All practice questions

Question 141Question →

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 142Question →

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 143Question →

An electronics technician connects three inductors in parallel across an AC filter circuit. The inductors have inductance values of $60\text{ mH}$ , $30\text{ mH}$ , and $20\text{ mH}$ , with negligible mutual inductance between them. What is the total equivalent inductance of the circuit in millihenries (mH)?

Show answer & explanation

Answer: 10

Answer

The total equivalent inductance of the parallel circuit is

10\text{ mH}

.

Inductors in parallel without mutual coupling combine according to the reciprocal formula

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

. Computing

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

, and inverting yields

10\text{ mH}

.

Step-by-Step Solution

1

State the reciprocal formula for inductors connected in parallel

\frac{1}{L_{\text{total}}} = \frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{L_3}

When inductors are in parallel and uncoupled, the total rate of current change is the sum of the individual branch rates, resulting in a reciprocal combination rule identical to parallel resistors.

2

Substitute the individual inductance values

\frac{1}{L_{\text{total}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{20}

Direct substitution of the known branch values:

L_1 = 60\text{ mH}

,

L_2 = 30\text{ mH}

, and

L_3 = 20\text{ mH}

.

3

Find the common denominator and sum the terms

\frac{1}{L_{\text{total}}} = \frac{1 + 2 + 3}{60} = \frac{6}{60} = \frac{1}{10}\text{ mH}^{-1}

Using the lowest common denominator (

60

) to perform fractional addition.

4

Invert the reciprocal to obtain equivalent inductance

L_{\text{total}} = 10\text{ mH}

Taking the reciprocal of

\frac{1}{10}\text{ mH}^{-1}

gives

10\text{ mH}

.

Key Concept

Parallel Inductance Calculation

Estimated Time:1m 0s

Question 144Question →

A quality assurance technician is inspecting a four-band carbon-composition resistor with the color bands in order: Yellow, Violet, Red, and Silver. To meet quality standards, the resistor must operate within its specified manufacturer tolerance. What is the minimum acceptable resistance value, in ohms ( $\Omega$ ), for this component?

Show answer & explanation

Answer: 4230

Answer

The minimum acceptable resistance value is

4230\text{ }\Omega

.

The first two bands (Yellow = 4, Violet = 7) establish the significant figures of 47. The third band (Red) indicates a multiplier of

10^2 = 100

, giving a nominal rating of

4700\text{ }\Omega

. The fourth band (Silver) specifies a

\pm 10\%

tolerance. Subtracting

10\%

of

4700\text{ }\Omega

(

470\text{ }\Omega

) from the nominal rating gives the minimum allowable threshold of

4230\text{ }\Omega

.

Step-by-Step Solution

1

Decode the first two color bands to obtain the significant numerical digits.

Yellow represents the digit

4

, and Violet represents the digit

7

, forming the base value of

47

.

Standard EIA four-band resistor color coding assigns the first two bands as significant figures of the resistance value.

2

Multiply the base significant figures by the power of ten indicated by the third band.

Red indicates a multiplier of

10^2

(

100

), so the nominal resistance is

47 \times 100 = 4700\text{ }\Omega

.

The third band functions as the decimal multiplier for four-band fixed resistors.

3

Identify the tolerance rating from the fourth band.

Silver represents a tolerance of

\pm 10\%

.

The fourth band specifies the allowable deviation range from the nominal resistance value.

4

Calculate the lower allowable resistance threshold.

Tolerance amount is

4700\text{ }\Omega \times 0.10 = 470\text{ }\Omega

. Minimum acceptable value is

4700\text{ }\Omega - 470\text{ }\Omega = 4230\text{ }\Omega

.

The minimum acceptable value represents the nominal value minus the total tolerance deviation.

Key Concept

Four-Band Resistor Color Code Decoding and Tolerance Range Calculation

Alternative Method

Multiply the nominal resistance directly by the minimum percentage factor:

4700\text{ }\Omega \times (1.00 - 0.10) = 4700\text{ }\Omega \times 0.90 = 4230\text{ }\Omega

.

