All practice questions

Question 781Question →

A direct current (DC) heating element is connected to a $120\text{ V}$ power supply. If an ammeter measures a current of $4\text{ A}$ flowing through the element, what is its electrical resistance?

480\text{ }\Omega

30\text{ }\Omega

0.033\text{ }\Omega

116\text{ }\Omega

Show answer & explanation

Answer: $30\text{ }\Omega$

Answer

30\text{ }\Omega

According to Ohm's law, electrical resistance (

R

) is defined as the ratio of voltage (

V

) across a component to the current (

I

) passing through it:

R = \frac{V}{I}

. Substituting

120\text{ V}

and

4\text{ A}

gives

R = \frac{120}{4} = 30\text{ }\Omega

.

Step-by-Step Solution

1

Identify the given values and the required quantity

Voltage

V = 120\text{ V}

, Current

I = 4\text{ A}

, Resistance

R

is unknown

Establishing the known circuit parameters determines which Ohm's law formula variation to use.

2

Apply Ohm's law rearranged to solve for resistance:

R = \frac{V}{I}

R = \frac{120\text{ V}}{4\text{ A}} = 30\text{ }\Omega

Resistance represents the opposition to current flow and is calculated as the ratio of electromotive force to electrical current.

Key Concept

Ohm's Law Resistance Calculation (

R = \frac{V}{I}

)

Estimated Time:45s

Question 782Question →

A technician installs an inductor (choke) in series with the output line of a direct current (DC) power supply to filter out residual alternating current (AC) ripple. What is the primary effect of the inductor on this electrical current?

It completely blocks steady direct current while allowing alternating current ripple to pass freely.

It steps up the fluctuating AC voltage into a higher voltage DC output through mutual induction.

It opposes rapid changes in current, resisting the AC ripple while allowing steady DC to pass with minimal resistance.

It dissipates both AC ripple and steady DC current equally as thermal heat by acting as a pure resistor.

Show answer & explanation

Answer: It opposes rapid changes in current, resisting the AC ripple while allowing steady DC to pass with minimal resistance.

Answer

The inductor opposes rapid current variations (the AC ripple) due to inductive reactance while allowing steady direct current (DC) to pass freely.

An inductor resists changes in current through self-induction. When placed in series with a DC power supply line, its inductive reactance (

X_L = 2\pi f L

) creates significant opposition to the fluctuating AC ripple component while providing near-zero opposition to the constant DC current.

Step-by-Step Solution

1

Identify the fundamental operating principle of an inductor.

Inductors store energy in a magnetic field and inherently oppose any change in electrical current (

V_L = L \frac{di}{dt}

).

According to Lenz's Law, when current changes through a coil, a self-induced counter-electromotive force (CEMF) is produced that resists that change.

2

Analyze the inductor's frequency-dependent opposition (inductive reactance,

X_L

).

For steady DC (

f = 0\text{ Hz}

),

X_L = 2\pi f L = 0\,\Omega

, so DC passes with negligible opposition. For AC ripple (

f > 0\text{ Hz}

),

X_L > 0\,\Omega

, which impedes the fluctuation.

Higher frequency signals encounter significant inductive reactance, attenuating the unwanted AC ripple while letting constant DC through to the load.

Key Concept

Inductors oppose changes in alternating current via inductive reactance while allowing steady direct current to flow with minimal opposition.

Estimated Time:45s

Question 783Question →

A solid copper wire used in an industrial application has a length of $200\text{ meters}$ and a cross-sectional area of $0.85 \, mm^2$ . The resistivity of copper at $20^\circ C$ is $1.7 \times 10^{-8} \, \Omega\cdot m$ , and its temperature coefficient of resistance is $0.004 \text{ per } ^\circ C$ . What is the resistance of this wire, in ohms, when its operating temperature reaches $70^\circ C$ ?

Show answer & explanation

Answer: 4.8

Answer

4.8

The solution requires a synthesis of two core resistance principles. First, the baseline resistance is found by dividing the length by the area (converted to square meters) and multiplying by the resistivity, yielding 4.0 ohms. Second, the 50°C temperature increase is multiplied by the coefficient (0.004) to find a 20% increase in total resistance, raising the final value to 4.8 ohms.

Step-by-Step Solution

1

Convert the cross-sectional area from

mm^2

to

m^2

.

0.85 \, mm^2 = 0.85 \times 10^{-6} \, m^2

Standard SI units must be used to cancel with the meters in the resistivity value.

2

Calculate the resistance at

20^\circ C

(

R_{20}

) using Pouillet's law (

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

).

