All practice questions

Question 741Question →

Match each circuit modification on the left with its corresponding effect on power dissipation relative to the baseline power ( $P_0$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Doubling voltage with constant resistance matches

4P_0

; Halving current with constant resistance matches

\frac{1}{4}P_0

; Doubling resistance with constant voltage matches

\frac{1}{2}P_0

; Tripling current with one-third resistance matches

3P_0

.

Each modification is correctly determined by applying Watt's Law formulas (

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

and

P = I^2 R

). Doubling voltage quadruples power (

2^2 = 4

), halving current reduces power to one-fourth (

(\frac{1}{2})^2 = \frac{1}{4}

), doubling resistance at constant voltage reduces power by half (

\frac{1}{2}

), and tripling current while reducing resistance to one-third increases power threefold (

3^2 \times \frac{1}{3} = 3

).

Step-by-Step Solution

1

Select the appropriate form of Watt's Law for each scenario depending on which parameters are held constant and which are modified.

Use

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

when voltage and resistance are given, and

P = I^2 R

when current and resistance are given.

Using the formula directly containing the relevant independent variables simplifies scaling calculations.

2

Calculate the scaling factor for doubling voltage with constant resistance.

P' = \frac{(2V)^2}{R} = 4\frac{V^2}{R} = 4P_0

.

Power is directly proportional to the square of the voltage across a fixed load.

3

Calculate the scaling factor for halving current with constant resistance.

P' = \left(\frac{1}{2}I\right)^2 R = \frac{1}{4}I^2 R = \frac{1}{4}P_0

.

Power is directly proportional to the square of current through a fixed load.

4

Calculate the scaling factor for doubling resistance at constant voltage.

P' = \frac{V^2}{2R} = \frac{1}{2}\frac{V^2}{R} = \frac{1}{2}P_0

.

At constant voltage, power is inversely proportional to resistance.

5

Calculate the scaling factor for tripling current while reducing resistance to one-third.

P' = (3I)^2 \cdot \left(\frac{1}{3}R\right) = 9I^2 \cdot \frac{1}{3}R = 3I^2 R = 3P_0

.

The squared current factor of 9 multiplied by the fractional resistance factor of 1/3 results in an overall factor of 3.

Key Concept

Watt's Law Power Scaling Relationships:

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

Estimated Time:1m 30s

Question 742Question →

When inspecting a standard 14-pin Dual In-line Package (DIP) integrated circuit with the reference notch oriented at the top, which statement correctly describes the standard pinout numbering convention used to identify pin 1 and pin 14?

Pin 1 is at the top-left beside the notch, and numbering progresses sequentially down the left side from pin 1 to 7, then restarts at the top-right with pin 8 and counts down to pin 14 at the bottom-right.

Pin 1 is at the top-left beside the notch, and numbering progresses counter-clockwise down the left side from pin 1 to 7, across to the bottom-right for pin 8, and up the right side to pin 14 at the top-right.

Pin 1 is at the top-right beside the notch, and numbering progresses clockwise down the right side from pin 1 to 7, across to the bottom-left for pin 8, and up the left side to pin 14 at the top-left.

Pin 1 is at the top-left, and numbering alternates horizontally across opposing pairs (pin 1 top-left, pin 2 top-right, pin 3 second-row left) down the package to pin 14 at the bottom-right.

Show answer & explanation

Answer: Pin 1 is at the top-left beside the notch, and numbering progresses counter-clockwise down the left side from pin 1 to 7, across to the bottom-right for pin 8, and up the right side to pin 14 at the top-right.

Answer

Pin 1 is located at the top-left immediately adjacent to the notch, and numbering continues counter-clockwise down the left column (pins 1 through 7) and upward along the right column (pins 8 through 14), placing pin 14 at the top-right directly opposite pin 1.

For all standard Dual In-line Package (DIP) integrated circuits, placing the index notch at the top establishes pin 1 at the top-left. Numbering then advances counter-clockwise down the left row (pins 1 to 7 for a 14-pin IC), transitions across to the bottom-right pin (pin 8), and ascends the right row to pin 14 at the top-right.

Step-by-Step Solution

1

Orient the integrated circuit package correctly.

The reference notch or alignment dot is positioned facing upward (top).

Standardized electronic manufacturing standards (JEDEC) require referencing the orientation marker before indexing lead pins.

2

Identify pin 1 based on the reference notch.

Pin 1 is the lead located immediately to the left of the notch in the top row.

By industry convention, pin 1 is always directly counter-clockwise adjacent to the top reference notch.

3

Trace the indexing sequence around the Dual In-line Package perimeter.

Pins 1 through 7 run down the left side, pin 8 is at the bottom-right, and pins 9 through 14 advance upward on the right side to pin 14 at the top-right.