Estimated Time:45s

Question 145Question →

A specialized weather station telemetry circuit is powered by a $96\text{ V}$ direct current (DC) power source connected in series with four resistive components: a surge-limiting resistor $R_1 = 18\ \Omega$ , a status indicator $R_2 = 27\ \Omega$ , a variable humidity sensor $R_3$ , and a reference resistor $R_4 = 15\ \Omega$ . A technician measures the voltage drop across the status indicator ( $R_2$ ) to be $21.6\text{ V}$ . What is the resistance of the humidity sensor $R_3$ in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 60

Answer

The resistance of the humidity sensor

R_3

is

60\ \Omega

.

In a series circuit, current is uniform throughout the entire loop. Using the measured voltage drop across

R_2

(

21.6\text{ V}

) and its known resistance (

27\ \Omega

), the loop current is

I = \frac{V_2}{R_2} = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}

. The total equivalent resistance connected across the

96\text{ V}

source is

R_T = \frac{V_{\text{source}}}{I} = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega

. Because series resistances are directly additive (

R_T = R_1 + R_2 + R_3 + R_4

), the unknown resistance is determined by subtracting the known values from the total:

R_3 = 120\ \Omega - (18\ \Omega + 27\ \Omega + 15\ \Omega) = 120\ \Omega - 60\ \Omega = 60\ \Omega

.

Step-by-Step Solution

1

Calculate loop current from the known component voltage drop and resistance.

I = \frac{21.6\text{ V}}{27\ \Omega} = 0.8\text{ A}

In a series circuit, current is identical through all components.

2

Calculate total circuit resistance using Ohm's Law and the source voltage.

R_T = \frac{96\text{ V}}{0.8\text{ A}} = 120\ \Omega

Total resistance equals source voltage divided by total circuit current.

3

Calculate the sum of all known resistances.

R_{\text{known}} = 18\ \Omega + 27\ \Omega + 15\ \Omega = 60\ \Omega

Resistances in a series circuit are directly additive.

4

Subtract the sum of known resistors from total circuit resistance to isolate

R_3

.

R_3 = 120\ \Omega - 60\ \Omega = 60\ \Omega

R_T = R_1 + R_2 + R_3 + R_4

, so

R_3 = R_T - (R_1 + R_2 + R_4)

.

Key Concept

Series Circuit Loop Current and Total Resistance Analysis

Alternative Method

Calculate individual voltage drops using loop current

I = 0.8\text{ A}

:

V_1 = 0.8\text{ A} \times 18\ \Omega = 14.4\text{ V}

and

V_4 = 0.8\text{ A} \times 15\ \Omega = 12.0\text{ V}

. By Kirchhoff's Voltage Law, the voltage across

R_3

is

V_3 = 96\text{ V} - (14.4\text{ V} + 21.6\text{ V} + 12.0\text{ V}) = 48.0\text{ V}

. Applying Ohm's Law directly to

R_3

yields

R_3 = \frac{V_3}{I} = \frac{48.0\text{ V}}{0.8\text{ A}} = 60\ \Omega

.

Estimated Time:1m 30s

Question 146Question →

A continuous spool of aluminum hookup wire has a total length of $80\text{ meters}$ and a measured electrical resistance of $2.4\ \Omega$ . If an electronics technician cuts a $20\text{-meter}$ piece from this spool for a project, what is the electrical resistance of the cut piece in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 0.6

Answer

0.6 Ω

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

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

). Since

20\text{ m}

is one-fourth (

\frac{1}{4}

or

0.25

) of the original

80\text{ m}

length, the resistance of the cut piece is one-fourth of the total resistance:

2.4\ \Omega \times 0.25 = 0.6\ \Omega

.

Step-by-Step Solution

1

Identify the physical relationship governing wire resistance and length.

Resistance is directly proportional to length:

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

, which means

R_2 = R_1 \times \frac{L_2}{L_1}

.

Because the material (resistivity

\rho

) and cross-sectional area (

A

) remain constant, resistance scales directly with length.

2

Find the proportion of the new length relative to the original length.

\frac{L_{\text{cut}}}{L_{\text{total}}} = \frac{20\text{ m}}{80\text{ m}} = 0.25

This establishes what fraction of the total opposition to electron flow the shorter segment represents.

3

Calculate the resistance of the 20-meter wire segment.

R_{\text{cut}} = 2.4\ \Omega \times 0.25 = 0.6\ \Omega

Multiplying the total resistance by the length ratio yields the exact resistance of the cut section.