R_{20} = (1.7 \times 10^{-8}) \times \frac{200}{0.85 \times 10^{-6}} = 4.0 \, \Omega

This establishes the baseline resistance of the wire before temperature effects are considered.

3

Calculate the change in temperature (

\Delta T

).

\Delta T = 70^\circ C - 20^\circ C = 50^\circ C

The temperature coefficient is applied to the difference between the operating temperature and the reference temperature.

4

Calculate the new resistance at

70^\circ C

using the temperature coefficient formula:

R_{new} = R_{ref}[1 + \alpha(\Delta T)]

.

R_{70} = 4.0 \times [1 + 0.004 \times 50] = 4.0 \times [1 + 0.2] = 4.0 \times 1.2 = 4.8 \, \Omega

As temperature increases, the resistance of a standard conductor like copper increases proportionally.

Key Concept

Calculating absolute resistance using dimensions and resistivity, followed by applying temperature coefficients.

Estimated Time:3m 0s

Question 784Question →

Match each direct current (DC) circuit scenario on the left with its corresponding calculated electrical quantity on the right using Ohm's law ( $V = I \times R$ ).

Click a left item, then click its matching right item

Show answer & explanation

Answer

The

120\text{ V}

load with

2.4\text{ k}\Omega

matches the resulting current of

50\text{ mA}

; the

5\text{ V}

circuit drawing

20\text{ mA}

matches the circuit resistance of

250\text{ }\Omega

; the

400\text{ mA}

current through

50\text{ }\Omega

matches the voltage drop of

20\text{ V}

; and the

36\text{ V}

loop carrying

3\text{ mA}

matches the loop resistance of

12\text{ k}\Omega

.

Each circuit condition correctly maps to its target electrical parameter by converting metric prefixes (

1\text{ k}\Omega = 1000\text{ }\Omega

,

1\text{ mA} = 0.001\text{ A}

) and applying the appropriate algebraic form of Ohm's law:

I = \frac{V}{R}

,

R = \frac{V}{I}

, and

V = I \times R

.

Step-by-Step Solution

1

Calculate current for the

120\text{ V}

,

2.4\text{ k}\Omega

circuit

I = \frac{120\text{ V}}{2400\text{ }\Omega} = 0.05\text{ A} = 50\text{ mA}

Ohm's law states

I = \frac{V}{R}

, requiring resistance in fundamental ohms (

1\text{ k}\Omega = 1000\text{ }\Omega

).

2

Calculate resistance for the

5\text{ V}

,

20\text{ mA}

circuit branch

R = \frac{5\text{ V}}{0.02\text{ A}} = 250\text{ }\Omega

Rearranging Ohm's law gives

R = \frac{V}{I}

, requiring current in amperes (

20\text{ mA} = 0.02\text{ A}

).

3

Calculate voltage drop across the

50\text{ }\Omega

element carrying

400\text{ mA}

V = 0.4\text{ A} \times 50\text{ }\Omega = 20\text{ V}

Ohm's law formula

V = I \times R

computes potential difference when current and resistance are known.

4

Calculate total resistance for the

36\text{ V}

,

3\text{ mA}

sensor loop

R = \frac{36\text{ V}}{0.003\text{ A}} = 12000\text{ }\Omega = 12\text{ k}\Omega

Applying

R = \frac{V}{I}

with

3\text{ mA} = 0.003\text{ A}

yields

12000\text{ }\Omega

, or

12\text{ k}\Omega

.

Key Concept

Ohm's law mathematically relates voltage (

V

), current (

I

), and resistance (

R

) via

V = I \times R

,

I = \frac{V}{R}

, and

R = \frac{V}{I}

, requiring consistent base metric units.

Estimated Time:1m 15s

Question 785Question →

In standard electronic schematic symbols for bipolar junction transistors (BJTs), the arrow is located on the emitter terminal and points outward away from the base for an NPN transistor, indicating the direction of conventional current flow.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because electronic schematic standards place the directional arrow exclusively on the emitter lead. For an NPN transistor, this arrow points outward (away from the base), which reflects the direction of conventional current flow across the forward-biased base-emitter junction.

Step-by-Step Solution

1

Identify the specific terminal that carries the directional arrow on a BJT schematic symbol.

The arrow is always placed on the emitter terminal, distinguishing it from the collector and base leads.

Standard electronic schematic conventions designate the emitter lead with an arrow to define transistor polarity and terminal identity.

2

Determine the orientation of the arrow for an NPN transistor configuration.

In an NPN transistor, the arrow points outward, away from the base.