DIP integrated circuits use a continuous counter-clockwise loop rather than parallel or alternating row numbering.

Key Concept

Dual In-line Package (DIP) Integrated Circuit Pin Numbering Convention

Estimated Time:1m 0s

Question 743Question →

In a direct current (DC) circuit, an ammeter measures the rate of charge flow past a specific point. If exactly $1.248 \times 10^{19}$ electrons flow past this point in $4\text{ seconds}$ , and $1\text{ Coulomb}$ is defined as approximately $6.24 \times 10^{18}$ electrons, which of the following correctly identifies the current magnitude and describes the standard convention for its flow direction relative to the actual electrons?

0.5\text{ A}

; conventional current flows from positive to negative, opposite to the actual electron flow.

0.05\text{ A}

; conventional current flows from positive to negative, opposite to the actual electron flow.

0.5\text{ A}

; conventional current flows from negative to positive, identical to the actual electron flow.

5.0\text{ A}

; conventional current flows from negative to positive, identical to the actual electron flow.

Show answer & explanation

Answer: $0.5\text{ A}$ ; conventional current flows from positive to negative, opposite to the actual electron flow.

Answer

The current is

0.5\text{ A}

, and conventional current is analyzed as flowing from positive to negative, which is opposite to the actual electron flow.

The total charge is found by dividing the number of electrons by the electrons per Coulomb:

1.248 \times 10^{19} / 6.24 \times 10^{18} = 2\text{ Coulombs}

. The current is the charge divided by time:

2\text{ C} / 4\text{ s} = 0.5\text{ A}

. Standard electrical convention dictates that current flows from positive to negative, which is physically opposite to the actual movement of negatively charged electrons.

Step-by-Step Solution

1

Convert the given number of electrons into Coulombs of charge.

1.248 \times 10^{19} \text{ electrons} / 6.24 \times 10^{18} \text{ electrons/C} = 2\text{ Coulombs}

.

Current is defined as the rate of charge flow in Coulombs per second, not electrons per second.

2

Calculate the current magnitude in Amperes.

I = Q / t = 2\text{ C} / 4\text{ s} = 0.5\text{ A}

.

One Ampere is equivalent to one Coulomb of charge passing a point in one second.

3

Determine the relationship between conventional current and actual electron flow.

Conventional current is directed positive to negative, opposite to actual electron flow (negative to positive).

Standard circuit analysis uses conventional current (positive charge flow), while solid conductors actually move negatively charged electrons.

Key Concept

Electrical Current Calculations and Flow Conventions

Question 744Question →

A cybersecurity analyst is monitoring the number of unauthorized access attempts blocked by a firewall each day. The mean number of blocked attempts for the entire $5$ -day work week is exactly $32$ . If the daily counts from Monday to Thursday were $28$ , $35$ , $27$ , and $34$ , how many access attempts were blocked on Friday?

Show answer & explanation

Answer: 36

Answer

The firewall blocked 36 access attempts on Friday.

The correct answer is found by determining the total sum required to achieve the target mean for

5

days (

32 \times 5 = 160

), and then subtracting the sum of the four known days (

124

) from that total. This yields a required value of

36

for the final day.

Step-by-Step Solution

1

Calculate the total sum of access attempts required for all

5

days.

32 \times 5 = 160

The mean is equal to the total sum divided by the number of values. Therefore, the total sum is the mean multiplied by the number of values (

N

).

2

Calculate the sum of the known access attempts from the first four days.

28 + 35 + 27 + 34 = 124

We must determine the combined total of the known days before we can find the missing day's value.

3

Subtract the sum of the four days from the required

5

-day total to find Friday's count.

160 - 124 = 36

The difference between the overall required total and the sum of the known values represents the exact value needed for the final day.

Key Concept

Calculating a missing value from a data set when given the arithmetic mean and the remaining values.

Question 745Question →

A direct current circuit contains two resistors connected in parallel across a power source. The first resistor has a resistance of $30\ \Omega$ and the second resistor has a resistance of $60\ \Omega$ . What is the total equivalent resistance of the circuit in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent resistance of the circuit is

20\ \Omega

.

For two resistors in parallel, total equivalent resistance is determined by

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

. Substituting

30\ \Omega

and

60\ \Omega

yields

\frac{1800}{90} = 20\ \Omega

.

Step-by-Step Solution

1

Select the appropriate equivalent resistance formula for parallel branches.

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

When two resistors are connected in parallel, the product-over-sum formula directly provides the total equivalent resistance.

2

Substitute the resistor values into the formula and solve.

R_{eq} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\ \Omega

Evaluate the arithmetic expression to calculate the equivalent circuit resistance.

