All practice questions

Question 861Question →

An aircraft technician monitors a steady direct-current (DC) sensor loop and determines that $24\text{ coulombs}$ of charge pass through a test conductor in $6\text{ seconds}$ . What is the current in this conductor, and in which direction do the electrons physically drift?

4\text{ A}

, drifting from the positive terminal toward the negative terminal

144\text{ A}

, drifting from the negative terminal toward the positive terminal

4\text{ A}

, drifting from the negative terminal toward the positive terminal

4\text{ A}

, oscillating back and forth continuously without net forward drift

Show answer & explanation

Answer: $4\text{ A}$ , drifting from the negative terminal toward the positive terminal

Answer

4\text{ A}

, drifting from the negative terminal toward the positive terminal

Electric current is the time rate of charge transfer (

I = \frac{Q}{t} = \frac{24\text{ C}}{6\text{ s}} = 4\text{ A}

). In electron flow theory, negatively charged electrons are repelled by the negative terminal and flow through the external circuit toward the positive terminal.

Step-by-Step Solution

1

Calculate the magnitude of the electric current using the relationship between charge, current, and time

I = \frac{Q}{t} = \frac{24\text{ C}}{6\text{ s}} = 4\text{ A}

Electric current is defined as the rate at which charge flows past a point in a conductor per unit time.

2

Determine the physical direction of electron flow in a direct-current circuit

Electrons move from negative potential toward positive potential.

Electrons carry a negative charge and are repelled by the negative terminal and attracted toward the positive terminal.

Key Concept

Electric current definition (

I = \frac{Q}{t}

) and physical electron drift direction in direct-current circuits

Estimated Time:45s

Question 862Question →

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

Show answer & explanation

Answer: 8

Answer

8 Ω

For two resistors in parallel with values of

10\ \Omega

and

40\ \Omega

, the total equivalent resistance is calculated using the product-over-sum formula:

R_{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.

Step-by-Step Solution

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.

Key Concept

Parallel Circuit Equivalent Resistance

Estimated Time:45s

Question 863Question →

A ceramic load resistor in a direct-current (DC) circuit has a resistance of $20\ \Omega$ and carries a steady current of $2\text{ A}$ . What is the total electrical power dissipated by this resistor?

40\text{ W}

80\text{ W}

10\text{ W}

800\text{ W}

Show answer & explanation

Answer: $80\text{ W}$

Answer

The total electrical power dissipated by the resistor is

80\text{ W}

.

According to Watt's Law, when current (

I

) and resistance (

R

) are known, electrical power is calculated using the formula

P = I^2 R

. Substituting

I = 2\text{ A}

and

R = 20\ \Omega

gives

P = (2\text{ A})^2 \times 20\ \Omega = 4 \times 20 = 80\text{ W}

.

Step-by-Step Solution

1

Identify the given electrical values and select the appropriate form of Watt's Law.

Given: current

I = 2\text{ A}

, resistance

R = 20\ \Omega

. Formula:

P = I^2 R

.

When current and resistance are known, the power dissipated is calculated using

P = I^2 R

.

2

Square the current value.

I^2 = (2\text{ A})^2 = 4\text{ A}^2

.

Watt's Law requires current to be squared before multiplying by resistance.

3

Multiply the squared current by the resistance to calculate power in watts.

P = 4\text{ A}^2 \times 20\ \Omega = 80\text{ W}

.

Multiplying amperes squared by ohms yields power in watts.

Key Concept

Calculating electrical power from current and resistance using Watt's Law (

P = I^2 R

)

Estimated Time:45s

Question 864Question →

When an individual internal transistor fails inside an integrated circuit, that specific component can be desoldered and replaced separately without replacing the whole package.

True

False

Show answer & explanation

Answer: False

Answer

False

Integrated circuits are fabricated on a single microscopic semiconductor die sealed inside plastic or ceramic, meaning internal components cannot be individually accessed or replaced.

