coil is controlled by a mechanical switch connected to a 24V DC power supply. Every time the switch is opened to turn off the relay, a visible spark arcs across the switch contacts, eventually causing them to fail. What electromagnetic principle primarily causes this arcing, and how does the coil's magnetic field behave to produce it?

Cevabı ve açıklamayı göster

Cevap: Self-induction; the sudden interruption of current causes the coil's magnetic field to collapse rapidly, inducing a high-voltage counter-EMF that bridges the switch gap.

Cevap

Self-induction; the sudden interruption of current causes the coil's magnetic field to collapse rapidly, inducing a high-voltage counter-EMF that bridges the switch gap.
The phenomenon described is known as inductive kickback or back-EMF, which is a direct result of self-induction. An inductor (like a relay coil) opposes any change in current. When the switch is opened, the current attempts to drop to zero instantly. The magnetic field surrounding the coil collapses rapidly. According to Faraday's Law, this rapid change in magnetic flux induces a very large voltage spike across the coil. This spike is often much higher than the original 24V supply and is strong enough to ionize the air in the small gap of the opening switch, creating an electrical arc.

Adım Adım Çözüm

1
Analyze the components and their state before the switch opens.
The relay coil acts as an inductor. When powered by 24V DC, a steady current flows, and energy is stored in a stable magnetic field around the coil.
Understanding the initial steady-state condition is required to determine what changes.
2
Determine the effect of opening the switch.
Opening the switch forces the current from the power supply to drop toward zero almost instantly.
Inductors oppose sudden changes in current flow.
3
Apply Faraday's Law of Induction to the changing field.
The rapid decrease in current causes the magnetic field to collapse at high speed. This high rate of change induces a very large voltage (counter-EMF) across the coil.
Induced voltage is directly proportional to the rate of change of the magnetic flux.
4
Relate the induced voltage to the physical arcing.
The induced high voltage (often hundreds of volts) easily exceeds the breakdown voltage of the tiny air gap between the separating switch contacts, causing the air to ionize and a spark to jump.
This explains the visible symptom (the spark) and the mechanism of failure.

Anahtar Kavram

Inductive kickback (flyback voltage) caused by self-induction
Tahmini Süre:1m 30s
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