All practice questions

Question 161Question →

In a direct current (DC) avionics test circuit, a $48\text{ V}$ power supply is connected in series with a $30\text{ }\Omega$ current-limiting resistor and a $50\text{ }\Omega$ load resistor. What is the electrical power dissipated by the $50\text{ }\Omega$ load resistor, in watts?

Show answer & explanation

Answer: 18

Answer

The power dissipated by the

50\text{ }\Omega

load resistor is

18\text{ W}

.

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

30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega

). With a

48\text{ V}

supply, the circuit current is

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

. The electrical power dissipated by the

50\text{ }\Omega

resistor is determined by

P = I^2 R = (0.6\text{ A})^2 \times 50\text{ }\Omega = 18\text{ W}

.

Step-by-Step Solution

1

Find the equivalent total resistance of the series circuit

R_{\text{total}} = 30\text{ }\Omega + 50\text{ }\Omega = 80\text{ }\Omega

Resistances in a series circuit are directly additive to determine total opposition to current flow.

2

Calculate the total circuit current using Ohm's Law

I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{48\text{ V}}{80\text{ }\Omega} = 0.6\text{ A}

Current is uniform throughout all components connected in a series loop.

3

Calculate the power dissipation in the load resistor using Watt's Law

P = I^2 \times R = (0.6\text{ A})^2 \times 50\text{ }\Omega = 0.36\text{ A}^2 \times 50\text{ }\Omega = 18\text{ W}

Power dissipated by a resistor is proportional to the square of the current flowing through it multiplied by its resistance.

Key Concept

Series circuit current and resistor power dissipation using combined Ohm's Law and Watt's Law (

P = I^2 R

or

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

)

Alternative Method

Find the voltage drop across the load resistor using the voltage divider rule:

V_{\text{load}} = 48\text{ V} \times \left(\frac{50\text{ }\Omega}{30\text{ }\Omega + 50\text{ }\Omega}\right) = 48 \times \frac{5}{8} = 30\text{ V}

. Then compute power using

P = \frac{V_{\text{load}}^2}{R_{\text{load}}} = \frac{30^2}{50} = \frac{900}{50} = 18\text{ W}

.

Estimated Time:1m 15s

Question 162Question →

A railway signal monitoring circuit is powered by a $72\text{ V}$ DC power supply. The circuit consists of three components connected in series: an $85\ \Omega$ relay coil, a $35\ \Omega$ surge-limiting resistor, and an unknown protective resistor $R_x$ . If the entire circuit draws a steady current of $0.4\text{ A}$ , what is the resistance of the protective resistor $R_x$ in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 60

Answer

The resistance of the protective resistor

R_x

is

60\ \Omega

.

Applying Ohm's Law (

R_T = \frac{V_T}{I}

) gives a total series resistance of

180\ \Omega

. Since resistances in series add directly (

R_T = 85\ \Omega + 35\ \Omega + R_x = 120\ \Omega + R_x

), solving for the unknown yields

R_x = 180\ \Omega - 120\ \Omega = 60\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law.

R_T = \frac{V_{\text{source}}}{I_{\text{total}}} = \frac{72\text{ V}}{0.4\text{ A}} = 180\ \Omega

In a series circuit, total resistance equals total source voltage divided by total circuit current.

2

Determine the combined resistance of the known series components.

R_{\text{known}} = R_{\text{relay}} + R_{\text{surge}} = 85\ \Omega + 35\ \Omega = 120\ \Omega

Resistances in a series circuit are directly additive.

3

Subtract the known series resistance from the total circuit resistance to solve for

R_x

.

R_x = R_T - R_{\text{known}} = 180\ \Omega - 120\ \Omega = 60\ \Omega

The sum of all individual series resistors must equal the total equivalent resistance (

R_T = R_1 + R_2 + R_x

).

Key Concept

In a direct-current (DC) series circuit, current is identical through every component (

I_T = I_1 = I_2 = I_3

), and the total equivalent resistance is the arithmetic sum of all individual resistances (

R_T = R_1 + R_2 + \dots + R_n

).

Alternative Method

Find the voltage drops across the known components first:

V_{\text{relay}} = 0.4\text{ A} \times 85\ \Omega = 34\text{ V}

and

V_{\text{surge}} = 0.4\text{ A} \times 35\ \Omega = 14\text{ V}

. Subtract these from the total source voltage using Kirchhoff's Voltage Law:

V_{Rx} = 72\text{ V} - (34\text{ V} + 14\text{ V}) = 24\text{ V}

. Then find

R_x = \frac{V_{Rx}}{I} = \frac{24\text{ V}}{0.4\text{ A}} = 60\ \Omega

.