Key Concept

Total equivalent resistance of parallel resistors

Alternative Method

Using the standard reciprocal formula:

\frac{1}{R_{eq}} = \frac{1}{30} + \frac{1}{60} = \frac{2}{60} + \frac{1}{60} = \frac{3}{60} = \frac{1}{20}

. Inverting both sides gives

R_{eq} = 20\ \Omega

.

Estimated Time:45s

Question 746Question →

Electronic systems utilize various solid-state semiconductor transistors tailored for specific switching, amplification, and sensing applications. Match each transistor type to its defining operating principle and physical characteristic.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Bipolar Junction Transistor (BJT) pairs with the current-controlled device using both electron and hole carriers; Junction Field-Effect Transistor (JFET) pairs with the voltage-controlled device using a reverse-biased PN junction; Metal-Oxide-Semiconductor FET (MOSFET) pairs with the insulated-gate component featuring high input impedance and an oxide dielectric; Phototransistor pairs with the optoelectronic component converting incident light into amplified collector current.

Each semiconductor device corresponds uniquely to its operating physics: BJTs are bipolar current-controlled devices; JFETs utilize reverse-biased PN junctions for channel modulation; MOSFETs employ an insulated dielectric gate for field-effect control; and phototransistors utilize optical stimulation to generate base-region carriers.

Step-by-Step Solution

1

Identify the carrier conduction and control mechanism of the BJT

The BJT is a current-controlled bipolar device relying on both majority and minority carriers (electrons and holes).

The term bipolar explicitly indicates the participation of both carrier polarities, controlled by base current.

2

Differentiate between the field-effect transistor junction types (JFET vs. MOSFET)

The JFET relies on a direct reverse-biased PN gate junction, while the MOSFET incorporates an insulated oxide layer that eliminates DC gate current entirely.

The metal-oxide insulator in a MOSFET provides near-infinite input impedance compared to the reverse-biased PN junction of a JFET.

3

Distinguish the optoelectronic sensor category (Phototransistor)

The phototransistor operates by replacing or augmenting electrical base current with photon-generated charge carriers.

Light striking the transparent casing generates electron-hole pairs at the junction to initiate conduction.

Key Concept

Classification, control mechanisms, and structural properties of major transistor families

Estimated Time:1m 0s

Question 747Question →

In a multi-stage radio receiver, an electronics technician places a coupling capacitor in series along the signal line between two amplifier stages. What is the primary function of this capacitor regarding the combined Direct Current (DC) bias and Alternating Current (AC) signal?

Allows the continuous DC bias to pass freely while completely stopping all fluctuating AC signal frequencies.

Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

Transforms the AC signal into a higher DC voltage by generating a strong magnetic field across its plates.

Conducts the steady DC voltage directly across its dielectric material while dissipating the AC signal as heat.

Show answer & explanation

Answer: Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

Answer

Permits the fluctuating AC signal to pass through while completely blocking the steady DC bias voltage.

A capacitor acts as an open circuit to steady Direct Current (DC) because the insulating dielectric stops continuous charge conduction once the plates reach the bias voltage level. Simultaneously, the alternating fluctuations of an AC signal continuously charge and discharge the plates, allowing the alternating signal to transfer cleanly to the next amplifier stage without shifting its DC bias point.

Step-by-Step Solution

1

Analyze capacitor behavior in a steady Direct Current (DC) state.

Because the dielectric layer is an electrical insulator, steady DC charges the plates until the potential difference equals the supply voltage, reducing steady-state current to zero (

X_C = \infty

at

0\text{ Hz}

). The capacitor acts as an open circuit to DC.

Determines how the DC operating bias of the first amplifier stage is isolated from the second stage.

2

Analyze capacitor behavior when exposed to an Alternating Current (AC) signal.

The alternating polarity causes continuous charging and discharging cycles, allowing the AC signal to transfer across the component with low capacitive reactance (

X_C = \frac{1}{2\pi f C}

).

Determines how dynamic information and audio/RF waveforms propagate across the circuit.

3

Evaluate the combined circuit function of a series coupling capacitor.

The component blocks the steady DC bias voltage while coupling the alternating signal waveform directly to the next amplifier stage.

Confirms the core operational role of coupling capacitors in electronic signal chains.

Key Concept

Capacitor DC Blocking and AC Coupling

Estimated Time:1m 0s

Question 748Question →

A marine navigation console contains three indicator loads connected in series across a $75\text{ V}$ direct current (DC) power supply. The individual resistances of the loads are $8\ \Omega$ , $12\ \Omega$ , and $5\ \Omega$ . What is the total equivalent resistance of the circuit, and what is the current flowing through each individual load? Complete the statements below.