Step-by-Step Solution

1

Identify the physical structure and manufacturing method of an integrated circuit.

An integrated circuit combines thousands or millions of microscopic components fabricated directly onto a single silicon chip sealed inside a protective housing.

Understanding IC construction determines whether internal elements can be accessed or individually serviced.

2

Evaluate the feasibility of repairing individual components within an integrated circuit.

Because the components are microscopic and permanently sealed on a monolithic die, individual parts cannot be desoldered or replaced.

Directly confirms that the entire IC package must be replaced when any internal component fails.

Key Concept

Monolithic construction and non-serviceable nature of integrated circuits

Estimated Time:45s

Question 865Question →

A maintenance kit contains $12$ metric wrenches, $15$ standard wrenches, $9$ screwdrivers, and an unknown number of pliers. If the probability of randomly selecting a screwdriver from the entire kit is $\frac{1}{6}$ , what is the probability of randomly selecting a standard wrench? Express your answer as a fraction in simplest form.

Show answer & explanation

Answer: $5/18; \frac{5}{18}; 5 / 18$

Answer

5/18

By using the given probability of selecting a screwdriver (1/6) and the known count of screwdrivers (9), we determine that the total number of items in the kit is 54 (because 9 is 1/6 of 54). The probability of randomly selecting a standard wrench is then the number of standard wrenches (15) divided by the total (54). Simplifying the fraction 15/54 by dividing both numbers by 3 yields 5/18.

Step-by-Step Solution

1

Set up an equation to find the total number of items in the kit.

\frac{9}{\text{Total}} = \frac{1}{6}

The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. Here, we know the probability and the exact count of screwdrivers.

2

Solve for the total number of items.

\text{Total} = 9 \times 6 = 54

Cross-multiplying isolates the total variable, revealing there are 54 items in the entire kit.

3

Calculate the probability of selecting a standard wrench.

\text{Probability} = \frac{15}{54}

There are 15 standard wrenches out of the 54 total items calculated.

4

Simplify the fraction.

\frac{15}{54} = \frac{5}{18}

Both 15 and 54 are divisible by 3. Dividing the numerator and denominator by 3 gives the simplest form.

Key Concept

Calculating probabilities by determining the total sample size from a known probability.

Question 866Question →

A technician is designing a control circuit to reverse the operating direction of a direct current (DC) motor by reversing the polarity of both supply leads simultaneously using a single manual switch. Which switch configuration is required to achieve this operation?

Single-Pole, Single-Throw (SPST)

Single-Pole, Double-Throw (SPDT)

Double-Pole, Double-Throw (DPDT)

Double-Pole, Single-Throw (DPST)

Show answer & explanation

Answer: Double-Pole, Double-Throw (DPDT)

Answer

Double-Pole, Double-Throw (DPDT)

A Double-Pole, Double-Throw (DPDT) switch contains two electrically isolated common terminals (poles) that simultaneously connect to one of two sets of output contacts (throws). By cross-wiring the two throw sets, toggling the switch swaps the positive and negative supply leads connected to the motor, reversing its rotational direction.

Step-by-Step Solution

1

Determine the number of isolated poles required

Two poles (Double-Pole) are needed

Reversing DC motor polarity requires controlling both the positive and negative supply lines simultaneously within isolated circuit paths.

2

Determine the number of positions/throws required per pole

Two throws (Double-Throw) are needed

Each pole must alternate between two different electrical connections (forward polarity and reversed polarity) rather than simply opening or closing.

3

Combine pole and throw specifications

A Double-Pole, Double-Throw (DPDT) switch is identified

When the outer terminals of a DPDT switch are cross-wired, flipping the switch reverses the voltage polarity applied across the motor terminals.

Key Concept

Switch nomenclature regarding poles (number of separate circuits controlled) and throws (number of output paths per pole) for polarity-reversal applications.

Estimated Time:1m 0s

Question 867Question →

In electronics, integrated circuits (ICs) are widely used to replace complex discrete component assemblies. Which of the following best describes the construction of a monolithic integrated circuit?