Estimated Time:1m 0s

Question 163Question →

A mobile radar subsystem is powered by a $24\text{ V}$ direct-current source. Its internal circuitry contains a parallel configuration with three distinct branches.

The total equivalent resistance of this parallel section is exactly $3\ \Omega$ .
A technician measures the current through the first branch to be $3\text{ A}$ .
The second branch has a known resistance of $24\ \Omega$ .

What is the resistance of the third branch in ohms?

Show answer & explanation

Answer: 6

Answer

The resistance of the third branch is

6\ \Omega

.

The correct value is derived by finding the total circuit current (

8\text{ A}

) and subtracting the known branch currents (

3\text{ A}

and

1\text{ A}

) to find the current in the third branch (

4\text{ A}

). Dividing the common

24\text{ V}

source by this

4\text{ A}

current yields

6\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit current using the total voltage and equivalent resistance.

I_{total} = 24\text{ V} / 3\ \Omega = 8\text{ A}

Ohm's law applies to the entire circuit, linking total voltage, total current, and equivalent resistance.

2

Determine the current flowing through the second branch.

I_2 = 24\text{ V} / 24\ \Omega = 1\text{ A}

In a parallel circuit, the full source voltage drops across every branch. Ohm's law (

I = V/R

) gives the branch current.

3

Calculate the current flowing through the third branch.

I_3 = 8\text{ A} - (3\text{ A} + 1\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current is the sum of the individual branch currents (

I_{total} = I_1 + I_2 + I_3

).

4

Calculate the resistance of the third branch.

R_3 = 24\text{ V} / 4\text{ A} = 6\ \Omega

Apply Ohm's law (

R = V/I

) specifically to the third branch using the source voltage and the calculated branch current.

Key Concept

Applying Ohm's law to both individual branches and the entire circuit, combined with parallel circuit rules for voltage and current.

Alternative Method

Use the parallel resistance reciprocal formula:

1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3

. First, find

R_1 = V/I_1 = 24\text{ V} / 3\text{ A} = 8\ \Omega

. Then, substitute the known values:

1/3 = 1/8 + 1/24 + 1/R_3

. Finding a common denominator of 24 gives

8/24 = 3/24 + 1/24 + 1/R_3

, which simplifies to

8/24 = 4/24 + 1/R_3

. Therefore,

1/R_3 = 4/24 = 1/6

, meaning

R_3 = 6\ \Omega

.

Estimated Time:1m 30s

Question 164Question →

An avionics technician is servicing a step-down transformer in an aircraft subsystem. The transformer receives an input voltage of $115\text{ V AC}$ across its primary coil and delivers an output voltage of $23\text{ V AC}$ across its secondary coil. If the primary winding has $750\text{ turns}$ , what is the number of turns on the secondary winding?

Show answer & explanation

Answer: 150

Answer

150 turns

The relationship between primary and secondary voltage and turns in a transformer is

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

. Given a primary voltage of

115\text{ V AC}

and a secondary voltage of

23\text{ V AC}

, the transformer has a step-down ratio of

5:1

(

115 / 23 = 5

). Dividing the primary turns (

750

) by

5

gives

150\text{ turns}

for the secondary winding.

Step-by-Step Solution

1

State the fundamental transformer turns ratio formula relating voltage and winding turns.

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

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

2

Isolate the secondary turns variable (

N_s

).

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

Algebraic rearrangement allows direct calculation of the secondary coil turns.

3

Substitute the given values (

V_p = 115\text{ V AC}

,

V_s = 23\text{ V AC}

,

N_p = 750\text{ turns}

) and simplify.

N_s = 750 \times \frac{23}{115} = 750 \times 0.2 = 150\text{ turns}

The voltage reduction factor is

5:1

, meaning the secondary winding must have one-fifth the number of turns of the primary winding.

Key Concept

Transformer turns and voltage ratio relationship

Estimated Time:1m 0s

Question 165Question →

A field communication station is powered by a $72\text{ V}$ direct-current supply connected across three parallel load branches. The resistance of the first branch is $18\ \Omega$ , the resistance of the second branch is $36\ \Omega$ , and the resistance of the third branch is $72\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), supplied to the circuit?

Show answer & explanation

Answer: 7

Answer

7 A

Each of the three branches is connected directly across the

72\text{ V}

source. Using Ohm's Law (

I = V/R

), the branch currents are

4\text{ A}

,

2\text{ A}

, and

1\text{ A}

. Adding these currents together yields the total current of

7\text{ A}

.