The arrow represents the direction of forward-biased conventional current flow, moving from the P-type base material out through the N-type emitter material.

3

Evaluate the accuracy of the full statement.

The statement correctly identifies both the terminal location (emitter) and the outward direction for an NPN transistor.

Both structural elements conform exactly to standard electrical schematic rules.

Key Concept

BJT schematic symbol identification and terminal arrow direction

Estimated Time:45s

Question 786Question →

In an electromechanical relay, which internal component creates the magnetic field needed to pull the movable armature and actuate the electrical contacts?

A thermal bimetallic strip

An electromagnetic wire coil

A dielectric ceramic barrier

An electrostatic storage plate

Show answer & explanation

Answer: An electromagnetic wire coil

Answer

An electromagnetic wire coil creates the magnetic field that pulls the armature to actuate the contacts in an electromechanical relay.

An electromechanical relay consists of an insulated wire coil wound around a ferromagnetic core. When a control current passes through this coil, it produces an electromagnetic field that pulls the hinged iron armature, physically shifting the contact positions to open or close the circuit.

Step-by-Step Solution

1

Identify the operating mechanism of an electromechanical relay.

An electromechanical relay uses an electrical control signal to produce mechanical motion via magnetic force.

Relays use electromagnetism to allow a low-power circuit to open or close a separate electrical circuit.

2

Determine which component generates the magnetic force within the relay.

An insulated wire wrapped around an iron core forms an electromagnetic coil that produces magnetic flux when energized.

When current flows through the coil, the resulting magnetic attraction pulls the movable iron armature toward the core, changing the position of the contacts.

Key Concept

Electromechanical relay coil operation and magnetic contact actuation

Estimated Time:45s

Question 787Question →

A technician wraps two separate copper wire coils, Coil 1 and Coil 2, around a single soft iron core. Coil 1 is connected to a steady 12V DC power supply, and the current has reached a steady state. Coil 2 is connected only to a sensitive digital voltmeter. If the technician rapidly pulls the soft iron core completely out of the coils while Coil 1 remains connected to the power supply, which of the following best describes the reading on the voltmeter during the exact moment the core is moving?

A voltage reading that spikes momentarily and then returns to zero, because the sudden decrease in magnetic permeability causes a rapid change in magnetic flux.

A continuous reading of zero volts, because a direct current (DC) power supply cannot produce the alternating magnetic field required for electromagnetic induction.

A steady 12V reading, because removing the iron core eliminates the magnetic resistance between the coils, allowing the full electromotive force to transfer.

A permanently higher voltage reading, because the opposing magnetic poles of the iron core are no longer present to attract and restrict the electron flow.

Show answer & explanation

Answer: A voltage reading that spikes momentarily and then returns to zero, because the sudden decrease in magnetic permeability causes a rapid change in magnetic flux.

Answer

The correct answer is a voltage reading that spikes momentarily and then returns to zero, because the sudden decrease in magnetic permeability causes a rapid change in magnetic flux.

Electromagnetic induction requires a changing magnetic flux over time. While a steady DC current normally produces a constant, unchanging magnetic field, the physical act of rapidly removing the highly permeable iron core drastically reduces the magnetic flux. This rapid change in flux induces a temporary voltage spike in the secondary coil. The induction ceases as soon as the core is completely removed, because the flux becomes constant again, returning the voltmeter reading to zero.

Step-by-Step Solution

1

Analyze the initial steady state of the circuit.

Coil 1 produces a constant magnetic flux that is concentrated by the soft iron core. Because the flux is not changing over time, the induced voltage in Coil 2 is zero.

Faraday's Law of Induction requires a changing magnetic field to induce an electromotive force (EMF).

2

Evaluate the physical change caused by the technician's action.

The technician removes the soft iron core, replacing a material with high magnetic permeability with air, which has very low magnetic permeability.

Magnetic permeability directly determines how well a material can support the formation of a magnetic field within itself.

3

Determine the effect of this physical change on the magnetic flux.

The total magnetic flux passing through both coils drops rapidly as the highly permeable core is removed.

Less permeability means weaker magnetic field strength for the same amount of current.

4

Apply Faraday's Law to the changing magnetic flux.

The rapidly decreasing magnetic flux induces a temporary voltage (a spike) in Coil 2.

An EMF is induced in a coil whenever there is a change in magnetic flux passing through it, proportional to the rate of change (

E = -N \times \frac{\Delta\Phi}{\Delta t}

).

5

Determine the final state after the core is completely removed.

The magnetic flux stabilizes at a new, much lower constant value. Since the flux is no longer changing, the induced voltage in Coil 2 returns to zero.