A protective device containing a metal strip that melts to interrupt excessive current flow

A complete electronic circuit composed of multiple interconnected components fabricated on a single semiconductor substrate

A non-conductive ceramic block designed strictly to isolate and prevent current from flowing between connections

A two-terminal semiconductor device that functions exclusively to permit current flow in only one direction

Show answer & explanation

Answer: A complete electronic circuit composed of multiple interconnected components fabricated on a single semiconductor substrate

Answer

A complete electronic circuit composed of multiple interconnected components fabricated on a single semiconductor substrate

A monolithic integrated circuit (IC), commonly called a microchip, consists of many microscopic active components (like transistors and diodes) and passive elements (like resistors and capacitors) manufactured together on a single semiconductor crystal substrate (silicon wafer).

Step-by-Step Solution

1

Identify the fundamental definition of an integrated circuit (IC)

An IC integrates multiple microscopic components (transistors, diodes, resistors, capacitors) into a functional system.

The prefix 'integrated' refers to combining individual circuit elements into a single package.

2

Determine the physical construction of a monolithic IC

Monolithic ICs are manufactured directly on a single piece of semiconductor material (substrate), typically silicon.

This contrasts with discrete component circuits where separate parts are wired together on a circuit board.

Key Concept

Monolithic integrated circuit construction and characteristics

Estimated Time:45s

Question 868Question →

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

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

The iron core stores electrostatic potential energy between the wire turns like a capacitor dielectric

The direct current is automatically converted into an alternating voltage inside the core

The electric current replaces the voltage in the circuit because they represent the same electrical quantity

Show answer & explanation

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

Answer

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.

Step-by-Step Solution

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

Key Concept

Electromagnetism in Coils with Ferromagnetic Cores

Estimated Time:45s

Question 869Question →

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?

Electrostatic discharge released from energy stored in the surrounding electric field

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

A continuous direct current forcing its way across the newly formed air gap

A sudden surge of forward current resulting from zero resistance across the contacts

Show answer & explanation

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

Answer

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.

Step-by-Step Solution

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 = -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.

Key Concept

Self-Induction and Counter-Electromotive Force (CEMF)

Estimated Time:1m 0s

Question 870Question →

An electronics technician is assembling an energy-storage circuit for an industrial sensor module. The circuit consists of two $100\text{ }\mu\text{F}$ capacitors connected in series, and this entire combination is wired in parallel with a single $150\text{ }\mu\text{F}$ capacitor. What is the total equivalent capacitance of this circuit in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 200

Answer

The total equivalent capacitance of the circuit is

200\text{ }\mu\text{F}

.

For capacitors in series, total capacitance decreases and is calculated as

\frac{1}{C_{\text{series}}} = \frac{1}{100\text{ }\mu\text{F}} + \frac{1}{100\text{ }\mu\text{F}} = \frac{2}{100\text{ }\mu\text{F}}

, giving

C_{\text{series}} = 50\text{ }\mu\text{F}

. When this combination is connected in parallel with the

150\text{ }\mu\text{F}

capacitor, parallel capacitances add directly:

C_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the two

100\text{ }\mu\text{F}

capacitors connected in series using the reciprocal formula

\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}

.

C_{\text{series}} = 50\text{ }\mu\text{F}

Capacitors in series combine inversely, meaning two identical capacitors in series have half the capacitance of one capacitor.

2

Add the capacitance of the parallel branch to the series equivalent:

C_{\text{total}} = C_{\text{series}} + C_3

.

C_{\text{total}} = 50\text{ }\mu\text{F} + 150\text{ }\mu\text{F} = 200\text{ }\mu\text{F}

Capacitors connected in parallel add directly because their effective plate areas combine, increasing overall charge storage capacity.

Key Concept

Series and parallel equivalent capacitance rules (series capacitors combine inversely; parallel capacitors add directly).