Step-by-Step Solution

1

Calculate the current in the first parallel branch using

I_1 = \frac{V}{R_1}

.

I_1 = \frac{72\text{ V}}{18\ \Omega} = 4\text{ A}

In a parallel DC circuit, the full supply voltage of

72\text{ V}

appears across every parallel branch.

2

Calculate the current in the second parallel branch using

I_2 = \frac{V}{R_2}

.

I_2 = \frac{72\text{ V}}{36\ \Omega} = 2\text{ A}

Applying Ohm's Law to the second branch with its specific resistance.

3

Calculate the current in the third parallel branch using

I_3 = \frac{V}{R_3}

.

I_3 = \frac{72\text{ V}}{72\ \Omega} = 1\text{ A}

Applying Ohm's Law to the third branch with its specific resistance.

4

Add all three individual branch currents together to determine the total current.

I_{\text{total}} = 4\text{ A} + 2\text{ A} + 1\text{ A} = 7\text{ A}

By Kirchhoff's Current Law, the total current leaving the power supply equals the sum of the currents through all parallel branches.

Key Concept

In a parallel circuit, the voltage across each parallel branch is identical to the source voltage, and the total circuit current is the sum of the individual branch currents (

I_{\text{total}} = I_1 + I_2 + \dots + I_n

).

Alternative Method

Find the total equivalent resistance (

R_{\text{eq}}

) first:

\frac{1}{R_{\text{eq}}} = \frac{1}{18} + \frac{1}{36} + \frac{1}{72} = \frac{4}{72} + \frac{2}{72} + \frac{1}{72} = \frac{7}{72}\ \Omega^{-1}

, so

R_{\text{eq}} = \frac{72}{7}\ \Omega

. Then find total current using

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{72}{72/7} = 7\text{ A}

.

Estimated Time:1m 0s

Question 166Question →

A radio technician is testing the output of an RF signal generator using an oscilloscope. The horizontal timebase control is set to $250\text{ }\mu\text{s/div}$ . On the screen graticule, a continuous sinusoidal waveform is observed where exactly $2.5\text{ complete cycles}$ span across $8\text{ horizontal divisions}$ . What is the frequency of this signal, in hertz ( $\text{Hz}$ )?

Show answer & explanation

Answer: 1250

Answer

1250 Hz

The total duration of the

8\text{ horizontal divisions}

at

250\text{ }\mu\text{s/div}

is

2000\text{ }\mu\text{s}

(

2\text{ ms}

). Dividing by the

2.5\text{ complete cycles}

gives a period of

T = 800\text{ }\mu\text{s} = 0.0008\text{ s}

per cycle. Taking the reciprocal gives the frequency:

f = 1 / 0.0008\text{ s} = 1250\text{ Hz}

.

Step-by-Step Solution

1

Calculate the total elapsed time across the observed graticule divisions.

Total time

= 8\text{ div} \times 250\text{ }\mu\text{s/div} = 2000\text{ }\mu\text{s}

The horizontal axis represents time, and the timebase setting indicates the duration per division.

2

Calculate the period (

T

) of a single cycle.

T = \frac{2000\text{ }\mu\text{s}}{2.5\text{ cycles}} = 800\text{ }\mu\text{s} = 800 \times 10^{-6}\text{ s} = 0.0008\text{ s}

The period is the time required to complete one single cycle.

3

Calculate the frequency (

f

) from the period.

f = \frac{1}{T} = \frac{1}{0.0008\text{ s}} = 1250\text{ Hz}

Frequency in hertz is the reciprocal of the period in seconds (

f = 1/T

).

Key Concept

Calculating waveform period and frequency from oscilloscope horizontal timebase settings and graticule divisions.

Estimated Time:1m 30s

Question 167Question →

An oscilloscope is used to measure an alternating signal. The horizontal sweep rate (Time/Div) is set to $8\text{ }\mu\text{s/div}$ , and one complete cycle of the waveform spans exactly $5\text{ horizontal divisions}$ across the graticule. What is the frequency of this signal in kilohertz ( $\text{kHz}$ )?

Show answer & explanation

Answer: 25

Answer

The frequency of the signal is

25\text{ kHz}

.

