All practice questions

Question 841Question →

In an electromechanical relay, the normally closed (NC) contacts maintain electrical continuity and allow current to flow through the load circuit when the control coil is de-energized.

True

False

Show answer & explanation

Answer: True

Answer

True

Relay contact states are defined relative to the de-energized (unpowered) coil condition. Normally Closed (NC) contacts are held shut by a mechanical return spring when the coil has no current, thereby maintaining a complete circuit and allowing current to pass.

Step-by-Step Solution

1

Identify the operating definition of 'normal' in relay terminology.

The normal state refers to the resting condition of the relay when no voltage or current is applied to the electromagnetic coil.

All contact designations (Normally Open and Normally Closed) are standardized based on the unpowered coil state.

2

Determine the physical contact position of Normally Closed (NC) contacts in the unpowered state.

The internal return spring holds the movable armature firmly against the NC contact terminal, forming a continuous electrical path.

Mechanical spring tension keeps the circuit closed until magnetic force from an energized coil pulls the armature away.

3

Evaluate circuit continuity and current flow capability.

Because the physical contacts touch, electrical continuity exists and current can flow through the load circuit.

Closed contacts provide a low-resistance path allowing electrical current to circulate.

Key Concept

Normally Closed (NC) vs. Normally Open (NO) Relay Contact States

Estimated Time:45s

Question 842Question →

A direct current (DC) cooling fan in an electronics enclosure operates at a voltage of $24\text{ V}$ and draws a current of $1.5\text{ A}$ . What is the total electrical power consumed by the fan?

16\text{ W}

25.5\text{ W}

36\text{ W}

54\text{ W}

Show answer & explanation

Answer: $36\text{ W}$

Answer

36\text{ W}

According to Watt's Law, electrical power in a DC circuit is determined by multiplying the voltage by the current (

P = V \times I

). Multiplying

24\text{ V}

by

1.5\text{ A}

results in

36\text{ W}

of power dissipation.

Step-by-Step Solution

1

Identify the given electrical quantities and the appropriate Watt's Law formula.

Voltage

V = 24\text{ V}

, Current

I = 1.5\text{ A}

, Formula:

P = V \times I

Electrical power in a DC circuit is directly defined as the product of electromotive force (voltage) and current.

2

Substitute the values into the formula and calculate the power.

P = 24\text{ V} \times 1.5\text{ A} = 36\text{ W}

Multiplying

24

by

1.5

gives

36

, with the standard unit of power being watts (W).

Key Concept

Watt's Law calculates electrical power (

P

) as the product of voltage (

V

) and current (

I

), expressed as

P = V \times I

.

Estimated Time:45s

Question 843Question →

When comparing two electrical wires of identical length, diameter, and temperature, which material property determines which wire has lower electrical resistance?

Applied voltage

Resistivity

Dielectric strength

Inductance

Show answer & explanation

Answer: Resistivity

Answer

Resistivity is the material property that determines the electrical resistance of equal-dimensioned conductors.

Resistivity (represented by

\rho

) is an intrinsic physical property of a material that measures its natural opposition to electric current flow. According to the formula

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

, when two conductors share identical length (

L

) and cross-sectional area (

A

) at constant temperature, the wire made of a material with lower resistivity will have lower total electrical resistance.

Step-by-Step Solution

1

Apply the fundamental formula for conductor resistance:

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

.

Total resistance (

R

) is directly proportional to conductor length (

L

), inversely proportional to cross-sectional area (

A

), and directly proportional to resistivity (

\rho

).

Because length, diameter (area), and temperature are held constant between the two wires, any difference in total resistance is strictly governed by the material's resistivity.

2

Identify the property representing the material's specific opposition to current.

Resistivity (

\rho

) is the intrinsic characteristic of the substance.

A material with lower resistivity (such as copper compared to iron) will always yield a lower resistance when shaped into identical physical dimensions.

Key Concept

Resistance and Resistivity

Estimated Time:45s

Question 844Question →

A standard circuit breaker trips and interrupts power to an appliance branch circuit due to an electrical overload. After the fault condition has been resolved, what action is required to restore power to the circuit?

Replace the unit with a new protective device because the internal conductor has permanently melted.

Increase the source voltage to force the internal sensing mechanism to disengage.