Estimated Time:1m 0s

Question 871Question →

A naval vessel alarm panel is powered by a $72\text{ V}$ DC supply connected in series with three resistive loads: an alert strobe ( $R_1 = 8\ \Omega$ ), an auxiliary buzzer ( $R_2 = 10\ \Omega$ ), and a line-balancing resistor ( $R_3 = 6\ \Omega$ ). What is the voltage drop across the auxiliary buzzer ( $R_2$ )?

30\text{ V}

24\text{ V}

18\text{ V}

3\text{ V}

Show answer & explanation

Answer: $30\text{ V}$

Answer

The voltage drop across the auxiliary buzzer (

R_2

) is

30\text{ V}

.

The correct answer is

30\text{ V}

. First, sum the resistances in series to obtain

R_T = 8\ \Omega + 10\ \Omega + 6\ \Omega = 24\ \Omega

. Next, determine the loop current using Ohm's Law:

I = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

. Finally, multiply the current by the resistance of the auxiliary buzzer (

10\ \Omega

) to get

V_2 = 3\text{ A} \times 10\ \Omega = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_T

) of the series circuit.

R_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 (

I

) flowing through the circuit.

I = \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 (

R_2

).

V_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.

Key Concept

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 \times R

).

Estimated Time:1m 0s

Question 872Question →

A technician is using a digital multimeter to test a standard cartridge fuse that has been removed from a circuit. If the fuse is intact and in good working condition, what reading should the multimeter display when set to measure resistance?

Infinite resistance or

\text{OL}

(Over Limit)

Approximately

0\ \Omega

(zero or near-zero resistance)

A voltage reading matching the circuit source voltage

A very high resistance in the megaohm (

\text{M}\Omega

) range

Show answer & explanation

Answer: Approximately $0\ \Omega$ (zero or near-zero resistance)

Answer

Approximately

0\ \Omega

(zero or near-zero resistance)

A functional, intact fuse is a continuous electrical conductor designed to offer almost zero opposition to current flow. When tested with a multimeter set to resistance (

\Omega

) or continuity, it will display a reading of approximately

0\ \Omega

, confirming complete circuit continuity.

Step-by-Step Solution

1

Identify the internal state of a good fuse.

An unbroken conductive metallic link inside the fuse casing.

Fuses are protective conductors designed to allow electrical current to pass with negligible opposition during normal operation.

2

Determine the expected ohmmeter reading across a continuous conductor.

The meter reads zero or near-zero ohms (

0\ \Omega

).

An ohmmeter sends a tiny test current through the component; because the path is unbroken and non-resistive, resistance is measured at virtually

0\ \Omega

.

Key Concept

Multimeter resistance and continuity testing across closed conductors versus open circuits

Estimated Time:45s

Question 873Question →

An industrial temperature sensor module with a calibrated internal resistance of $4.5\text{ k}\Omega$ draws a steady direct current of $8\text{ mA}$ during normal operation. What is the voltage drop across the sensor module in volts ( $\text{V}$ )?

Show answer & explanation

Answer: 36

Answer

The voltage drop across the sensor module is

36\text{ V}

.

According to Ohm's Law, voltage (

V

) equals current (

I

) multiplied by resistance (

R

). Given

I = 8\text{ mA} = 0.008\text{ A}

and

R = 4.5\text{ k}\Omega = 4500\text{ }\Omega

, the calculation is

V = 0.008\text{ A} \times 4500\text{ }\Omega = 36\text{ V}

.

Step-by-Step Solution

1

Convert the given metric prefixes to standard base units (amperes and ohms).

I = 8\text{ mA} = 8 \times 10^{-3}\text{ A} = 0.008\text{ A}

and

R = 4.5\text{ k}\Omega = 4.5 \times 10^{3}\text{ }\Omega = 4500\text{ }\Omega

Ohm's law requires fundamental SI units (volts, amperes, and ohms) for accurate mathematical evaluation.

2

Select the appropriate form of Ohm's Law to calculate voltage (

V

).