One full waveform cycle occupies

5

horizontal divisions at

8\text{ }\mu\text{s/division}

, giving a period of

T = 40\text{ }\mu\text{s}

(

40 \times 10^{-6}\text{ s}

). The frequency is the reciprocal of the period (

f = \frac{1}{T} = \frac{1}{40 \times 10^{-6}\text{ s}} = 25{}000\text{ Hz}

), which equals

25\text{ kHz}

.

Step-by-Step Solution

1

Multiply the number of horizontal graticule divisions per cycle by the time base setting to find the period (

T

).

T = 5\text{ divisions} \times 8\text{ }\mu\text{s/div} = 40\text{ }\mu\text{s} = 40 \times 10^{-6}\text{ s}

Horizontal divisions represent the time dimension on an oscilloscope display.

2

Calculate the frequency (

f

) in hertz using the reciprocal relationship

f = \frac{1}{T}

.

f = \frac{1}{40 \times 10^{-6}\text{ s}} = 25{}000\text{ Hz}

Frequency is defined as the number of cycles per second.

3

Convert the resulting frequency from hertz (

\text{Hz}

) to kilohertz (

\text{kHz}

).

\frac{25{}000\text{ Hz}}{1{}000} = 25\text{ kHz}

The question specifies the final answer unit as kilohertz.

Key Concept

Calculating waveform period and frequency from oscilloscope horizontal Time/Div settings and graticule division counts

Estimated Time:1m 0s

Question 168Question →

An industrial direct-current (DC) indicator lamp dissipates $1.8\text{ W}$ of power when a current of $150\text{ mA}$ flows through it. What is the voltage drop across the lamp, in volts?

Show answer & explanation

Answer: 12

Answer

The voltage drop across the indicator lamp is

12\text{ V}

.

According to Watt's Law, electrical power is the product of voltage and current (

P = V \times I

). Rearranging the formula to solve for voltage gives

V = \frac{P}{I}

. Converting

150\text{ mA}

to

0.15\text{ A}

and evaluating

\frac{1.8\text{ W}}{0.15\text{ A}}

yields

12\text{ V}

.

Step-by-Step Solution

1

Convert the current from milliamperes to amperes.

I = 150\text{ mA} = 0.15\text{ A}

Standard electrical formulas require current in base units of amperes (A).

2

Rearrange Watt's Law to solve for voltage.

V = \frac{P}{I}

Watt's Law defines power as

P = V \times I

, so isolating voltage yields

V = \frac{P}{I}

.

3

Substitute the known values into the equation to find the voltage.

V = \frac{1.8\text{ W}}{0.15\text{ A}} = 12\text{ V}

Dividing the power dissipation (

1.8\text{ W}

) by the current (

0.15\text{ A}

) gives the operating voltage.

Key Concept

Watt's Law describes the mathematical relationship between electrical power, voltage, and current:

P = V \times I

. When power and current are known, voltage is calculated using

V = \frac{P}{I}

, provided current is expressed in amperes.

Estimated Time:1m 0s

Question 169Question →

An electronic circuit uses a standard four-band resistor with the color bands Yellow, Violet, Brown, and Gold reading from left to right. What is the nominal resistance value of this resistor in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 470

Answer

The nominal resistance value is

470\ \Omega

.

In the standard electronic color code system, Yellow represents 4, Violet represents 7, Brown represents a multiplier of 10 (

10^1

), and Gold indicates a tolerance of

\pm 5\%

. Combining the digits gives 47, and multiplying by 10 produces

470\ \Omega

.

Step-by-Step Solution

1

Decode the first two color bands into the first two digits

Yellow corresponds to 4 and Violet corresponds to 7, giving the base number 47

The first two bands on a four-band resistor represent the first and second significant figures of the resistance value

2

Decode the third band into the decimal multiplier

Brown corresponds to a multiplier of

10^1

(

10

)

The third band indicates the power of ten by which the base number must be multiplied

3

Calculate the total nominal resistance

47 \times 10 = 470\ \Omega

Multiplying the base digits by the multiplier gives the rated nominal resistance in ohms

Key Concept

Four-band resistor color code decoding

Estimated Time:30s

Question 170Question →

A maintenance technician is tracking the number of work orders completed each day. Over a $5$ -day work week, the technician wants to achieve a mean of $14$ completed work orders per day. If the technician completed $12$ , $16$ , $11$ , and $13$ work orders on the first four days, how many work orders must be completed on the fifth day to exactly meet the target mean?

Show answer & explanation

Answer: 18

Answer

The technician must complete 18 work orders on the fifth day.