Fill in the blanks below

The total equivalent resistance of the series circuit is

\Omega

, and the current flowing through each load is

\text{A}

.

Show answer & explanation

Answer

The total equivalent resistance is

25\ \Omega

and the current through each load is

3\ \text{A}

.

In a series circuit, the total resistance is found by directly summing all individual component resistances (

8\ \Omega + 12\ \Omega + 5\ \Omega = 25\ \Omega

). Using Ohm's Law (

I = V / R_T

), the total circuit current is

75\text{ V} / 25\ \Omega = 3\text{ A}

. In any series circuit, the current is constant throughout the entire loop, meaning every connected component experiences the exact same

3\text{ A}

of current.

Step-by-Step Solution

1

Calculate the total equivalent resistance of the series circuit by adding the individual resistance values.

R_T = R_1 + R_2 + R_3 = 8\ \Omega + 12\ \Omega + 5\ \Omega = 25\ \Omega

In a series circuit, total resistance equals the arithmetic sum of all individual resistances in the loop.

2

Calculate the total circuit current using Ohm's Law (

I = V_T / R_T

).

I = \frac{75\text{ V}}{25\ \Omega} = 3\text{ A}

Ohm's Law defines current as total applied electromotive force divided by total series resistance.

3

Determine the current flowing through each individual component.

I_1 = I_2 = I_3 = I_T = 3\text{ A}

Because there is only one continuous path for charge flow in a series circuit, current is identical at all points.

Key Concept

Series Circuit Resistance and Current Characteristics

Estimated Time:1m 0s

Question 749Question →

A technician is testing a direct current (DC) relay coil connected to a $48\text{ V}$ power supply. An ammeter placed in series with the circuit measures a steady current of $150\text{ mA}$ . Based on Ohm's law, what is the internal electrical resistance of the relay coil in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 320

Answer

The internal electrical resistance of the relay coil is

320\text{ }\Omega

.

According to Ohm's law, resistance is found by dividing voltage by current (

R = \frac{V}{I}

). Converting

150\text{ mA}

to

0.15\text{ A}

and dividing the applied voltage of

48\text{ V}

by

0.15\text{ A}

yields exactly

320\text{ }\Omega

.

Step-by-Step Solution

1

Convert the current measurement from milliamperes to base amperes

I = 150\text{ mA} = 150 \times 10^{-3}\text{ A} = 0.15\text{ A}

Standard electrical calculations using Ohm's law require current in amperes.

2

Rearrange Ohm's law formula (

V = I \times R

) to isolate resistance (

R

)

R = \frac{V}{I}

The circuit voltage and current are known, and resistance is the required unknown quantity.

3

Substitute the given voltage and converted current into the formula and solve

R = \frac{48\text{ V}}{0.15\text{ A}} = 320\text{ }\Omega

Dividing

48

by

0.15

(or multiplying

48 \times \frac{20}{3}

) computes the exact resistance value.

Key Concept

Ohm's law states that electrical resistance (

R

) is the ratio of voltage (

V

) across a component to the current (

I

) flowing through it:

R = \frac{V}{I}

.

Estimated Time:1m 0s

Question 750Question →

Consider the physical principles governing electrical resistance ( $R = \rho \frac{L}{A}$ ) and material resistivity. Match each specific physical modification of a cylindrical wire conductor with its resulting net effect on the wire's total electrical resistance. Which pairings accurately reflect these proportional and thermal relationships?

Click a left item, then click its matching right item

Show answer & explanation

Answer

Stretching a wire to double length at constant volume pairs with increasing resistance by a factor of 4; doubling wire diameter pairs with decreasing resistance to one-fourth; doubling both length and radius pairs with decreasing resistance to one-half; and heating a copper conductor pairs with increasing resistance due to higher resistivity.

Each match strictly obeys the physical governing equation for resistance

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

. For cylindrical geometry,

A = \pi r^2 = \frac{\pi d^2}{4}

. Uniform stretching alters length and cross-sectional area inversely (

2L \times 0.5A \to 4R

). Diameter changes scale quadratically (

2d \to 4A \to 0.25R

). Simultaneous doubling of length and radius yields a ratio of

2/4 = 0.5R

. Thermal elevation in conductors increases resistivity (

\rho

) due to enhanced electron-phonon collisions.

Step-by-Step Solution

1

Analyze volume conservation under mechanical stretching

Stretching to double length (

L' = 2L

) causes cross-sectional area to halve (

A' = A/2

), giving

R' = \rho \frac{2L}{A/2} = 4R

.

Since volume

V = A \cdot L

remains constant, stretching alters both length and cross-sectional area simultaneously.