Induction only occurs during the period of change.

Key Concept

Faraday's Law of Induction and Magnetic Permeability

Question 788Question →

A direct current (DC) relay coil with a fixed resistance draws $30\text{ mA}$ when connected across an $18\text{ V}$ power supply. When the same relay coil is connected to a different DC power supply, an ammeter measures a current flow of $75\text{ mA}$ through the coil. What is the voltage of the second power supply?

7.2\text{ V}

45\text{ V}

4.5\text{ V}

75\text{ V}

Show answer & explanation

Answer: $45\text{ V}$

Answer

45\text{ V}

The correct answer is determined by using Ohm's law (

R = \frac{V}{I}

) to find the coil's fixed resistance from the initial state (

R = \frac{18\text{ V}}{0.030\text{ A}} = 600\,\Omega

) and then determining the new voltage (

V = 0.075\text{ A} \times 600\,\Omega = 45\text{ V}

).

Step-by-Step Solution

1

Convert the initial current from milliamperes to amperes and calculate the resistance of the relay coil using Ohm's law (

R = \frac{V}{I}

).

I_1 = 30\text{ mA} = 0.030\text{ A}

;

R = \frac{18\text{ V}}{0.030\text{ A}} = 600\,\Omega

The resistance of the coil remains constant regardless of the applied supply voltage.

2

Convert the second current reading to amperes and calculate the new supply voltage using Ohm's law (

V = I \times R

).

I_2 = 75\text{ mA} = 0.075\text{ A}

;

V_2 = 0.075\text{ A} \times 600\,\Omega = 45\text{ V}

Ohm's law states that the potential difference across a resistor equals current multiplied by resistance.

Key Concept

Direct proportionality between voltage and current across a fixed resistance (

V = I \times R

) and metric unit conversion.

Estimated Time:45s

Question 789Question →

An emergency communications console is powered by a $120\text{ V}$ DC source connected in series with three resistive elements: $R_1 = 15\ \Omega$ , $R_2 = 25\ \Omega$ , and an unknown load resistor $R_3$ . If a voltmeter placed across $R_1$ reads $18\text{ V}$ , what is the resistance value of $R_3$ ?

85\ \Omega

60\ \Omega

100\ \Omega

104\ \Omega

Show answer & explanation

Answer: $60\ \Omega$

Answer

The resistance value of

R_3

is

60\ \Omega

.

In a series circuit, the current is identical at all points. Using Ohm's Law on the first resistor gives a loop current of

I = \frac{18\text{ V}}{15\ \Omega} = 1.2\text{ A}

. The total circuit resistance is

R_T = \frac{V_T}{I} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

. Since series resistances are additive (

R_T = R_1 + R_2 + R_3

), the unknown resistance is

R_3 = 100\ \Omega - (15\ \Omega + 25\ \Omega) = 60\ \Omega

.

Step-by-Step Solution

1

Calculate the common series circuit current using the known voltage drop and resistance of the first resistor.

I = \frac{V_1}{R_1} = \frac{18\text{ V}}{15\ \Omega} = 1.2\text{ A}

In a series circuit, current remains constant through every component.

2

Calculate the total equivalent resistance of the entire circuit using Ohm's Law with the total source voltage.

R_T = \frac{V_T}{I} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

Total resistance equals total supply voltage divided by the total circuit current.

3

Determine the resistance of the unknown resistor by subtracting the known resistors from the total series resistance.

R_3 = R_T - R_1 - R_2 = 100\ \Omega - 15\ \Omega - 25\ \Omega = 60\ \Omega

The total resistance in a series circuit is the linear sum of all individual resistances:

R_T = R_1 + R_2 + R_3

.

Key Concept

Series Circuit Analysis and Ohm's Law Multi-Step Synthesis

Estimated Time:1m 30s

Question 790Question →

An electrical technician is analyzing a perfectly ohmic heating element in a circuit. During initial operation, when an unknown voltage $V$ is applied, the element draws a current of $45 \text{ mA}$ . The technician notes that if the applied voltage is increased by exactly $500 \text{ mV}$ , the current drawn by the element increases by $2.5 \text{ mA}$ . Assuming the element's resistance remains completely constant throughout the test, what was the original applied voltage $V$ ?

9.5 V

9 V

225 V

0.9 V

Show answer & explanation

Answer: 9 V

Answer

9 V

For an ohmic component, resistance (

R

) is constant. This means the change in voltage is directly proportional to the change in current. The resistance can be found using the changes:

R = \Delta V / \Delta I

. After converting to base units,

R = 0.5 \text{ V} / 0.0025 \text{ A} = 200 \, \Omega

. Using Ohm's Law (

V = I \times R

), the original voltage is the original current (

0.045 \text{ A}

) multiplied by this resistance, resulting in

0.045 \text{ A} \times 200 \, \Omega = 9 \text{ V}

.