V = I \times R

Voltage across a resistive load equals the product of the current flowing through it and its resistance.

3

Substitute the converted values into the formula and solve.

V = 0.008\text{ A} \times 4500\text{ }\Omega = 36\text{ V}

Multiplying

8 \times 10^{-3}\text{ A}

by

4.5 \times 10^{3}\text{ }\Omega

simplifies directly to

8 \times 4.5 = 36\text{ V}

because the metric prefixes milli (

10^{-3}

) and kilo (

10^{3}

) cancel each other.

Key Concept

Ohm's Law: Voltage Calculation (

V = I \times R

)

Alternative Method

Recognize that the prefix multiplier milli- (

10^{-3}

) and kilo- (

10^{3}

) cancel out when multiplied together:

(8\text{ mA}) \times (4.5\text{ k}\Omega) = 8 \times 4.5\text{ V} = 36\text{ V}

.

Estimated Time:1m 0s

Question 874Question →

A $10\text{ k}\Omega$ linear potentiometer is wired as a variable rheostat by tying the center wiper (terminal 2) directly to outside terminal 3. If the control shaft is set to $25\%$ of its total travel starting from terminal 1 toward terminal 3, what is the effective resistance measured between terminal 1 and the combined terminal 2/3 connection?

7.5\text{ k}\Omega

2.5\text{ k}\Omega

1.875\text{ k}\Omega

25\text{ k}\Omega

Show answer & explanation

Answer: $2.5\text{ k}\Omega$

Answer

2.5\text{ k}\Omega

A potentiometer wired with its wiper connected to one outer terminal functions as a two-terminal variable resistor (rheostat). Setting the shaft to

25\%

rotation from terminal 1 creates a

2.5\text{ k}\Omega

resistance path between terminal 1 and terminal 2. Because terminal 2 is strapped to terminal 3, the remaining

7.5\text{ k}\Omega

section of the resistive element is completely bypassed by the short, leaving an effective resistance of

2.5\text{ k}\Omega

.

Step-by-Step Solution

1

Determine the resistance of the active resistive track between terminal 1 and the wiper (terminal 2).

R_{1-2} = 0.25 \times 10\text{ k}\Omega = 2.5\text{ k}\Omega

Linear potentiometers divide resistance directly in proportion to shaft travel percentage from the reference end.

2

Analyze the state of the remaining track segment between the wiper (terminal 2) and terminal 3.

R_{2-3} = 7.5\text{ k}\Omega

is short-circuited.

Connecting terminal 2 directly to terminal 3 creates a zero-resistance path across this section, bypassing it entirely.

3

Calculate the total effective resistance between terminal 1 and the joined 2/3 terminal.

R_{\text{eff}} = 2.5\text{ k}\Omega

Current flows through the unshorted

R_{1-2}

segment and exits via the combined wiper/terminal 3 connection.

Key Concept

Rheostat Wiring and Potentiometer Track Division

Estimated Time:45s

Question 875Question →

A semiconductor diode is a two-terminal electronic component that acts as a one-way valve for electric current. For current to conduct freely through a standard diode in a direct current (DC) circuit, how must the diode be connected relative to the power source?

The cathode must be connected toward the positive voltage terminal and the anode toward the negative terminal.

The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

Both the anode and cathode must be connected to the positive terminal of the power source.

The diode must be connected to an alternating current source so voltage polarities constantly reverse.

Show answer & explanation

Answer: The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

Answer

The anode must be connected toward the positive voltage terminal and the cathode toward the negative terminal.

A diode conducts electric current when it is placed in forward bias. In a direct current circuit, forward bias is established by connecting the anode toward the positive terminal of the power supply and the cathode toward the negative terminal.

Step-by-Step Solution

1

Identify the two terminals of a standard diode and the condition needed for conduction.

A diode has an anode (p-type material) and a cathode (n-type material), and it conducts current only when forward-biased.