Key Concept

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

R \propto L

) when cross-sectional area and material remain constant.

Estimated Time:45s

Question 147Question →

A structural mechanic is testing the tensile strength of a batch of aluminum rivets. The mean tensile strength of the first $4$ rivets tested is $410$ MPa. If the quality standard requires the mean strength of $5$ rivets to be exactly $415$ MPa, what tensile strength must the $5$ th rivet have?

Show answer & explanation

Answer: 435

Answer

The 5th rivet must have a tensile strength of 435 MPa.

The correct answer is derived by finding the total required sum for 5 rivets (2075) and subtracting the existing sum of the first 4 rivets (1640).

Step-by-Step Solution

1

Calculate the total tensile strength of the first 4 rivets.

1640 MPa

The sum of a data set is equal to its mean multiplied by the number of values.

2

Calculate the required total tensile strength for all 5 rivets.

2075 MPa

To achieve a target mean of 415 MPa with 5 items, the overall sum must equal the new mean multiplied by the new count.

3

Find the difference between the target total and the current total.

435 MPa

Subtracting the sum of the first 4 items from the required sum of all 5 items isolates the value needed for the 5th item.

Key Concept

Calculating a missing data value given an initial mean and a target mean.

Alternative Method

Use the deviation method: The first 4 rivets have a mean of 410 MPa, which is 5 MPa below the target mean of 415 MPa. This creates a total deficit of 4 × 5 = 20 MPa. The 5th rivet must provide enough strength to cover its own requirement (415 MPa) plus the 20 MPa deficit from the others. Therefore, 415 + 20 = 435 MPa.

Estimated Time:1m 0s

Question 148Question →

A DC cooling fan installed in an equipment enclosure has an internal resistance of $16\ \Omega$ and draws a steady current of $1.5\text{ A}$ during normal operation. According to Ohm's law, what is the voltage drop across the fan in volts ( $\text{V}$ )?

Show answer & explanation

Answer: 24

Answer

24 V

According to Ohm's law, voltage (

V

) is found by multiplying current (

I

) by resistance (

R

). With a current of

1.5\text{ A}

and a resistance of

16\ \Omega

, the voltage is

1.5 \times 16 = 24\text{ V}

.

Step-by-Step Solution

1

Identify the known electrical quantities from the problem statement.

I = 1.5\text{ A}

and

R = 16\ \Omega

Ohm's law requires current and resistance values to calculate voltage.

2

State Ohm's law formula arranged to solve for voltage (

V

).

V = I \times R

The potential difference across a resistive element is directly proportional to the product of current and resistance.

3

Multiply the current by the resistance to find the voltage.

V = 1.5\text{ A} \times 16\ \Omega = 24\text{ V}

Performing the multiplication yields the potential difference in volts.

Key Concept

Ohm's Law (

V = I \times R

)

Estimated Time:30s

Question 149Question →

A diesel generator preheat circuit is powered by a $24\text{ V}$ DC source connected in series with a ballast resistor $R_1 = 1.8\ \Omega$ and two identical glow plugs ( $R_2$ and $R_3$ ). If an ammeter in the circuit measures a steady current of $4.0\text{ A}$ , what is the resistance of a single glow plug in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 2.1

Answer

The resistance of a single glow plug is

2.1\ \Omega

.

Applying Ohm's Law gives a total series loop resistance of

R_{\text{total}} = 24\text{ V} / 4.0\text{ A} = 6.0\ \Omega

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

R_{\text{total}} = R_1 + R_2 + R_3

). Subtracting the ballast resistor (

1.8\ \Omega

) leaves

4.2\ \Omega

across the two identical glow plugs. Dividing

4.2\ \Omega

equally between them gives

2.1\ \Omega

per glow plug.

Step-by-Step Solution

1

Calculate the total circuit resistance using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{24\text{ V}}{4.0\text{ A}} = 6.0\ \Omega

In a closed DC loop, total resistance equals total source voltage divided by total current.

2

Subtract the known ballast resistance from the total resistance to determine the combined resistance of both glow plugs.

R_2 + R_3 = R_{\text{total}} - R_1 = 6.0\ \Omega - 1.8\ \Omega = 4.2\ \Omega

Resistances in a series circuit are directly additive (

R_{\text{total}} = R_1 + R_2 + R_3

).