Step-by-Step Solution

1

Identify the relationship between the change in voltage (ΔV) and the change in current (ΔI).

R = \frac{\Delta V}{\Delta I}

For a perfectly ohmic component, the resistance is constant. This means the slope of the voltage-current relationship defines the resistance.

2

Convert the change values into standard base units (Volts and Amperes) and calculate the resistance.

\Delta V = 500 \text{ mV} = 0.5 \text{ V}

, and

\Delta I = 2.5 \text{ mA} = 0.0025 \text{ A}

.

R = \frac{0.5 \text{ V}}{0.0025 \text{ A}} = 200 \, \Omega

Standardizing metric prefixes prevents decimal shift errors before applying Ohm's Law.

3

Apply Ohm's Law to the original state to find the initial voltage.

V = I_{original} \times R = 0.045 \text{ A} \times 200 \, \Omega = 9 \text{ V}

Multiplying the originally measured current by the established component resistance yields the original voltage applied.

Key Concept

Applying Ohm's Law using proportional changes (ΔV and ΔI) while accurately managing metric prefix conversions.

Question 791Question →

An environmental scientist measures the daily water flow (in gallons per minute) from a natural spring over five days. The recorded values are $22$ , $18$ , $25$ , $20$ , and $17$ . If the scientist needs the average (mean) flow rate across a full six-day period to be exactly $21$ gallons per minute, what must the flow rate be on the sixth day?

18

3

24

20

Show answer & explanation

Answer: $24$

Answer

24

To find the missing value for the sixth day, first determine the total sum required for all six days. Since the target mean is

21

, the required total sum is

21 \times 6 = 126

. Next, find the sum of the five days already recorded:

22 + 18 + 25 + 20 + 17 = 102

. The required flow rate for the sixth day is the difference between these two sums:

126 - 102 = 24

.

Step-by-Step Solution

1

Calculate the sum of the water flow for the first five days.

22 + 18 + 25 + 20 + 17 = 102

This establishes how much total water flow has already been recorded.

2

Calculate the required total sum for the full six days to achieve the target mean.

21 \times 6 = 126

The formula for the mean is Total Sum divided by the Number of Values. Therefore, Total Sum = Mean × Number of Values.

3

Subtract the sum of the first five days from the target total sum.

126 - 102 = 24

The difference between the target six-day total and the current five-day total gives the required value for the sixth day.

Key Concept

Calculating a missing data point in a set when the target arithmetic mean is provided.

Question 792Question →

An avionics technician is assembling a filtering network using three capacitors connected in series. The capacitors have ratings of $30\text{ }\mu\text{F}$ , $20\text{ }\mu\text{F}$ , and $60\text{ }\mu\text{F}$ . What is the total equivalent capacitance of this series network in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 10

Answer

The total equivalent capacitance of the series circuit is

10\text{ }\mu\text{F}

.

For capacitors connected in series, the total capacitance is found using the formula

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

. Substituting

30\text{ }\mu\text{F}

,

20\text{ }\mu\text{F}

, and

60\text{ }\mu\text{F}

gives

\frac{2}{60} + \frac{3}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10\text{ }\mu\text{F}}

. Inverting this fraction yields the correct total capacitance of

10\text{ }\mu\text{F}

, which is less than the smallest individual capacitor (

20\text{ }\mu\text{F}

).

Step-by-Step Solution

1

State the series capacitance relationship

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

Capacitors in series combine inversely because the total dielectric separation increases, reducing total capacitance.

2

Calculate the sum of the reciprocals using a common denominator

\frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}

Finding a least common multiple of

60

allows direct fractional addition.

3

Invert the result to find

C_{\text{total}}

C_{\text{total}} = 10\text{ }\mu\text{F}

The final equivalent capacitance is the inverse of the sum of the reciprocals.

Key Concept

Series capacitance follows the reciprocal sum rule, resulting in a total capacitance smaller than any individual capacitor in the string.

Estimated Time:1m 0s

Question 793Question →

In standard metallic conductors such as copper wire, electric current consists of the directed movement of free valence electrons drifting through the material.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because the physical flow of electric current in metallic conductors like copper is carried exclusively by free valence electrons drifting through the atomic lattice under an applied electric field.

Step-by-Step Solution

1

Identify the charge carriers present in solid metallic conductors.