2

Calculate the geometric effect of doubling diameter

Doubling diameter (

d' = 2d

) quadruples area (

A' = 4A

), resulting in

R' = \rho \frac{L}{4A} = 0.25R

.

Cross-sectional area varies quadratically with diameter (

A = \frac{\pi d^2}{4}

).

3

Calculate simultaneous scaling of length and radius

Doubling length (

2\times

) and radius (

4\times

area) produces

R' = \frac{2}{4}R = 0.5R

.

The length factor of 2 in the numerator is divided by the area factor of 4 in the denominator.

4

Evaluate the thermal effect on metallic conductors

Increasing temperature increases copper's resistivity

\rho

, raising resistance without altering geometry.

Metals possess a positive temperature coefficient of resistance where lattice vibrations hinder electron drift.

Key Concept

Conductor resistance is directly proportional to length and resistivity, and inversely proportional to the square of radius or diameter (

R = \rho \frac{L}{\pi r^2}

), with metallic resistivity increasing as temperature rises.

Estimated Time:1m 30s

Question 751Question →

Match each magnetic component type to its primary operational characteristic and circuit application.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Air-core inductor matches with eliminating core saturation and hysteresis losses; Ferrite-core toroidal inductor matches with confining magnetic flux within a closed circular path to minimize EMI; Center-tapped step-down transformer matches with supplying two equal-magnitude, out-of-phase AC voltages; Variable autotransformer matches with utilizing a single continuous tapped winding without galvanic isolation.

Each component is correctly matched according to its electromagnetic construction: air-core inductors avoid high-frequency magnetic core losses; toroidal ferrite inductors contain magnetic flux to limit EMI; center-tapped transformers yield balanced dual out-of-phase secondary voltages; and autotransformers perform voltage adjustments via a single shared winding without galvanic isolation.

Step-by-Step Solution

1

Analyze core material and geometry impacts on the inductor types.

Air lacks ferromagnetic material, meaning it cannot saturate or exhibit core losses at RF frequencies (identifying the air-core inductor). A toroid geometry forms a closed magnetic loop that keeps magnetic fields contained to prevent EMI (identifying the toroidal ferrite inductor).

Inductor core selection balances permeability, saturation limits, operating frequency, and stray field radiation.

2

Analyze winding arrangements and isolation properties for the transformer types.

A center-tapped secondary winding produces two equal-amplitude outputs that are

180^\\circ

out of phase relative to the center tap (identifying the center-tapped transformer). An autotransformer uses a single tapped winding shared between input and output, which eliminates electrical isolation (identifying the variable autotransformer).

Winding configuration determines output phase symmetry and whether galvanic isolation exists between input and output circuits.

Key Concept

Operational characteristics, core geometries, and winding configurations of inductors and transformers

Estimated Time:1m 30s

Question 752Question →

Which of the following best describes the primary function of an inductor in an electronic circuit?

Opposing changes in electric current by storing energy in a magnetic field

Opposing changes in voltage by storing energy in an electrostatic field

Converting alternating current (AC) directly into steady direct current (DC)

Stepping up direct current (DC) voltage without requiring a fluctuating magnetic flux

Show answer & explanation

Answer: Opposing changes in electric current by storing energy in a magnetic field

Answer

Opposing changes in electric current by storing energy in a magnetic field

An inductor is an electrical component designed to store energy within an electromagnetic field generated by current flowing through its coils. Due to self-inductance, it produces an opposing electromotive force whenever current changes, thereby resisting rapid current variations.

Step-by-Step Solution

1

Identify the fundamental physical construction and behavior of an inductor.

An inductor is made of wire coiled around a core (air, iron, or ferrite).

When electric current passes through the coil, Ampere's law dictates that a magnetic field develops around the conductor.

2

Determine how the component reacts to dynamic circuit conditions.

According to Lenz's Law and Faraday's Law of Induction, any change in current produces a counter-electromotive force (back EMF) that opposes that change.

This property gives inductors their key application as chokes and energy storage devices in magnetic fields.

Key Concept

Inductors store energy in a magnetic field and oppose changes in electric current.

Estimated Time:40s

Question 753Question →

An electronics technician tests a direct current (DC) alarm buzzer with a measured resistance of $40\text{ }\Omega$ . If an ammeter connected in the circuit measures a current flow of $0.5\text{ A}$ , what is the voltage applied across the buzzer?

80\text{ V}

20\text{ V}

0.0125\text{ V}

20\text{ mV}

Show answer & explanation

Answer: $20\text{ V}$

Answer

20\text{ V}

According to Ohm's law, voltage across a component equals the current flowing through it multiplied by its resistance (

V = I \times R

). Multiplying

0.5\text{ A}

by

40\text{ }\Omega

gives exactly

20\text{ V}

.