To find a missing value when the mean is known, first determine the total sum required by multiplying the target mean (14) by the total number of data points (5), which equals 70. Next, add the known values from the first four days (12 + 16 + 11 + 13 = 52). Finally, subtract the sum of the known values from the total required sum (70 - 52 = 18). Therefore, 18 work orders must be completed on the fifth day.

Step-by-Step Solution

1

Determine the total sum required for all 5 days.

14 \times 5 = 70

The mean is the total sum divided by the number of values. To find the required total sum, multiply the target mean by the total number of days.

2

Calculate the sum of the work orders completed so far.

12 + 16 + 11 + 13 = 52

This gives the total number of work orders already completed in the first four days.

3

Find the difference between the required total and the completed total.

70 - 52 = 18

Subtracting the sum of the first four days from the required 5-day total yields the exact value needed for the fifth day.

Key Concept

Calculating a missing value from a given mean requires multiplying the mean by the total number of items to find the required sum, then subtracting the known values.

Question 171Question →

An electronics technician is fabricating a heating coil using nichrome resistance wire. The wire has a cross-sectional area of $0.2\text{ mm}^2$ , a total length of $6\text{ meters}$ , and a material resistivity of $1.10\ \Omega\cdot\text{mm}^2/\text{m}$ . What is the total electrical resistance of the heating coil in ohms ( $\Omega$ ) across its entire length?

Show answer & explanation

Answer: 33

Answer

The total resistance of the nichrome wire heating coil is

33\ \Omega

.

The total resistance of any conductor is governed by

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

. Substituting the length of

6\text{ m}

, cross-sectional area of

0.2\text{ mm}^2

, and resistivity of

1.10\ \Omega\cdot\text{mm}^2/\text{m}

yields

R = \frac{1.10 \times 6}{0.2} = 33\ \Omega

.

Step-by-Step Solution

1

State the governing physical formula relating electrical resistance to conductor dimensions and material resistivity.

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

, where

R

is resistance in ohms,

\rho

is resistivity,

L

is length, and

A

is cross-sectional area.

Electrical resistance is directly proportional to wire length and resistivity, and inversely proportional to cross-sectional area.

2

Substitute the given numerical values into the formula with compatible units.

R = \frac{1.10\ \Omega\cdot\text{mm}^2/\text{m} \times 6\text{ m}}{0.2\text{ mm}^2}

The unit of resistivity (

\Omega\cdot\text{mm}^2/\text{m}

) aligns directly with meters for length and square millimeters for area, canceling out spatial units.

3

Calculate the numerator product and divide by the cross-sectional area.

R = \frac{6.60}{0.2} = 33\ \Omega

Dividing

6.60

by

0.2

gives the final total resistance in ohms.

Key Concept

Calculation of electrical resistance from material resistivity, conductor length, and cross-sectional area (

R = \rho L / A

)

Estimated Time:1m 0s

Question 172Question →

A marine auxiliary direct-current power circuit supplies a constant $48\text{ V}$ across three parallel branches. The first branch has a resistance of $16\ \Omega$ , the second branch has a resistance of $24\ \Omega$ , and the third branch has a resistance of $48\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), drawn from the power supply?

Show answer & explanation

Answer: 6

Answer

The total current drawn from the power supply is

6\text{ A}

.

Each parallel branch receives the full source voltage of

48\text{ V}

. Dividing the voltage by each branch resistance yields branch currents of

3\text{ A}

,

2\text{ A}

, and

1\text{ A}

. Summing these branch currents gives a total circuit current of

6\text{ A}

.

Step-by-Step Solution

1

Calculate the current passing through the first branch

I_1 = \frac{48\text{ V}}{16\ \Omega} = 3\text{ A}

The full source voltage of

48\text{ V}

appears across every parallel branch.

2

Calculate the current passing through the second branch

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

Applying Ohm's Law (

I = \frac{V}{R}

) with the same supply voltage across the

24\ \Omega

branch.

3

Calculate the current passing through the third branch

I_3 = \frac{48\text{ V}}{48\ \Omega} = 1\text{ A}

Applying Ohm's Law (

I = \frac{V}{R}

) across the

48\ \Omega

branch.

4

Sum all individual branch currents to determine total current

I_{\text{total}} = 3\text{ A} + 2\text{ A} + 1\text{ A} = 6\text{ A}

In a parallel circuit, the total current equals the sum of all branch currents (

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

).

Key Concept

In a parallel DC circuit, the voltage across each branch is identical to the source voltage, and the total current supplied is the sum of the individual branch currents.