Understanding diode terminal polarity is necessary to determine proper circuit orientation.

2

Determine the voltage orientation required to forward-bias the diode.

The anode must be at a higher electrical potential than the cathode.

Applying a positive voltage to the anode and a negative voltage to the cathode overcomes the internal barrier potential, narrowing the depletion region and permitting current flow.

Key Concept

A semiconductor diode allows current to flow when it is forward-biased, which occurs when the anode is connected toward the positive supply voltage and the cathode toward the negative side.

Estimated Time:45s

Question 876Question →

An industrial security monitoring loop is powered by a $36\text{ V}$ DC power supply connected in series with a current-limiting resistor $R_1 = 150\ \Omega$ , an optical sensor module $R_2$ , and a line termination resistor $R_3 = 250\ \Omega$ . When the circuit is active, a voltmeter measures a $12\text{ V}$ drop across the optical sensor module $R_2$ . What is the electrical resistance of the optical sensor module $R_2$ ?

133\ \Omega

200\ \Omega

400\ \Omega

600\ \Omega

Show answer & explanation

Answer: $200\ \Omega$

Answer

The electrical resistance of the optical sensor module

R_2

is

200\ \Omega

.

The correct answer is determined by first finding the remaining voltage across the known resistors (

36\text{ V} - 12\text{ V} = 24\text{ V}

). Dividing this by their combined resistance (

150\ \Omega + 250\ \Omega = 400\ \Omega

) yields a constant loop current of

0.06\text{ A}

. Dividing the sensor's

12\text{ V}

drop by

0.06\text{ A}

gives

200\ \Omega

.

Step-by-Step Solution

1

Calculate the combined voltage drop across the known series resistors

R_1

and

R_3

.

V_{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.

R_{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 = \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

R_2

using its measured voltage drop and the loop current.

R_2 = \frac{V_2}{I} = \frac{12\text{ V}}{0.06\text{ A}} = 200\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) to the individual component provides its exact resistance.

Key Concept

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

Alternative Method

Using the voltage divider formula:

\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

.

Estimated Time:1m 30s

Question 877Question →

In basic electronics, distinguishing between charge behavior, measurement units, and current flow conventions is essential. Match each electrical concept on the left with its correct definition or governing rule on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electron Flow Theory matches the movement of negative charge carriers from negative to positive; Conventional Current Theory matches the historical assumption of flow from positive to negative; Electric Current matches the rate of charge flow over time; One Ampere matches one coulomb of charge per second.

Each electrical term accurately maps to its standard scientific definition: Electron Flow describes negative electrons drifting toward the positive terminal; Conventional Current describes the standard positive-to-negative circuit model; Electric Current represents the rate of charge flow (

I = Q/t

); and One Ampere specifically equals one coulomb passing a given point in one second (

1\text{ C/s}

).

Step-by-Step Solution

1

Identify the physical reality of electron drift versus historical convention.

Electrons are negatively charged, so they drift from negative to positive (Electron Flow). Conventional current historically assumes flow from positive to negative.

Clear differentiation between historical convention and physical subatomic movement is foundational in electrical fundamentals.

2

Define the quantity of electric current.

Current (

I

) represents the rate of charge (

Q

) flow per unit time (

t

), yielding

I = \frac{Q}{t}

.

Current measures the flow rate of charge through a conductor, not electrical pressure or potential difference.

3

Relate the ampere to fundamental SI electrical units.

1\text{ ampere} = 1\text{ coulomb per second}

(

1\text{ A} = 1\text{ C/s}

).

The ampere quantifies current flow as a unit of coulombs transferred over time in seconds.

Key Concept

Fundamental definitions of electric current, charge rate, ampere units, and directional flow models.

Alternative Method

Recall the fundamental dimensional relationships: Current is

I = \frac{\Delta Q}{\Delta t}

, so its unit (

1\text{ A}

) must be equal to

\frac{1\text{ C}}{1\text{ s}}

. Then remember that because electrons possess a negative charge, like charges repel and opposite charges attract, forcing them to move away from the negative terminal toward the positive terminal.