Manually reset the breaker operating handle back to the closed (ON) position.

Rewire the branch circuit so the protective device is connected in parallel with the load.

Show answer & explanation

Answer: Manually reset the breaker operating handle back to the closed (ON) position.

Answer

Manually reset the breaker operating handle back to the closed (ON) position.

Circuit breakers are reusable protective devices engineered with bimetallic strips or electromagnetic coils that open electrical contacts during overcurrent events. Unlike fuses that melt and must be discarded, a circuit breaker simply trips its mechanical latch and can be reset manually to the ON position once the electrical fault is cleared.

Step-by-Step Solution

1

Identify the type of protective device specified in the scenario.

The device is a circuit breaker, which is a reusable electromechanical switch.

Differentiating between sacrificial devices (fuses) and resettable devices (circuit breakers) determines the correct maintenance action.

2

Determine the procedure for restoring an open circuit breaker after a trip.

The internal trip mechanism mechanically separates the contacts, which can be re-engaged by toggling the external handle.

Circuit breakers do not destroy their internal elements during normal tripping and do not need to be replaced.

Key Concept

Circuit Breaker Operation and Resetting

Estimated Time:45s

Question 845Question →

Match each circuit protective device on the left with its primary operating characteristic or application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Fast-acting fuse pairs with opening rapidly to protect sensitive semiconductor electronics; Slow-blow fuse pairs with tolerating brief inrush currents from inductive motors; Magnetic circuit breaker pairs with using an electromagnetic coil to trip during heavy short circuits; Metal-oxide varistor pairs with clamping transient high-voltage surges.

Each protective device is engineered for a specific fault type or application: fast-acting fuses react immediately to protect delicate electronics; slow-blow fuses accommodate brief inductive surges; magnetic breakers use electromagnetism for instant short-circuit tripping; and metal-oxide varistors suppress high-voltage spikes.

Step-by-Step Solution

1

Identify the characteristic of a fast-acting fuse

Matches with immediate element melting to protect delicate solid-state electronics from sudden overcurrent.

Semiconductors cannot withstand overcurrent for even brief durations.

2

Identify the characteristic of a slow-blow fuse

Matches with tolerating brief inrush currents during electric motor startup.

Inductive loads draw high starting current that would blow standard fuses prematurely.

3

Identify the characteristic of a magnetic circuit breaker

Matches with utilizing an electromagnetic coil to trip instantaneously during high-current short circuits.

Electromagnetic pull is proportional to current and trips the contact latch mechanism without relying on heat.

4

Identify the characteristic of a metal-oxide varistor (MOV)

Matches with clamping high-voltage transient spikes.

MOVs are voltage-dependent resistors designed specifically for surge suppression rather than direct overcurrent interruption.

Key Concept

Operating principles and applications of circuit protection devices including thermal, magnetic, time-delay, and voltage suppression elements

Estimated Time:45s

Question 846Question →

A circuit is constructed with a $10\text{ V}$ DC power supply connected in series with a $220\ \Omega$ current-limiting resistor. After the resistor, the circuit splits into two parallel branches that eventually reconnect to the negative terminal of the power supply. The first parallel branch contains a Zener diode (rated at $V_Z = 6.2\text{ V}$ ) installed in a reverse-biased orientation. The second parallel branch contains a standard silicon diode installed in a forward-biased orientation. Based on this configuration, what is the approximate voltage measured across the parallel diode combination, and which component dictates this voltage?

0.7\text{ V}

, dictated by the forward-biased standard silicon diode clamping the voltage.

6.2\text{ V}

, dictated by the reverse-biased Zener diode regulating the branch voltage.

6.9\text{ V}

, dictated by the combined additive voltage drops of both diodes in the circuit.

10\text{ V}

, dictated by the power supply because the reverse-biased Zener diode acts as an open circuit.

Show answer & explanation

Answer: $0.7\text{ V}$ , dictated by the forward-biased standard silicon diode clamping the voltage.

Answer

The correct answer is

0.7\text{ V}

, dictated by the forward-biased standard silicon diode clamping the voltage.

The correct answer accurately identifies that in a parallel circuit, the component with the lowest forward voltage drop or breakdown voltage dictates the voltage across the entire parallel block. Since the standard silicon diode is forward-biased, it drops approximately 0.7 V. This prevents the voltage across the parallel branches from ever reaching the 6.2 V required to cause the Zener diode to break down and conduct. The Zener diode remains effectively an open circuit, and the standard diode clamps the branch voltage at 0.7 V.