Alternative Method

Find the equivalent resistance (

R_{\text{eq}}

) first using the reciprocal formula:

\frac{1}{R_{\text{eq}}} = \frac{1}{16} + \frac{1}{24} + \frac{1}{48} = \frac{3}{48} + \frac{2}{48} + \frac{1}{48} = \frac{6}{48} = \frac{1}{8}\implies R_{\text{eq}} = 8\ \Omega

. Then, calculate total current via Ohm's Law:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{48\text{ V}}{8\ \Omega} = 6\text{ A}

.

Estimated Time:1m 0s

Question 173Question →

A solid copper conductor with a length of $150\text{ meters}$ and a uniform cross-sectional area of $3.0\text{ mm}^2$ has a measured resistance of $0.85\ \Omega$ . An electrical technician replaces this conductor with another copper conductor of the same composition having a length of $90\text{ meters}$ and a cross-sectional area of $1.5\text{ mm}^2$ . What is the electrical resistance of the replacement conductor, in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 1.02

Answer

1.02 ohms (

\Omega

)

Resistance depends directly on conductor length and inversely on cross-sectional area. The length decreases by a factor of

\frac{90}{150} = 0.60

, while the cross-sectional area decreases by a factor of

\frac{1.5}{3.0} = 0.50

, which doubles the resistance contribution (factor of

2.0

). Multiplying the original resistance by both factors gives

0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega

.

Step-by-Step Solution

1

State the governing formula for electrical resistance of a uniform conductor.

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

, where

\rho

is material resistivity,

L

is conductor length, and

A

is cross-sectional area.

Since both conductors are made of identical copper, resistivity

\rho

remains constant, making resistance directly proportional to length and inversely proportional to cross-sectional area.

2

Determine the scaling factor resulting from the change in length.

Length factor

= \frac{L_2}{L_1} = \frac{90\text{ m}}{150\text{ m}} = 0.60

.

Decreasing conductor length proportionally reduces resistance to 60% of its initial value.

3

Determine the scaling factor resulting from the change in cross-sectional area.

Area factor

= \frac{A_1}{A_2} = \frac{3.0\text{ mm}^2}{1.5\text{ mm}^2} = 2.0

.

Halving the cross-sectional area doubles the resistance due to inverse proportionality.

4

Calculate the final resistance of the replacement wire.

R_2 = 0.85\ \Omega \times 0.60 \times 2.0 = 1.02\ \Omega

.

Combine the independent effects of length and cross-sectional area changes on the original resistance.

Key Concept

The electrical resistance of a conductor is directly proportional to its length and inversely proportional to its cross-sectional area (

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

).

Estimated Time:1m 0s

Question 174Question →

An emergency egress lighting system operates from a $50\text{ V}$ direct-current (DC) power source connected across two parallel branch circuits. Branch 1 contains an LED driver with an effective resistance of $25\ \Omega$ . If the total current drawn from the power source is $3\text{ A}$ , what is the resistance, in ohms ( $\Omega$ ), of Branch 2?

Show answer & explanation

Answer: 50

Answer

The resistance of Branch 2 is

50\ \Omega

.

In a parallel circuit, each branch receives the full source voltage of

50\text{ V}

. Branch 1 draws

I_1 = \frac{50\text{ V}}{25\ \Omega} = 2\text{ A}

. Since the total current is

3\text{ A}

, Branch 2 must draw

I_2 = 3\text{ A} - 2\text{ A} = 1\text{ A}

. Applying Ohm's Law to Branch 2 gives

R_2 = \frac{50\text{ V}}{1\text{ A}} = 50\ \Omega

.

Step-by-Step Solution

1

Determine the voltage across each parallel branch.

V_1 = V_2 = 50\text{ V}

In any parallel circuit configuration, the full source voltage appears equally across every individual parallel branch.

2

Calculate the current flowing through Branch 1 using Ohm's Law (

I = \frac{V}{R}

).

I_1 = \frac{50\text{ V}}{25\ \Omega} = 2\text{ A}

Ohm's Law allows solving for branch current when branch voltage and resistance are known.

3

Calculate the remaining current flowing through Branch 2 by subtracting Branch 1 current from total line current.

I_2 = 3\text{ A} - 2\text{ A} = 1\text{ A}

According to Kirchhoff's Current Law, total current entering a parallel network must equal the sum of all individual branch currents (

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

).

4

Calculate the resistance of Branch 2 using Ohm's Law (

R = \frac{V}{I}

).

R_2 = \frac{50\text{ V}}{1\text{ A}} = 50\ \Omega

Dividing the branch voltage by the branch current yields the branch's effective resistance.