Step-by-Step Solution

1

Identify the given values and unknown variable from the problem statement.

Current (

I

) =

0.5\text{ A}

, Resistance (

R

) =

40\text{ }\Omega

, Voltage (

V

) = ?

Establishing the known quantities determines which algebraic arrangement of Ohm's law is required.

2

Select and apply the standard Ohm's law formula for voltage.

V = I \times R

Voltage is equal to the product of current in amperes and resistance in ohms.

3

Substitute the given numerical values into the formula and calculate.

V = 0.5\text{ A} \times 40\text{ }\Omega = 20\text{ V}

Multiplying the decimal current by the integer resistance yields the total potential difference in volts.

Key Concept

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

V = I \times R

).

Estimated Time:45s

Question 754Question →

A technician connects a diagnostic meter into a vehicle wire harness to measure the quantity of electric charge passing through a conductor per second. Which electrical property is the technician measuring?

Electromotive force

Electric current

Electrical resistance

Alternating frequency

Show answer & explanation

Answer: Electric current

Answer

Electric current represents the measured rate of electrical charge passing through a conductor per second.

Electric current is defined as the rate at which electrical charge passes a given cross-section of a conductor, mathematically represented by

I = \frac{Q}{t}

, where one ampere equals one coulomb of charge per second.

Step-by-Step Solution

1

Analyze the physical measurement described in the question.

The measurement monitors charge (

Q

) moving past a point over elapsed time (

t

).

Identifying the relationship between electrical charge and time determines the correct physical quantity.

2

Match the formula

I = \frac{Q}{t}

to its standard electrical property.

The rate of charge flow per second defines electric current, measured in amperes (

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

).

Electric current is the formal definition of moving electric charge carriers through a conductor.

Key Concept

Electric current is the rate of flow of electric charge over time, expressed as

I = \frac{Q}{t}

and measured in amperes.

Estimated Time:35s

Question 755Question →

An electric heating element is rated to dissipate $1.8\text{ kW}$ of power when connected across a standard $120\text{ V}$ DC supply line. If a control circuit reduces the direct current flowing through the heating element to $5\text{ A}$ while its internal resistance remains constant, what is the new power dissipated by the element?

600\text{ W}

200\text{ W}

40\text{ W}

0.2\text{ W}

Show answer & explanation

Answer: $200\text{ W}$

Answer

200\text{ W}

The correct answer of

200\text{ W}

is found by first converting

1.8\text{ kW}

to

1800\text{ W}

. Using the initial ratings, the element's resistance is

R = \frac{V^2}{P} = \frac{120^2}{1800} = 8\text{ }\Omega

. With the current controlled to

5\text{ A}

, the dissipated power is

P = I^2 R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

.

Step-by-Step Solution

1

Convert the rated power from kilowatts to watts

P_1 = 1.8\text{ kW} = 1800\text{ W}

Metric prefixes must be converted to base SI units before performing Watt's Law calculations.

2

Calculate the constant internal resistance (

R

) of the heating element using the initial rating

R = \frac{V_1^2}{P_1} = \frac{(120\text{ V})^2}{1800\text{ W}} = \frac{14400}{1800} = 8\text{ }\Omega

The heating element's resistance remains unchanged and serves as the bridge to calculate power under the new operating current.

3

Calculate the new power dissipation with the reduced current (

I_2 = 5\text{ A}

)

P_2 = I_2^2 \times R = (5\text{ A})^2 \times 8\text{ }\Omega = 25 \times 8 = 200\text{ W}

Watt's Law specifies that power dissipation across a known resistance is directly proportional to the square of the current.

Key Concept

Watt's Law Power and Resistance Relationships (

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

and

P = I^2 R

)

Alternative Method

Determine the initial current

I_1 = \frac{P_1}{V_1} = \frac{1800\text{ W}}{120\text{ V}} = 15\text{ A}

. Since

P \propto I^2

for a constant resistance, scaling the power by the ratio of the currents squared gives

P_2 = P_1 \times \left(\frac{I_2}{I_1}\right)^2 = 1800\text{ W} \times \left(\frac{5\text{ A}}{15\text{ A}}\right)^2 = 1800 \times \left(\frac{1}{3}\right)^2 = \frac{1800}{9} = 200\text{ W}

.

Estimated Time:1m 30s

Question 756Question →

A technician needs to wire a single DC power source so that an operator can toggle power between an amber standby indicator and a green run indicator, ensuring only one light is energized at any given time. Which switch type should be selected?