Step-by-Step Solution

1

Determine the biasing state of each diode in the parallel branches based on the power supply orientation.

The Zener diode is reverse-biased (requires 6.2 V to reach breakdown), and the standard silicon diode is forward-biased (requires ~0.7 V to conduct).

A diode's behavior depends fundamentally on whether it is forward or reverse-biased.

2

Analyze the voltage rules for parallel circuits.

The voltage across all components connected in parallel must be identical.

This fundamental circuit law means both the Zener diode branch and the standard diode branch must share the exact same voltage drop.

3

Determine the clamping voltage of the parallel combination.

The forward-biased silicon diode begins conducting heavily at 0.7 V, clamping the voltage across both branches at this level.

Because the silicon diode provides a low-resistance path at 0.7 V, the voltage cannot rise high enough to reach the Zener diode's 6.2 V breakdown threshold. The Zener diode remains effectively off (open circuit).

Key Concept

Diode Biasing and Parallel Circuit Voltage Clamping

Estimated Time:1m 30s

Question 847Question →

A power supply circuit utilizes a step-down transformer to convert an alternating current (AC) input of $100\text{ V}$ to an output of $25\text{ V}$ . If the primary winding contains $400\text{ turns}$ , how many turns are on the secondary winding?

1{}600\text{ turns}

100\text{ turns}

25\text{ turns}

10\text{ turns}

Show answer & explanation

Answer: $100\text{ turns}$

Answer

100\text{ turns}

The correct answer is

100\text{ turns}

. According to the transformer turns ratio law, the ratio of secondary voltage to primary voltage equals the ratio of secondary turns to primary turns:

\frac{V_s}{V_p} = \frac{N_s}{N_p}

. Rearranging to solve for the secondary turns yields

N_s = N_p \times \left(\frac{V_s}{V_p}\right) = 400 \times \left(\frac{25}{100}\right) = 100\text{ turns}

. Since this is a step-down transformer (

100\text{ V}

down to

25\text{ V}

, a

4:1

step-down ratio), the secondary coil must have one-fourth the number of turns of the primary coil.

Step-by-Step Solution

1

Identify the given transformer parameters

Primary voltage

V_p = 100\text{ V}

, secondary voltage

V_s = 25\text{ V}

, primary turns

N_p = 400\text{ turns}

Extracting the known values is necessary to set up the turns ratio equation.

2

State the transformer voltage and turns ratio formula

\frac{V_p}{V_s} = \frac{N_p}{N_s} \implies N_s = N_p \times \left(\frac{V_s}{V_p}\right)

In an ideal transformer, the voltage across each winding is directly proportional to the number of turns.

3

Substitute the values and calculate the secondary winding turns

N_s = 400 \times \left(\frac{25\text{ V}}{100\text{ V}}\right) = 400 \times 0.25 = 100\text{ turns}

Performing the proportional calculation gives the exact number of secondary turns required.

Key Concept

Transformer Turns Ratio and Voltage Transformation

Estimated Time:45s

Question 848Question →

A piece of copper hookup wire with a length of $15\text{ m}$ has an electrical resistance of $0.6\ \Omega$ . If the wire is replaced with a $45\text{ m}$ piece of the exact same wire type and gauge at the same temperature, what is the resistance of the new wire in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 1.8

Answer

1.8 \Omega

According to the fundamental resistance relationship

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

, resistance is directly proportional to the length

L

of the conductor when resistivity

\rho

and cross-sectional area

A

remain unchanged. Tripling the wire length from

15\text{ m}

to

45\text{ m}

multiplies the resistance by a factor of 3, resulting in

0.6\ \Omega \times 3 = 1.8\ \Omega

.

Step-by-Step Solution

1

Calculate the proportional change in conductor length.

The new length is 3 times the original length (

45\text{ m} \div 15\text{ m} = 3

).

Identifying the scaling factor allows direct application of the resistance formula.

2

Apply the direct proportionality between conductor length and electrical resistance (

R \propto L

).

R_{\text{new}} = R_{\text{old}} \times 3 = 0.6\ \Omega \times 3 = 1.8\ \Omega

.