3

Divide the combined glow plug resistance equally between the two identical units.

R_{\text{plug}} = \frac{4.2\ \Omega}{2} = 2.1\ \Omega

Because the two glow plugs are identical, each provides exactly half of the remaining series resistance.

Key Concept

Series Total Resistance and Ohm's Law

Estimated Time:1m 0s

Question 150Question →

A technician is testing a direct current (DC) cooling fan motor installed in an equipment rack. When powered by a $48\text{ V}$ DC supply, the motor draws a steady current of $250\text{ mA}$ . What is the electrical resistance of the cooling fan motor in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 192

Answer

The electrical resistance of the cooling fan motor is

192\text{ }\Omega

.

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

R = \frac{V}{I}

). Converting

250\text{ mA}

to

0.25\text{ A}

and dividing

48\text{ V}

by

0.25\text{ A}

gives exactly

192\text{ }\Omega

.

Step-by-Step Solution

1

Convert the current from milliamperes to amperes

I = 250\text{ mA} = 0.25\text{ A}

Ohm's law calculations require current to be expressed in the base SI unit of amperes.

2

Calculate resistance using Ohm's law formula (

R = \frac{V}{I}

)

R = \frac{48\text{ V}}{0.25\text{ A}} = 192\text{ }\Omega

Dividing the supply voltage by the total circuit current yields the resistance in ohms.

Key Concept

Calculating resistance using Ohm's law (

R = \frac{V}{I}

) combined with metric prefix conversion from milliamperes to amperes.

Estimated Time:1m 0s

Question 151Question →

A $48\text{ V}$ direct-current telecommunications power bus supplies electricity to three parallel branches. The first branch draws a current of $2\text{ A}$ , the second branch contains a load with a resistance of $24\ \Omega$ , and the total current supplied by the power bus is $7\text{ A}$ . What is the electrical resistance of the third branch, in ohms?

Show answer & explanation

Answer: 16

Answer

The resistance of the third branch is

16\ \Omega

.

Because the circuit is parallel, the full supply voltage of

48\text{ V}

is present across every branch. The second branch with

24\ \Omega

draws

I_2 = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

. Subtracting the known currents of Branch 1 (

2\text{ A}

) and Branch 2 (

2\text{ A}

) from the total current of

7\text{ A}

leaves

3\text{ A}

flowing through Branch 3. Using Ohm's Law,

R_3 = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

.

Step-by-Step Solution

1

Calculate the current flowing through Branch 2

I_2 = \frac{V}{R_2} = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

In a parallel circuit, the voltage across each parallel branch is identical to the supply voltage (

V_1 = V_2 = V_3 = 48\text{ V}

).

2

Calculate the current flowing through Branch 3

I_3 = I_{\text{total}} - I_1 - I_2 = 7\text{ A} - 2\text{ A} - 2\text{ A} = 3\text{ A}

The total current entering a parallel network equals the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + I_3

).

3

Calculate the resistance of Branch 3

R_3 = \frac{V}{I_3} = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) using the common

48\text{ V}

supply voltage and the calculated

3\text{ A}

branch current yields the branch resistance.

Key Concept

In a parallel DC circuit, voltage remains constant across all branches, total current is the algebraic sum of individual branch currents, and branch resistance is determined by Ohm's Law (

R = \frac{V}{I}

).

Question 152Question →

A $24\text{ V}$ direct current (DC) power supply delivers a steady current of $3\text{ A}$ to a resistive heating element. What is the total electrical power consumed by the heating element in watts ( $\text{W}$ )?

Show answer & explanation

Answer: 72

Answer

The total electrical power consumed is

72\text{ W}

.

Electrical power in a direct current circuit is calculated using the formula

P = V \times I

. Substituting the given voltage of

24\text{ V}

and current of

3\text{ A}

yields

P = 24 \times 3 = 72\text{ W}

.

Step-by-Step Solution

1

Identify the given values from the circuit problem

Voltage

V = 24\text{ V}

and current

I = 3\text{ A}

Electrical power calculations require knowing at least two circuit parameters.

2

Select the appropriate electrical power formula

P = V \times I

Watt's Law defines electrical power as the product of voltage across a component and the current flowing through it.

3

Substitute the values and calculate the final result

P = 24\text{ V} \times 3\text{ A} = 72\text{ W}

Multiplying volts by amperes gives power directly in watts.