Double-Pole, Single-Throw (DPST) switch

Single-Pole, Double-Throw (SPDT) switch

Single-Pole, Single-Throw (SPST) switch

Inline thermal circuit breaker

Show answer & explanation

Answer: Single-Pole, Double-Throw (SPDT) switch

Answer

Single-Pole, Double-Throw (SPDT) switch

A Single-Pole, Double-Throw (SPDT) switch features one common movable contact (the pole) that connects to either of two fixed output contacts (the throws). When flipped, it breaks contact with the first load and makes contact with the second, allowing selective operation between two indicators.

Step-by-Step Solution

1

Determine the number of input circuits (poles) required.

One input circuit (Single-Pole)

The circuit uses a single DC power feed connected to the switch's common terminal.

2

Determine the number of output positions (throws) required.

Two output paths (Double-Throw)

The operator must alternate power between two distinct loads (amber lamp and green lamp).

3

Combine the pole and throw specifications to identify the proper component.

SPDT switch

A single-pole, double-throw switch routes one common line to one of two selectable output terminals.

Key Concept

Switch Poles and Throws Nomenclature

Estimated Time:45s

Question 757Question →

In a test circuit, a $1.2\text{ k}\Omega$ resistor is connected across an adjustable DC power supply. When the supply voltage is increased by $18\text{ V}$ , the current through the resistor reaches $35\text{ mA}$ . What was the initial current flowing through the resistor in milliamperes ( $\text{mA}$ )?

Show answer & explanation

Answer: 20

Answer

The initial current flowing through the resistor was

20\text{ mA}

.

Applying Ohm's Law (

I = \frac{V}{R}

), an

18\text{ V}

voltage increase across a

1.2\text{ k}\Omega

(

1{}200\ \Omega

) resistor causes a current increase of

\Delta I = \frac{18\text{ V}}{1{}200\ \Omega} = 0.015\text{ A} = 15\text{ mA}

. Subtracting this

15\text{ mA}

change from the final current of

35\text{ mA}

yields an initial current of

20\text{ mA}

.

Step-by-Step Solution

1

Convert resistance from kilohms (

k\Omega

) to ohms (

\Omega

).

R = 1.2\text{ k}\Omega = 1{}200\ \Omega

Ohm's Law calculations require standard base SI units (

V

in volts,

I

in amperes,

R

in ohms).

2

Calculate the increase in current (

\Delta I

) resulting from the

18\text{ V}

voltage increase.

\Delta I = \frac{\Delta V}{R} = \frac{18\text{ V}}{1{}200\ \Omega} = 0.015\text{ A} = 15\text{ mA}

According to Ohm's Law (

I = \frac{V}{R}

), an increase in voltage produces a directly proportional increase in current across a constant resistance.

3

Find the initial current by subtracting the current increase from the final current.

I_{\text{initial}} = 35\text{ mA} - 15\text{ mA} = 20\text{ mA}

The final current represents the sum of the initial current and the incremental current produced by the added voltage.

Key Concept

Ohm's Law defines the direct relationship between voltage and current across a constant resistance (

I = \frac{V}{R}

), meaning a change in voltage causes a proportional change in current.

Alternative Method

Calculate the final voltage first:

V_{\text{final}} = I_{\text{final}} \times R = 0.035\text{ A} \times 1{}200\ \Omega = 42\text{ V}

. Then subtract the

18\text{ V}

increase to find the initial voltage:

V_{\text{initial}} = 42\text{ V} - 18\text{ V} = 24\text{ V}

. Finally, compute the initial current:

I_{\text{initial}} = \frac{24\text{ V}}{1{}200\ \Omega} = 0.020\text{ A} = 20\text{ mA}

.

Estimated Time:50s

Question 758Question →

Match each capacitor circuit application with its primary functional role in electronic systems.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Coupling Capacitor matches transferring AC signal voltages between amplifier stages while blocking DC bias levels; Filter Capacitor matches smoothing pulsating direct current output from a rectifier; Bypass Capacitor matches diverting AC noise or signals directly to ground around a component; Tuning Capacitor matches providing adjustable capacitance to set resonant frequency in RF circuits.

Each application corresponds directly to fundamental capacitive behaviors: coupling capacitors leverage DC-blocking and AC-passing in series with the signal path; filter capacitors utilize charge storage to smooth rectified DC voltage ripple; bypass capacitors create a low-reactance parallel path to ground for AC signals; and tuning capacitors provide variable capacitance for frequency selection in resonant LC circuits.

Step-by-Step Solution

1

Determine the function of a coupling capacitor in multi-stage signal amplification.

Coupling capacitors block DC bias voltages while passing AC information signals between amplifier stages.

Capacitors provide infinite opposition (reactance) to zero-frequency DC while exhibiting low capacitive reactance to high-frequency AC signals.