Estimated Time:45s

Question 878Question →

In electrical and electronic systems, understanding the fundamental physical properties governing current flow is essential. Match each resistance-related concept on the left with its correct physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electrical Resistance matches the total opposition of a specific component in ohms; Material Resistivity matches the intrinsic geometry-independent material property; American Wire Gauge matches the inverse sizing scale where smaller numbers mean thicker wire; and Positive Temperature Coefficient matches the increase in resistance as temperature rises.

Each electrical concept matches its core definition: Electrical Resistance (

R

) is the total opposition to current flow in ohms; Material Resistivity (

\rho

) is the intrinsic material property; American Wire Gauge (AWG) is the inverse sizing standard where smaller numbers indicate thicker conductors; and a Positive Temperature Coefficient defines the increase of resistance with rising temperature.

Step-by-Step Solution

1

Distinguish between total electrical resistance and intrinsic material resistivity.

Electrical resistance (

R

) is the overall opposition in a specific conductor measured in ohms (

\Omega

), while resistivity (

\rho

) is the inherent property of the material itself.

Total resistance depends on both the material's resistivity and its physical dimensions according to

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

.

2

Analyze standard conductor sizing and thermal behavior principles.

AWG numbers inversely scale with conductor diameter (smaller number = thicker wire), and a positive temperature coefficient means resistance rises with heat.

Wire gauge dictates cross-sectional area, and thermal excitation in metals increases electron collisions, raising resistance.

Key Concept

Core physical factors and definitions governing electrical resistance and resistivity in conductors

Estimated Time:45s

Question 879Question →

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

36\ \Omega

12\ \Omega

4\ \Omega

3\ \Omega

Show answer & explanation

Answer: $4\ \Omega$

Answer

4\ \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 (

R_{\text{total}} = R / N

). Here,

12\ \Omega / 3 = 4\ \Omega

.

Step-by-Step Solution

1

Identify the number of parallel branches and branch resistance

Branch resistance

R = 12\ \Omega

, number of branches

N = 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

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

R_{\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.

Key Concept

Equal-Resistor Parallel Shortcut Rule (

R_{\text{total}} = R / N

)

Estimated Time:45s

Question 880Question →

In direct-current (DC) circuit analysis, different forms of Watt's Law are used depending on which electrical quantities are known. Match each set of known circuit parameters on the left with the correct formula on the right used to calculate the target value directly.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Each given set of circuit parameters matches the direct algebraic form of Watt's Law designed for those specific variables: Voltage and Current to

P = V \times I

; Current and Resistance to

P = I^2 \times R

; Voltage and Resistance to

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

; and Power and Voltage to

I = \frac{P}{V}

.

Each match correctly pairs the given known variables with the exact Watt's Law equation variant required to solve directly for the unknown quantity without needing extraneous calculation steps.

Step-by-Step Solution

1

Identify the base definition of electrical power in terms of voltage and current.

P = V \times I

Power in watts equals electromotive force in volts multiplied by electron current flow in amperes.

2

Derive the power formula when only current and resistance are given using Ohm's Law (

V = I \times R

).

P = (I \times R) \times I = I^2 \times R

Replacing voltage with the product of current and resistance calculates power without needing voltage explicitly.

3

Derive the power formula when only voltage and resistance are given using Ohm's Law (

I = \frac{V}{R}

).

P = V \times \left(\frac{V}{R}\right) = \frac{V^2}{R}

Replacing current with voltage divided by resistance calculates power without needing current explicitly.

4

Rearrange the primary power formula to solve for current when power and voltage are known.

I = \frac{P}{V}

Dividing both sides of

P = V \times I

by voltage isolates the current parameter.

Key Concept

Watt's Law and its derived variations relate electrical power (

P

), voltage (

V

), current (

I

), and resistance (

R

).

Estimated Time:45s

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