Key Concept

Electrical power is the rate at which electrical energy is transferred in a circuit, calculated using Watt's Law (

P = V \times I

).

Estimated Time:45s

Question 153Question →

A secure military facility uses three independent detection systems to monitor its perimeter. The probability that the primary radar detects an incoming drone is 0.80. The probability that the infrared system detects it is 0.90. The probability that the acoustic sensors detect it is 0.75. If a drone approaches the perimeter, what is the probability that exactly two of the three systems detect it? Express your answer as a decimal.

Show answer & explanation

Answer: 0.375

Answer

0.375

The correct answer is found by calculating the compound probability for each of the three specific scenarios where exactly two systems succeed and one fails, and then adding those mutually exclusive probabilities together.

Step-by-Step Solution

1

Identify the mutually exclusive scenarios that result in exactly two systems detecting the drone.

Scenario 1: Radar and Infrared detect, Acoustic fails. Scenario 2: Radar and Acoustic detect, Infrared fails. Scenario 3: Infrared and Acoustic detect, Radar fails.

Because these three systems act independently, we must calculate the compound probability of each specific combination of two successes and one failure.

2

Calculate the probability of Scenario 1 (Radar and Infrared detect, Acoustic fails).

0.80 \times 0.90 \times (1 - 0.75) = 0.80 \times 0.90 \times 0.25 = 0.180

Multiply the probabilities of the two specific successful detections by the probability of the specific failure.

3

Calculate the probability of Scenario 2 (Radar and Acoustic detect, Infrared fails).

0.80 \times (1 - 0.90) \times 0.75 = 0.80 \times 0.10 \times 0.75 = 0.060

Multiply the probabilities of the two specific successful detections by the probability of the specific failure.

4

Calculate the probability of Scenario 3 (Infrared and Acoustic detect, Radar fails).

(1 - 0.80) \times 0.90 \times 0.75 = 0.20 \times 0.90 \times 0.75 = 0.135

Multiply the probabilities of the two specific successful detections by the probability of the specific failure.

5

Sum the probabilities of the three mutually exclusive scenarios.

0.180 + 0.060 + 0.135 = 0.375

To find the total probability of any of these mutually exclusive events occurring, add their individual joint probabilities together.

Key Concept

Calculating the probability of an exact number of successes among independent events by identifying all possible combinations of success and failure.

Estimated Time:2m 30s

Question 154Question →

An oscilloscope displays a periodic alternating voltage waveform on its screen. One complete cycle of the waveform occupies $4\text{ horizontal divisions}$ across the graticule. If the oscilloscope's horizontal time base (Time/Div) control is set to $15\text{ ms/div}$ , what is the period of the waveform in milliseconds?

Show answer & explanation

Answer: 60

Answer

60\text{ ms}

The period (

T

) of a signal displayed on an oscilloscope is found by multiplying the number of horizontal divisions for one complete cycle by the Time/Div control setting:

4\text{ divisions} \times 15\text{ ms/div} = 60\text{ ms}

.

Step-by-Step Solution

1

Identify the number of horizontal graticule divisions occupied by one full waveform cycle.

4\text{ divisions}

On an oscilloscope display, the horizontal axis represents time, and one full cycle corresponds to the period (

T

).

2

Multiply the division count by the horizontal scale factor (

15\text{ ms/div}

).

4\text{ div} \times 15\text{ ms/div} = 60\text{ ms}

Multiplying the span in divisions by the time allocated per division gives the total duration of one cycle.

Key Concept

Calculating waveform period from oscilloscope horizontal division settings

Estimated Time:45s

Question 155Question →

A $36\text{ V}$ DC control circuit supplies power to a single heating element. An inline ammeter measures a steady current flow of $3\text{ A}$ through the circuit. What is the electrical resistance of the heating element in ohms?

Show answer & explanation

Answer: 12

Answer

12 ohms

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

R = \frac{V}{I}

). Dividing the source voltage of

36\text{ V}

by the current of

3\text{ A}

yields

12\ \Omega

.

Step-by-Step Solution

1

Identify the known electrical values provided in the problem

Voltage (

V

) =

36\text{ V}

, Current (

I

) =

3\text{ A}

Ohm's Law relates potential difference, current flow, and circuit resistance.