Since material resistivity (

\rho

) and cross-sectional area (

A

) are held constant, tripling the length triples the total resistance.

Key Concept

The electrical resistance of a uniform conductor is directly proportional to its length (

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

).

Estimated Time:45s

Question 849Question →

In an external circuit connected to a direct-current (DC) power source, electron flow theory states that negative charge carriers travel from the positive terminal to the negative terminal.

True

False

Show answer & explanation

Answer: False

Answer

False

Electron flow theory describes the physical movement of negatively charged electrons, which travel from the negative terminal through the external circuit load toward the positive terminal. Traveling from positive to negative describes conventional current flow.

Step-by-Step Solution

1

Identify the electrical current model referenced in the statement.

The statement explicitly cites 'electron flow theory' in an external circuit.

Electronics distinguishes between electron flow theory and conventional current flow theory.

2

Determine the physical direction of electrons based on fundamental electrostatic laws.

Electrons possess a negative charge, causing them to be repelled by the negative terminal and drawn toward the positive terminal.

Like charges repel each other while opposite charges attract.

3

Evaluate the accuracy of the statement's claimed direction.

Movement from positive to negative defines conventional current, making the claim regarding electron flow theory false.

Electron flow moves from negative to positive externally.

Key Concept

Direction of Electron Flow vs. Conventional Current

Estimated Time:35s

Question 850Question →

An electronics technician is replacing components on a circuit board and selects an electrolytic capacitor for the DC power supply filtering stage. Why must the technician observe strict polarity when installing an electrolytic capacitor into a DC circuit?

They store electrical energy in a magnetic field that collapses and damages surrounding components if polarity is reversed.

They possess an electrochemical insulating oxide layer that breaks down and can cause the component to overheat or rupture under reverse voltage.

They are constructed to allow direct current to flow continuously while blocking all alternating current signals.

They operate through direct electrical conduction between the two internal plates without an insulating dielectric barrier.

Show answer & explanation

Answer: They possess an electrochemical insulating oxide layer that breaks down and can cause the component to overheat or rupture under reverse voltage.

Answer

Electrolytic capacitors have an electrochemical oxide dielectric layer that breaks down and causes the component to overheat or rupture if connected with reversed polarity.

Electrolytic capacitors are polarized components because their extremely thin dielectric layer (aluminum or tantalum oxide) is formed and maintained electrochemically in one specific voltage direction. If connected in reverse polarity, this insulating layer breaks down rapidly, allowing heavy leakage current to generate boiling electrolyte gases, resulting in device failure, venting, or explosion.

Step-by-Step Solution

1

Identify the internal construction of electrolytic capacitors.

Electrolytic capacitors utilize an extremely thin dielectric film of aluminum oxide or tantalum oxide maintained by a chemical electrolyte.

Understanding the unique dielectric mechanism explains why electrolytic capacitors provide high capacitance in a compact size but possess polarity restrictions.

2

Analyze the electrical behavior of the dielectric layer under reverse polarity.

When reversed voltage is applied, the electrochemical reaction breaks down and de-forms the insulating oxide layer.

The formation and insulation of the oxide layer are unidirectional.

3

Evaluate the practical consequence of reverse installation in a circuit.

With the dielectric compromised, large leakage currents flow through the electrolyte, producing excessive heat, gas buildup, and catastrophic rupture or short circuit.

This necessitates strict observance of the positive and negative terminal markings during assembly and maintenance.

Key Concept

Electrolytic Capacitor Polarity and Dielectric Integrity

Estimated Time:1m 0s

Question 851Question →

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 $12\text{ V}$ 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?

The meter registers a steady, continuous deflection during steady current, then drops smoothly to zero without deflecting when the switch is opened.

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.

The meter reads a constant voltage proportional to the supply voltage at all times because the iron core directly transfers electrical potential between coils.

The meter reads zero during steady current and remains at zero when the switch opens because the magnetic field stores charge like an electrostatic capacitor.

Show answer & explanation

Answer: 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.

Answer

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 (

V_{\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.

Step-by-Step Solution

1

Analyze the electromagnetic state during steady-state DC current flow.

Magnetic flux is constant (

\frac{\Delta \Phi}{\Delta t} = 0

), resulting in

0\text{ V}

induced in the secondary coil.