Metals have an atomic structure characterized by a lattice of fixed positive ions surrounded by loosely held, mobile valence electrons.

Determining which atomic particles are free to move establishes the physical mechanism of conduction.

2

Evaluate the particle movement that constitutes electrical current.

When a voltage source is applied, these free valence electrons drift collectively through the conductor, which defines the physical flow of electric current.

Atomic nuclei remain locked in the crystal lattice while only the free electrons travel through the wire.

Key Concept

Physical electric current in solid metallic conductors is the collective drift of free valence electrons.

Estimated Time:30s

Question 794Question →

Four identical $40\ \Omega$ heating elements are connected in parallel across a $120\text{ V}$ power line. What is the total equivalent resistance of this circuit?

160\ \Omega

10\ \Omega

40\ \Omega

20\ \Omega

Show answer & explanation

Answer: $10\ \Omega$

Answer

The total equivalent resistance of the circuit is

10\ \Omega

.

For

N

identical resistors of equal resistance

R

connected in parallel, the total equivalent resistance is simply the resistance of one branch divided by the total number of branches:

R_{\text{eq}} = \frac{R}{N} = \frac{40\ \Omega}{4} = 10\ \Omega

.

Step-by-Step Solution

1

Identify the number of parallel branches (

N

) and individual branch resistance (

R

).

N = 4

identical branches, each with

R = 40\ \Omega

.

Recognizing identical parallel branches allows the use of the equal-resistor shortcut formula.

2

Apply the parallel equivalent resistance formula for identical resistors:

R_{\text{eq}} = \frac{R}{N}

.

R_{\text{eq}} = \frac{40\ \Omega}{4} = 10\ \Omega

.

Connecting resistors in parallel creates multiple paths for electron flow, reducing the total circuit resistance below that of any individual branch.

Key Concept

Total Equivalent Resistance of Equal Parallel Resistors

Estimated Time:45s

Question 795Question →

A technician connects a primary coil wound around a soft iron core to a constant $12\text{ V}$ direct current (DC) supply, establishing a steady magnetic field through an adjacent secondary coil. Which of the following modifications will produce a continuous, sustained output voltage across the secondary coil?

Increasing the steady DC supply voltage from

12\text{ V}

to

24\text{ V}

while holding the current constant

Replacing the DC supply with an alternating current (AC) source to continuously vary the magnetic flux

Reversing the DC terminal polarity so that steady electron flow moves in the opposite direction

Doubling the number of turns in the secondary winding while maintaining the steady DC input

Show answer & explanation

Answer: Replacing the DC supply with an alternating current (AC) source to continuously vary the magnetic flux

Answer

Replacing the DC supply with an alternating current (AC) source to continuously vary the magnetic flux

According to Faraday's law of induction, an electromotive force (EMF) is only induced when there is relative motion or a changing magnetic flux linking the conductor. An alternating current (AC) continuously cycles in amplitude and direction, generating an expanding and collapsing magnetic field that sustains a continuous output voltage in the secondary coil.

Step-by-Step Solution

1

Apply Faraday's Law of Electromagnetic Induction

The induced electromotive force is given by

\mathcal{E} = -N \frac{d\Phi}{dt}

, where

\frac{d\Phi}{dt}

represents the rate of change of magnetic flux.

Induction strictly depends on a dynamic, changing magnetic field rather than a static magnetic field.

2

Evaluate the magnetic field produced by steady DC

A steady direct current creates a constant magnetic flux (

\frac{d\Phi}{dt} = 0

), producing

0\text{ V}

across the secondary winding in steady state.

Static direct currents cannot sustain continuous transformer action or mutual induction.

3

Determine the condition for continuous, sustained induction

Supplying alternating current (AC) causes the magnetic flux to expand and collapse continuously, providing a perpetual non-zero

\frac{d\Phi}{dt}

.

Continuous induction requires a continuously oscillating or varying current source.

Key Concept

Electromagnetic Induction and Time-Varying Magnetic Flux

Estimated Time:1m 0s

Question 796Question →

A control transformer in an HVAC system steps down an input line voltage of $240\text{ V}$ AC to an output voltage of $24\text{ V}$ AC to safely power a thermostat circuit. If the primary winding consists of $800$ turns of wire, what is the number of turns on the secondary winding?

Show answer & explanation

Answer: 80

Answer

80 turns

The secondary winding requires 80 turns. According to the transformer turns ratio relationship,

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

. Since the secondary voltage (

24\text{ V}

) is one-tenth of the primary input voltage (

240\text{ V}

), the secondary coil must have one-tenth the number of turns of the primary coil:

800 \times 0.10 = 80

turns.