2

Determine the function of a filter capacitor in a DC power supply.

Filter capacitors reduce voltage ripple by charging at peak rectified voltages and discharging during valleys.

This energy storage smooths pulsating DC output into a steady DC voltage for electronic circuits.

3

Determine the function of a bypass capacitor across a circuit element.

Bypass capacitors shunt alternating current signals to ground around a component without disturbing the DC operating point.

The capacitor provides an alternate low-reactance path for AC currents.

4

Determine the function of a tuning capacitor in radio frequency systems.

Tuning capacitors vary circuit capacitance to alter the resonant frequency of an LC circuit.

Resonant frequency varies inversely with the square root of capacitance, allowing station or frequency selection.

Key Concept

Operational roles and circuit applications of capacitors across DC power filtering, signal coupling, AC bypassing, and RF tuning

Estimated Time:1m 0s

Question 759Question →

A technician connects a diagnostic tachometer to a running four-stroke automotive engine and measures a crankshaft rotational speed of $2{}400\text{ RPM}$ . Under these conditions, at what rotational speed does the engine's camshaft spin?

2{}400\text{ RPM}

, because the camshaft and crankshaft maintain a direct one-to-one rotational speed ratio

1{}200\text{ RPM}

, because the camshaft rotates at one-half the speed of the crankshaft

4{}800\text{ RPM}

, because the camshaft must rotate at twice the speed of the crankshaft to operate each valve

600\text{ RPM}

, because the camshaft completes only one rotation for every four complete crankshaft cycles

Show answer & explanation

Answer: $1{}200\text{ RPM}$ , because the camshaft rotates at one-half the speed of the crankshaft

Answer

The camshaft rotates at

1{}200\text{ RPM}

because it turns at half the speed of the crankshaft in a four-stroke engine.

In a four-stroke engine, each cylinder requires two complete crankshaft rotations (

720^\circ

) to complete all four strokes. Because each intake and exhaust valve opens only once per four-stroke cycle, the camshaft must turn at exactly half the speed of the crankshaft. Dividing the crankshaft speed of

2{}400\text{ RPM}

by

2

gives a camshaft speed of

1{}200\text{ RPM}

.

Step-by-Step Solution

1

Determine the rotational relationship between the crankshaft and camshaft in a four-stroke cycle.

A complete four-stroke cycle (intake, compression, power, and exhaust) requires two full rotations of the crankshaft (

720^\circ

). During this complete cycle, each valve must open exactly once, meaning the camshaft completes one full rotation for every two crankshaft rotations (a

1:2

speed ratio).

Establishing the mechanical ratio is required to calculate the valvetrain operating speed.

2

Calculate the camshaft rotational speed using the established

1:2

ratio.

\text{Camshaft RPM} = \frac{\text{Crankshaft RPM}}{2} = \frac{2{}400\text{ RPM}}{2} = 1{}200\text{ RPM}

.

Dividing the crankshaft RPM by two provides the exact rotational speed of the camshaft.

Key Concept

In a four-stroke engine, a complete operational cycle requires two full crankshaft revolutions (

720^\circ

) for every one camshaft revolution (

360^\circ

), establishing a fixed

1:2

speed ratio.

Estimated Time:45s

Question 760Question →

An automotive technician tests a lighting circuit consisting of two auxiliary lamps wired in parallel. The resistance of the first lamp is $10\ \Omega$ , and the resistance of the second lamp is $15\ \Omega$ . What is the total equivalent resistance of this parallel circuit?

25\ \Omega

6\ \Omega

12.5\ \Omega

5\ \Omega

Show answer & explanation

Answer: $6\ \Omega$

Answer

The total equivalent resistance of the parallel circuit is

6\ \Omega

.

The equivalent resistance of two resistors connected in parallel is calculated using

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

. Substituting

10\ \Omega

and

15\ \Omega

gives

\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega

.

Step-by-Step Solution

1

Identify the formula for 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}

Resistors in parallel provide multiple paths for current, resulting in an equivalent resistance smaller than any individual branch.

2

Substitute the given resistances (

R_1 = 10\ \Omega

and

R_2 = 15\ \Omega

) into the product-over-sum formula

R_{\text{eq}} = \frac{10 \times 15}{10 + 15} = \frac{150}{25}

Using the product-over-sum shortcut provides a direct calculation for two parallel branches.

3

Calculate the final equivalent resistance

R_{\text{eq}} = 6\ \Omega

Dividing

150

by

25

yields the total resistance of

6\ \Omega

.

Key Concept

Equivalent resistance in a parallel circuit is always less than the smallest individual branch resistance and is calculated using the reciprocal or product-over-sum rule.

Estimated Time:45s

Answer

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