According to Faraday's Law of Electromagnetic Induction (

V_{\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 (

\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.

Key Concept

Faraday's Law of Electromagnetic Induction and transient flux changes

Estimated Time:1m 30s

Question 852Question →

A technician is troubleshooting a test circuit connected to a $32\text{V}$ DC power supply. The circuit consists of a fixed resistor $R_1$ in series with a parallel block. The parallel block contains a $12\text{ k}\Omega$ fixed resistor and a variable resistor (rheostat) $R_v$ . The resistor $R_1$ has four color bands: Green, Blue, Red, and Gold. The technician adjusts the rheostat $R_v$ until the total current drawn from the power supply is exactly $4\text{ mA}$ .

What is the resistance setting of the rheostat $R_v$ in kilohms (k $\Omega$ )?

Show answer & explanation

Answer: 3

Answer

3

The total resistance required by the circuit to draw

4\text{ mA}

from a

32\text{V}

source is

8\text{ k}\Omega

. The series resistor

R_1

(Green, Blue, Red) contributes

5.6\text{ k}\Omega

. This leaves exactly

2.4\text{ k}\Omega

required for the parallel section. For a

12\text{ k}\Omega

resistor in parallel with an unknown

R_v

to yield an equivalent resistance of

2.4\text{ k}\Omega

,

R_v

must be set to

3\text{ k}\Omega

.

Step-by-Step Solution

1

Decode the color code of resistor

R_1

.

R_1 = 5.6\text{ k}\Omega

The color bands are Green (5), Blue (6), and Red (multiplier of

10^2

or 100). Thus,

56 \times 100 = 5,600\ \Omega

, which is

5.6\text{ k}\Omega

. The Gold band indicates a 5% tolerance, which applies to real-world variance but does not alter our nominal calculation.

2

Calculate the total equivalent resistance (

R_{eq}

) of the circuit.

R_{eq} = 8\text{ k}\Omega

Using Ohm's Law (

R = \frac{V}{I}

), divide the source voltage (

32\text{V}

) by the total current (

4\text{ mA}

or

0.004\text{ A}

) to find the total resistance required to draw that current.

3

Determine the required equivalent resistance of the parallel block.

R_{parallel} = 2.4\text{ k}\Omega

Since

R_1

is in series with the parallel block, subtract

R_1

from the total circuit resistance to find the resistance of the parallel segment:

8\text{ k}\Omega - 5.6\text{ k}\Omega = 2.4\text{ k}\Omega

.

4

Solve for the rheostat's resistance (

R_v

).

R_v = 3\text{ k}\Omega

The parallel block contains a

12\text{ k}\Omega

resistor and

R_v

. Using the parallel resistance formula

\frac{1}{R_p} = \frac{1}{R_a} + \frac{1}{R_b}

, we get

\frac{1}{2.4} = \frac{1}{12} + \frac{1}{R_v}

. Rearranging gives

\frac{1}{R_v} = \frac{1}{2.4} - \frac{1}{12} = \frac{10}{24} - \frac{2}{24} = \frac{8}{24} = \frac{1}{3}

, meaning

R_v = 3\text{ k}\Omega

.

Key Concept

Integrating Ohm's law with series-parallel equivalent resistance and resistor color code decoding.

Estimated Time:3m 0s

Question 853Question →

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.

Fill in the blanks below

A

50\text{ V}

DC source is connected in series with a

15\ \Omega

resistor and a second resistor

R_2

. If the circuit draws a current of

2\text{ A}

, the total circuit resistance is

\Omega

, and the resistance of

R_2

is

\Omega

.

Show answer & explanation

Answer

The total circuit resistance is

25\ \Omega

, and the resistance of

R_2

is

10\ \Omega

.

Applying Ohm's Law to the total circuit yields a total resistance of

R_T = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

. In a series circuit, total resistance equals the sum of the individual component resistances (

R_T = R_1 + R_2

). Subtracting the known resistor value (

15\ \Omega

) from the total resistance (

25\ \Omega

) results in

R_2 = 10\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law with source voltage and total current.

R_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

R_2

using the series circuit resistance addition formula (

R_T = R_1 + R_2

).

R_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.