Step-by-Step Solution

1

Identify known electrical parameters from the problem statement.

V_p = 240\text{ V}

,

V_s = 24\text{ V}

, and

N_p = 800\text{ turns}

.

Establishing the input and output parameters is necessary to set up the transformer relationship.

2

Set up the ideal transformer turns ratio equation relating voltage and turns.

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

The voltage induced across transformer windings is directly proportional to the number of turns in each coil.

3

Rearrange the equation to isolate the unknown secondary winding turns (

N_s

).

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

Algebraic isolation allows direct computation of secondary turns.

4

Substitute the known numerical values into the formula and calculate the result.

N_s = 800 \times \left(\frac{24\text{ V}}{240\text{ V}}\right) = 800 \times 0.10 = 80\text{ turns}

Carrying out the arithmetic yields the exact number of secondary turns required.

Key Concept

Transformer Turns Ratio and Voltage Transformation

Alternative Method

Calculate the turns-per-volt ratio of the primary coil:

\frac{800\text{ turns}}{240\text{ V}} = \frac{10}{3}\text{ turns/V}

. Then multiply by the desired secondary voltage:

24\text{ V} \times \frac{10}{3}\text{ turns/V} = 80\text{ turns}

.

Estimated Time:45s

Question 797Question →

A direct current (DC) solenoid valve in an industrial pneumatic system has an internal coil resistance of $15\text{ }\Omega$ . When connected to a $30\text{ V}$ DC power supply, what is the current flowing through the solenoid coil in amperes ( $\text{A}$ )?

Show answer & explanation

Answer: 2

Answer

The current flowing through the solenoid coil is

2\text{ A}

.

According to Ohm's law, electrical current (

I

) equals voltage (

V

) divided by resistance (

R

). Substituting the given values yields

I = \frac{30\text{ V}}{15\text{ }\Omega} = 2\text{ A}

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage

V = 30\text{ V}

, Resistance

R = 15\text{ }\Omega

Extract the given values to identify which Ohm's law formula variation is required.

2

State Ohm's law formula for current

I = \frac{V}{R}

Current is directly proportional to applied voltage and inversely proportional to total circuit resistance.

3

Substitute the values and evaluate

I = \frac{30\text{ V}}{15\text{ }\Omega} = 2\text{ A}

Dividing

30

by

15

yields the circuit current in amperes.

Key Concept

Ohm's Law (

I = \frac{V}{R}

)

Estimated Time:45s

Question 798Question →

In electronics and solid-state physics, materials are fundamentally classified by the number of valence electrons in their outermost atomic shell and their conductivity characteristics. An element contains exactly four valence electrons in its outer shell, forming covalent bonds in a crystalline lattice that exhibits moderate resistivity at room temperature and increased conductivity as temperature rises. Which of the following substances possesses these specific atomic and electrical properties?

Copper

Silicon

Mica

Aluminum

Show answer & explanation

Answer: Silicon

Answer

Silicon, because it is a semiconductor with four valence electrons in its outer shell and a negative temperature coefficient of resistance.

Silicon is an elemental semiconductor belonging to Group IV of the periodic table, meaning each atom has four valence electrons. In a solid crystal lattice, these four electrons form covalent bonds with adjacent atoms. At room temperature, thermal excitation frees a small fraction of these electrons across the narrow energy gap into the conduction band, giving silicon an intermediate level of conductivity that increases as thermal energy rises.

Step-by-Step Solution

1

Identify the material category based on the outer shell valence electron count.

An atom with exactly four valence electrons is classified as a semiconductor (Group IV element).

Good conductors typically have 1 to 3 valence electrons, semiconductors have exactly 4, and insulators generally have 5 to 8 tightly held valence electrons.

2

Evaluate the electrical conduction behavior described in the scenario.

The material exhibits moderate resistivity and a negative temperature coefficient of resistance (conductivity increases with temperature).

In intrinsic semiconductors, added thermal energy breaks covalent bonds, releasing electron-hole pairs into the conduction band to facilitate charge flow.

3

Match the specified atomic and electrical characteristics to the given material choices.

Silicon is a Group IV semiconductor, whereas copper and aluminum are conductors and mica is an insulator.

Silicon has an atomic number of 14 with an electron configuration of 2-8-4, giving it precisely four valence electrons.

Key Concept

Semiconductor atomic structure and valence electron properties

Estimated Time:1m 15s

Question 799Question →

In an electronic power system, two ideal transformers are cascaded (connected in sequence).