2

Rearrange Ohm's Law to express resistance in terms of voltage and current

R = \frac{V}{I}

Resistance is defined as the ratio of voltage across a component to the current flowing through it.

3

Substitute the numerical values into the equation and solve

R = \frac{36\text{ V}}{3\text{ A}} = 12\ \Omega

Dividing 36 by 3 gives the resistance directly in ohms.

Key Concept

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance (

V = I \times R

, which rearranges to

R = \frac{V}{I}

).

Estimated Time:30s

Question 156Question →

An electronics technician is assembling a tuning filter network. The circuit consists of two inductors connected in series with values of $18\text{ mH}$ and $42\text{ mH}$ . This series combination is then connected in parallel with a third inductor rated at $40\text{ mH}$ . Assuming there is no mutual inductance between the coils, what is the total equivalent inductance of the circuit in millihenries ( $\text{mH}$ )?

Show answer & explanation

Answer: 24

Answer

The total equivalent inductance of the circuit is

24\text{ mH}

.

Combining the two series inductors (

18\text{ mH} + 42\text{ mH}

) yields an equivalent branch inductance of

60\text{ mH}

. Connecting this branch in parallel with the

40\text{ mH}

inductor gives a total inductance of

\frac{60 \times 40}{60 + 40} = \frac{2400}{100} = 24\text{ mH}

.

Step-by-Step Solution

1

Calculate the equivalent inductance of the series branch (

L_{\text{series}}

).

L_{\text{series}} = 18\text{ mH} + 42\text{ mH} = 60\text{ mH}

Inductors in series combine by direct addition, similar to series resistors (

L_{\text{series}} = L_1 + L_2

).

2

Calculate the total inductance of the series branch in parallel with the third inductor.

L_{\text{total}} = \frac{60\text{ mH} \times 40\text{ mH}}{60\text{ mH} + 40\text{ mH}} = \frac{2400}{100} = 24\text{ mH}

Inductors in parallel combine using the reciprocal or product-over-sum formula (

L_{\text{total}} = \frac{L_A \times L_B}{L_A + L_B}

).

Key Concept

Calculation of total inductance for series-parallel inductor combinations without mutual coupling.

Estimated Time:1m 15s

Question 157Question →

An auxiliary power transformer on a vessel steps down an input voltage of $440\text{ V AC}$ to $110\text{ V AC}$ . If the secondary winding is connected to a lighting circuit drawing $14\text{ A}$ , what is the primary current in amperes ( $\text{A}$ ), assuming an ideal transformer with zero power losses?

Show answer & explanation

Answer: 3.5

Answer

The primary current is

3.5\text{ A}

.

In an ideal transformer with negligible losses, the primary apparent power equals the secondary apparent power (

V_p \times I_p = V_s \times I_s

). Because the transformer reduces voltage from

440\text{ V AC}

to

110\text{ V AC}

(a

4:1

step-down ratio), the primary winding requires only one-fourth of the secondary current. Multiplying

14\text{ A}

by

\frac{110}{440}

yields

3.5\text{ A}

.

Step-by-Step Solution

1

State the ideal transformer power balance equation

V_p \times I_p = V_s \times I_s

An ideal transformer operates at 100% efficiency, meaning input power (

P_p

) equals output power (

P_s

).

2

Isolate the unknown primary current variable (

I_p

)

I_p = I_s \times \left(\frac{V_s}{V_p}\right)

Dividing both sides by the primary voltage (

V_p

) provides the direct formula for primary current.

3

Substitute the given numerical values and calculate

I_p = 14\text{ A} \times \left(\frac{110\text{ V}}{440\text{ V}}\right) = 14 \times \frac{1}{4} = 3.5\text{ A}

The voltage is stepped down by a factor of

4

, which requires the current to be stepped down by the inverse factor of

4

in the primary circuit relative to the secondary.

Key Concept

Transformer Voltage and Current Transformation Ratio

Estimated Time:1m 0s

Question 158Question →

A workshop bench grinder is connected to a $120\text{ V}$ electrical supply line and draws a steady current of $5\text{ A}$ . What is the total electrical power consumed by the grinder in watts ( $\text{W}$ )?