Key Concept

Parallel circuit branch voltage equality, Kirchhoff's Current Law (

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

), and Ohm's Law (

R = \frac{V}{I}

).

Estimated Time:1m 0s

Question 175Question →

An electromagnetic security gate lock operates on an $18\text{ V}$ direct-current (DC) power source. When fully energized, the lock draws a steady current of $250\text{ mA}$ . What is the electrical resistance of the lock in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 72

Answer

The electrical resistance of the electromagnetic lock is

72\ \Omega

.

According to Ohm's Law (

V = I \times R

), resistance is found by dividing the potential difference by the current (

R = \frac{V}{I}

). Converting

250\text{ mA}

to

0.25\text{ A}

and dividing

18\text{ V}

by

0.25\text{ A}

yields exactly

72\ \Omega

.

Step-by-Step Solution

1

Convert the current measurement from milliamperes (

\text{mA}

) to amperes (

\text{A}

).

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

Ohm's Law calculations require current in base SI units (amperes) to yield resistance in ohms.

2

Rearrange Ohm's Law (

V = I \times R

) to isolate resistance (

R

).

R = \frac{V}{I}

Resistance is equal to the applied electromotive force (voltage) divided by the resulting current flow.

3

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

R = \frac{18\text{ V}}{0.25\text{ A}} = 72\ \Omega

Dividing

18\text{ V}

by

0.25\text{ A}

gives the total circuit resistance in ohms.

Key Concept

Ohm's Law states that resistance (

R

) equals voltage (

V

) divided by current (

I

), represented by the formula

R = \frac{V}{I}

.

Estimated Time:45s

Question 176Question →

A cylindrical water storage tank is filled to a depth of $4\text{ meters}$ , creating a hydrostatic pressure of $40,000\text{ Pascals}$ at the tank's flat bottom. If water is drained from the tank until the depth is exactly $1\text{ meter}$ , what is the new hydrostatic pressure at the bottom in Pascals? (Assume the density of water and gravity remain constant.)

Show answer & explanation

Answer: 10000

Answer

10000

Because hydrostatic pressure is directly proportional to depth (

P = \rho g h

), reducing the depth to one-fourth of its original value (

1\text{ meter}

instead of

4\text{ meters}

) also reduces the pressure exactly to one-fourth of its original value (

40,000 / 4 = 10,000

).

Step-by-Step Solution

1

Identify the relationship between hydrostatic pressure and fluid depth.

Hydrostatic pressure is directly proportional to depth (

P \propto h

) when density and gravity are constant.

This fundamental principle allows us to solve the problem using a simple ratio.

2

Calculate the pressure per unit of depth.

\frac{40,000\text{ Pascals}}{4\text{ meters}} = 10,000\text{ Pascals per meter}

.

Finding the rate of pressure change per meter simplifies the calculation for the new depth.

3

Multiply the pressure per meter by the new depth.

10,000\text{ Pascals/meter} \times 1\text{ meter} = 10,000\text{ Pascals}

.

This yields the final pressure at the reduced depth.

Key Concept

Hydrostatic pressure is directly proportional to the depth of the fluid.

Question 177Question →

A specialized alloy heating wire has a length of $40\text{ meters}$ , a uniform cross-sectional area of $0.8\text{ mm}^2$ , and a material resistivity of $2.5 \times 10^{-7}\ \Omega\cdot\text{m}$ . What is the total electrical resistance of this wire in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 12.5

Answer

12.5 Ω

The electrical resistance of a uniform conductor is determined by

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

. Converting

0.8\text{ mm}^2

to

8 \times 10^{-7}\text{ m}^2

and substituting

L = 40\text{ m}

along with

\rho = 2.5 \times 10^{-7}\ \Omega\cdot\text{m}

yields

\frac{(2.5 \times 10^{-7})(40)}{8 \times 10^{-7}} = 12.5\ \Omega

.

Step-by-Step Solution

1

Convert the cross-sectional area from square millimeters to square meters.

A = 0.8\text{ mm}^2 = 0.8 \times 10^{-6}\text{ m}^2 = 8 \times 10^{-7}\text{ m}^2

Resistivity is given in ohm-meters (

\Omega\cdot\text{m}

), requiring all geometric dimensions to be in standard SI units (meters and square meters).

2

Apply the fundamental resistance formula relating resistivity, length, and cross-sectional area.

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

The total resistance of a uniform conductor is directly proportional to its resistivity and length, and inversely proportional to its cross-sectional area.