Key Concept

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

R_T = R_1 + R_2 + \dots

), and the circuit current is constant through all components as governed by Ohm's Law (

R_T = V_T / I

).

Estimated Time:1m 0s

Question 854Question →

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

True

False

Show answer & explanation

Answer: False

Answer

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.

Step-by-Step Solution

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 (

\Delta \Phi / \Delta t

). A steady DC source produces a constant magnetic field (

\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).

Key Concept

Electromagnetic induction in transformers requires a time-varying magnetic flux, meaning a steady direct current (DC) cannot maintain continuous secondary induction.

Estimated Time:45s

Question 855Question →

Match each resistor configuration or color-coded component on the left with its corresponding electrical characteristic and operating principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The three-terminal potentiometer matches the voltage divider function; the rheostat matches the series current limiter with wiper-to-end protection; the Yellow-Violet-Orange-Gold resistor matches

47\text{ k}\Omega \pm 5\%

; and the Brown-Gray-Green-Silver resistor matches

1.8\text{ M}\Omega \pm 10\%

.

Each item correctly links the component physical description or color code to its precise electrical definition: potentiometers provide variable potential division, rheostats adjust circuit current, Yellow-Violet-Orange-Gold evaluates to

47\text{ k}\Omega \pm 5\%

, and Brown-Gray-Green-Silver evaluates to

1.8\text{ M}\Omega \pm 10\%

.

Step-by-Step Solution

1

Analyze the three-terminal potentiometer and rheostat wiring configurations

A potentiometer connected across a power source acts as a continuous voltage divider (

0\text{ V}

to

V_{\text{in}}

) at the wiper terminal. A rheostat uses a variable two-terminal configuration in series with the load to regulate current.

Variable resistors differ in circuit function based on whether they are wired as three-terminal potential dividers or two-terminal current adjusters.

2

Decode the four-band color code for Yellow-Violet-Orange-Gold

Yellow = 4, Violet = 7, Orange = multiplier

10^3 = 1,000

, Gold =

\pm 5\%

. Value

= 47 \times 1,000\ \Omega = 47\text{ k}\Omega \pm 5\%

.

Standard EIA 4-band color coding assigns significant digits to the first two bands, a decimal multiplier to the third band, and tolerance percentage to the fourth band.

3

Decode the four-band color code for Brown-Gray-Green-Silver

Brown = 1, Gray = 8, Green = multiplier

10^5 = 100,000

, Silver =

\pm 10\%

. Value

= 18 \times 100,000\ \Omega = 1.8\text{ M}\Omega \pm 10\%

.

Standard EIA 4-band color coding converts

1,800,000\ \Omega

using standard metric prefixes to

1.8\text{ M}\Omega

with a silver tolerance of

\pm 10\%

.

Key Concept

Operating principles and terminal configurations of fixed and variable resistors (potentiometers, rheostats, and standard color codes)

Alternative Method

To quickly decode 4-band resistors, write down the two digits (e.g., 47 for Yellow-Violet or 18 for Brown-Gray), append the number of zeros indicated by the multiplier color (3 zeros for Orange = 47,000, 5 zeros for Green = 1,800,000), and simplify to metric units (kilo or mega).

Estimated Time:1m 0s

Question 856Question →

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.

True

False

Show answer & explanation

Answer: True

Answer

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.

Step-by-Step Solution

1

Identify the factors that determine magnetic flux density in a solenoid

Magnetic flux density is governed by the relation

B = \mu \frac{N}{L} I

, where

\mu

is core permeability,

N

is number of turns,

L

is coil length, and

I

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.

Key Concept

Effect of ferromagnetic core permeability on electromagnetic field strength

Estimated Time:45s

Question 857Question →

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

25\ \Omega

12.5\ \Omega

4\ \Omega

100\ \Omega

Show answer & explanation

Answer: $4\ \Omega$

Answer

4\ \Omega

In a parallel circuit with two branches, the equivalent resistance is calculated using the product-over-sum formula

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}

. Substituting

20\ \Omega

and

5\ \Omega

gives

\frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega

. Total resistance in parallel is always less than the lowest individual resistor value.

Step-by-Step Solution

1

Identify the formula for the equivalent resistance of two parallel resistors.

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2}

or

\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 (

R_1 = 20\ \Omega

and

R_2 = 5\ \Omega

) into the product-over-sum formula.