Transformer 1 receives an input voltage of $1{}200\text{ V AC}$ on its primary coil, which has $3{}000\text{ turns}$ . The secondary coil of Transformer 1 has $600\text{ turns}$ , and its output voltage is fed directly into the primary coil of Transformer 2.

Transformer 2 further steps down this voltage to supply exactly $48\text{ V AC}$ to an electronic load.

If the secondary coil of Transformer 2 contains $50\text{ turns}$ , how many turns must the primary coil of Transformer 2 have?

Show answer & explanation

Answer: 250

Answer

250 turns

The correct answer is derived by sequentially applying the transformer turns ratio formula (

V_p/V_s = N_p/N_s

) across the cascaded system. Transformer 1 operates with a 5:1 step-down ratio, reducing the

1{}200\text{ V}

input to

240\text{ V}

. Transformer 2 receives this

240\text{ V}

and must further step it down to

48\text{ V}

, which also requires a 5:1 ratio. Multiplying the

50

secondary turns by this ratio yields

250

primary turns.

Step-by-Step Solution

1

Calculate the secondary voltage of Transformer 1.

V_{s1} = 240\text{ V AC}

Using the turns ratio formula

V_{s1} / V_{p1} = N_{s1} / N_{p1}

, we find

V_{s1} = 1{}200 \times (600 / 3{}000) = 1{}200 \times 0.2 = 240\text{ V AC}

.

2

Identify the primary voltage of Transformer 2.

V_{p2} = 240\text{ V AC}

Because the transformers are cascaded in series, the output voltage of Transformer 1 serves directly as the input voltage for Transformer 2.

3

Set up the turns ratio equation for Transformer 2.

N_{p2} / 50 = 240 / 48

Applying the turns ratio formula

N_{p2} / N_{s2} = V_{p2} / V_{s2}

to the second transformer using its specific input and output voltages.

4

Solve for the primary turns of Transformer 2.

N_{p2} = 250\text{ turns}

N_{p2} = 50 \times (240 / 48) = 50 \times 5 = 250\text{ turns}

.

Key Concept

Cascaded Transformer Turns Ratios

Question 800Question →

An electronics technician is testing a power supply circuit featuring a four-diode full-wave bridge rectifier driven by a $60\text{ Hz}$ AC secondary winding. During testing under load, one of the four rectifier diodes fails open. Which of the following best describes the electrical output observed across the load resistor?

The circuit delivers pulsating DC with a ripple frequency of

60\text{ Hz}

because only one half-cycle path remains functional.

The circuit continues full-wave rectification at a

120\text{ Hz}

ripple frequency because the three remaining diodes reroute current through both alternations.

The output across the load reverts entirely to an alternating current (AC) waveform at

60\text{ Hz}

.

The output voltage and current drop completely to zero across the entire AC cycle due to an open circuit.

Show answer & explanation

Answer: The circuit delivers pulsating DC with a ripple frequency of $60\text{ Hz}$ because only one half-cycle path remains functional.

Answer

The circuit operates as a half-wave rectifier, supplying pulsating DC with a ripple frequency of

60\text{ Hz}

because only one alternation pair conducts.

A standard bridge rectifier uses two diodes per half-cycle to route current through the load in the same direction. When one diode opens, the series loop for that half-cycle is broken, producing zero output during that alternation. The other pair continues to conduct normally during the opposite half-cycle, resulting in half-wave rectification where pulses occur once per AC period at

60\text{ Hz}

.

Step-by-Step Solution

1

Analyze normal full-wave bridge rectifier conduction paths.

In normal operation, two diodes conduct in series on the positive half-cycle, and the opposite two diodes conduct in series on the negative half-cycle, yielding full-wave pulsating DC at twice the input line frequency (

2 \times 60\text{ Hz} = 120\text{ Hz}

).

Understanding normal operation establishes the baseline for analyzing component failure.

2

Determine the impact of one diode failing open on both AC half-cycles.

The half-cycle requiring the open diode cannot complete its series loop, resulting in zero conduction during that alternation. The opposite half-cycle still has both diodes forward-biased and fully operational.

Diodes operate as one-way directional switches in specific series pairs per AC polarity.

3

Identify the resulting waveform characteristics across the load.

The circuit delivers unidirectional current pulses during only one half-cycle of the input AC sine wave, which defines half-wave rectification with a fundamental ripple frequency equal to the input line frequency (

60\text{ Hz}

).

A single pulse per full AC cycle corresponds directly to the source frequency of 60 Hz.

Key Concept

Diode bridge rectification mechanics and single-diode open fault behavior

Estimated Time:1m 30s

Answer

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