3

Substitute the known values into the equation and solve for resistance.

R = \frac{(2.5 \times 10^{-7}\ \Omega\cdot\text{m})(40\text{ m})}{8 \times 10^{-7}\text{ m}^2} = \frac{100 \times 10^{-7}}{8 \times 10^{-7}} = 12.5\ \Omega

Canceling the common power factor

10^{-7}

simplifies the expression to

100 / 8 = 12.5\ \Omega

.

Key Concept

Calculating conductor resistance from material resistivity, length, and cross-sectional area using

R = \rho L / A

.

Estimated Time:1m 15s

Question 178Question →

A $24\text{ V}$ direct-current (DC) power bus supplies two cooling fans connected in parallel inside a communications cabinet. The primary fan draws a steady branch current of $3\text{ A}$ , while the secondary fan has an internal resistance of $12\text{ }\Omega$ . What is the total current, in amperes ( $\text{A}$ ), delivered by the power bus to the parallel circuit?

Show answer & explanation

Answer: 5

Answer

The total current delivered by the power bus is

5\text{ A}

.

In a parallel DC circuit, the source voltage (

24\text{ V}

) is applied directly across each branch. The secondary branch current is found using Ohm's law:

I_2 = \frac{24\text{ V}}{12\text{ }\Omega} = 2\text{ A}

. Adding the known primary branch current (

3\text{ A}

) to the secondary branch current (

2\text{ A}

) yields the total source current of

5\text{ A}

.

Step-by-Step Solution

1

Determine the voltage across the secondary branch

Voltage across the secondary fan is

24\text{ V}

In any parallel circuit, every branch experiences the identical full source voltage (

V_{\text{total}} = V_1 = V_2

).

2

Calculate the secondary branch current (

I_2

) using Ohm's law

I_2 = \frac{24\text{ V}}{12\text{ }\Omega} = 2\text{ A}

Ohm's law states that current equals voltage divided by resistance (

I = \frac{V}{R}

).

3

Sum all individual branch currents to determine total circuit current

I_{\text{total}} = 3\text{ A} + 2\text{ A} = 5\text{ A}

Total current in a parallel circuit is the sum of the individual branch currents (

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

).

Key Concept

Parallel Circuit Current and Ohm's Law

Estimated Time:1m 0s

Question 179Question →

An aircraft instrument panel illumination circuit is connected to a $28\text{ V}$ direct-current (DC) power source. A multimeter measures a steady current flow of $350\text{ mA}$ through the circuit. What is the total electrical resistance of the circuit in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 80

Answer

The total electrical resistance of the circuit is

80\ \Omega

.

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

R = \frac{V}{I}

). Converting

350\text{ mA}

to amperes yields

0.35\text{ A}

. Dividing

28\text{ V}

by

0.35\text{ A}

gives the correct resistance of

80\ \Omega

.

Step-by-Step Solution

1

Convert the current from milliamperes to amperes

I = 350\text{ mA} = 0.35\text{ A}

Ohm's Law calculations require current to be expressed in standard amperes.

2

Apply Ohm's Law formula solved for resistance

R = \frac{V}{I} = \frac{28\text{ V}}{0.35\text{ A}} = 80\ \Omega

Dividing the applied electromotive force by the current flow yields total circuit resistance.

Key Concept

Calculating resistance using Ohm's Law (

R = \frac{V}{I}

) with metric prefix conversion (

1\text{ A} = 1{}000\text{ mA}

).

Estimated Time:45s

Question 180Question →

A quartermaster is organizing a drawer of $60$ uniform patches. The drawer contains $15$ rank insignia patches, $21$ unit patches, and the remaining patches are flag patches. If one patch is randomly selected from the drawer, what is the probability of selecting a flag patch? Express your answer as a decimal.

Show answer & explanation

Answer: 0.4

Answer

0.4

The probability is found by dividing the number of flag patches (

24

) by the total number of patches (

60

). The fraction

\frac{24}{60}

simplifies to

\frac{2}{5}

, which is

0.4

when expressed as a decimal.

Step-by-Step Solution

1

Calculate the number of flag patches in the drawer.

60 - (15 + 21) = 60 - 36 = 24

flag patches.

To determine the probability, the exact number of the desired item (flag patches) must first be known.

2

Set up the probability fraction.

\frac{24}{60}

The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes.

3

Convert the fraction to a decimal.

\frac{24}{60} = \frac{2}{5} = 0.4

The question explicitly specifies that the final answer must be expressed as a decimal.

Key Concept

Single Event Probability

Answer

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