R_{\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.

R_{\text{eq}} = 4\ \Omega

Computing

\frac{100}{25}

yields the final equivalent resistance.

Key Concept

Total Equivalent Resistance in Parallel Circuits

Estimated Time:45s

Question 858Question →

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.

True

False

Show answer & explanation

Answer: True

Answer

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.

Step-by-Step Solution

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 = -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 (

\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.

Key Concept

Lenz's Law and Self-Induction in Inductors

Estimated Time:45s

Question 859Question →

A direct current (DC) circuit consists of a $48\text{ V}$ power supply connected in series with three resistors: $R_1 = 12\ \Omega$ , $R_2 = 24\ \Omega$ , and $R_3 = 24\ \Omega$ . In its baseline operating state, the total circuit resistance is $R_T = 60\ \Omega$ and the circuit loop current is $I = 0.8\text{ A}$ .

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair: (1) shorting resistor

R_3

with total resistance dropping to

36\ \Omega

, current rising to

1.33\text{ A}

, and

0\text{ V}

across

R_3

; (2) an open at

R_2

with loop current dropping to

0\text{ A}

and

48\text{ V}

appearing across

R_2

; (3) adding a

20\ \Omega

series resistor with total resistance rising to

80\ \Omega

, current dropping to

0.6\text{ A}

, and

7.2\text{ V}

across

R_1

; and (4) raising the source to

72\text{ V}

with total resistance remaining

60\ \Omega

, current rising to

1.2\text{ A}

, and

28.8\text{ V}

across

R_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 (

R_T = R_1 + R_2 + \dots + R_n

), and current is constant throughout the single loop (

I_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.

Step-by-Step Solution

1

Analyze the short circuit condition across

R_3

R_T = 12 + 24 = 36\ \Omega

;

I = \frac{48}{36} \approx 1.33\text{ A}

;

V_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

R_2

I = 0\text{ A}

;

V_1 = 0\text{ V}

;

V_3 = 0\text{ V}

;

V_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 (

R_4 = 20\ \Omega

)

R_T = 60 + 20 = 80\ \Omega

;

I = \frac{48}{80} = 0.6\text{ A}

;

V_1 = 0.6 \times 12 = 7.2\text{ V}

In series circuits, equivalent resistance is additive (

R_T = \sum R

). Increasing total resistance reduces total current according to Ohm's Law (

I = \frac{V}{R_T}

), which proportionally reduces individual component voltage drops.

4

Analyze the voltage source increase to

72\text{ V}

R_T = 60\ \Omega

;

I = \frac{72}{60} = 1.2\text{ A}

;

V_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.

Key Concept

Series Circuit Governing Rules: Additive Resistance, Current Continuity, Ohm's Law, and Fault Dynamics

Estimated Time:2m 0s

Question 860Question →

A marine communications transceiver utilizes a direct current (DC) bias network. A voltmeter connected across a fixed $2.5\text{ k}\Omega$ resistor measures a steady voltage drop of $15\text{ V}$ . What is the current flowing through this resistor?

37.5\text{ mA}

167\text{ mA}

6\text{ mA}

0.6\text{ mA}

Show answer & explanation

Answer: $6\text{ mA}$

Answer

The current flowing through the resistor is

6\text{ mA}

.

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

I = V / R

). Dividing

15\text{ V}

by

2.5\text{ k}\Omega

(

2500\text{ }\Omega

) yields

0.006\text{ A}

, which converts directly to

6\text{ mA}

.

Step-by-Step Solution

1

Convert the given resistance from kilo-ohms to fundamental SI units (ohms).

R = 2.5\text{ k}\Omega = 2.5 \times 10^3\text{ }\Omega = 2500\text{ }\Omega

Standard calculations in Ohm's law require values in fundamental units (volts, amperes, ohms).

2

Apply Ohm's law formula to solve for current (

I = V / R

).

I = \frac{15\text{ V}}{2500\text{ }\Omega} = 0.006\text{ A}

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

3

Convert the resulting current from amperes to milliamperes.

0.006\text{ A} = 0.006 \times 1000\text{ mA} = 6\text{ mA}

Matching the standard unit formatting in electronic measurement options.

Key Concept

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

I = V / R

).

Estimated Time:1m 0s

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