Show answer & explanation

Answer: 600

Answer

600 W

Electrical power is calculated using the formula

P = V \times I

, where

P

is power in watts,

V

is electromotive force in volts, and

I

is current in amperes. Substituting the given values gives

P = 120\text{ V} \times 5\text{ A} = 600\text{ W}

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage

V = 120\text{ V}

and Current

I = 5\text{ A}

Determining the known values is necessary to select the correct power formula.

2

Apply Watt's Law formula for power

P = V \times I

Watt's Law defines electrical power in watts as the product of potential difference in volts and current in amperes.

3

Calculate the total power

P = 120\text{ V} \times 5\text{ A} = 600\text{ W}

Multiplying voltage by current yields the total power consumed by the electrical device.

Key Concept

Electrical power (

P

) in a circuit is the rate at which electrical energy is converted into another form of energy, calculated using Watt's Law as

P = V \times I

.

Estimated Time:45s

Question 159Question →

A linear potentiometer has a total resistance of $20\text{ k}\Omega$ between its two outer terminals (Terminal 1 and Terminal 3). If the center wiper (Terminal 2) is adjusted so that the resistance between Terminal 1 and Terminal 2 is $6.5\text{ k}\Omega$ , what is the resistance in kilohms ( $\text{k}\Omega$ ) between Terminal 2 and Terminal 3?

Show answer & explanation

Answer: 13.5

Answer

The resistance between Terminal 2 and Terminal 3 is

13.5\text{ k}\Omega

.

In a three-terminal potentiometer, the resistive element connects between the two outer terminals (1 and 3), providing a fixed total resistance of

20\text{ k}\Omega

. The adjustable wiper (terminal 2) contacts this element and divides it into two series portions such that

R_{1-2} + R_{2-3} = R_{1-3}

. Subtracting the known section (

6.5\text{ k}\Omega

) from the total (

20\text{ k}\Omega

) gives

13.5\text{ k}\Omega

for the remaining section between Terminal 2 and Terminal 3.

Step-by-Step Solution

1

Set up the resistance relationship for the three terminals of a potentiometer.

R_{\text{total}} = R_{1-2} + R_{2-3}

The wiper divides the continuous internal resistive strip into two series sections whose resistances add up to the total end-to-end resistance.

2

Subtract the resistance of the first section from the total end-to-end resistance.

R_{2-3} = 20\text{ k}\Omega - 6.5\text{ k}\Omega = 13.5\text{ k}\Omega

Isolating the unknown segment gives the remaining resistance along the track to the opposite outer terminal.

Key Concept

Potentiometer Terminal Resistance Division

Estimated Time:30s

Question 160Question →

A direct-current (DC) power supply maintains a constant potential difference of $15\text{ V}$ across a circuit resistor that dissipates $4.5\text{ W}$ of electrical power. What is the electrical resistance of the resistor in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 50

Answer

The resistance of the resistor is

50\ \Omega

.

According to Watt's Law, power dissipated by a resistor with a known voltage across it is given by

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

. Solving for resistance gives

R = \frac{V^2}{P} = \frac{15^2}{4.5} = \frac{225}{4.5} = 50\ \Omega

.

Step-by-Step Solution

1

State the known values and identify the target variable.

Voltage

V = 15\text{ V}

, Power

P = 4.5\text{ W}

, Resistance

R = ?

Establishing the known electrical parameters determines the appropriate form of Watt's Law to apply.

2

Select and rearrange the Watt's Law power equation that relates power, voltage, and resistance.

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

Since voltage and power are known, using

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

allows direct solving for resistance.

3

Calculate the squared voltage.

V^2 = 15^2 = 225\text{ V}^2

Evaluating the numerator is the first calculation step in solving for

R

.

4

Divide the squared voltage by the power dissipation.

R = \frac{225}{4.5} = 50\ \Omega

Carrying out the division yields the final resistance value in ohms.

Key Concept

Watt's Law relates electrical power, voltage, and resistance through the formula

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

, which rearranges algebraically to

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

.

Alternative Method

First find the circuit current using

I = \frac{P}{V} = \frac{4.5\text{ W}}{15\text{ V}} = 0.3\text{ A}

. Then apply Ohm's Law to determine the resistance:

R = \frac{V}{I} = \frac{15\text{ V}}{0.3\text{ A}} = 50\ \Omega

.

Estimated Time:1m 0s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept