All practice questions

Question 121Question →

A technician connects a direct current (DC) indicator lamp across a $10\text{ V}$ power supply. A digital multimeter connected in series measures a current of $0.5\text{ A}$ flowing through the circuit. What is the electrical resistance of the indicator lamp in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 20

Answer

The resistance of the indicator lamp is

20\text{ }\Omega

.

According to Ohm's Law, resistance is determined by dividing voltage by current (

R = \frac{V}{I}

). Substituting

10\text{ V}

and

0.5\text{ A}

gives

R = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega

.

Step-by-Step Solution

1

Identify the given circuit parameters from the problem statement.

Voltage

V = 10\text{ V}

, Current

I = 0.5\text{ A}

Applying Ohm's law requires identifying the known electromotive force and circuit current.

2

Select and arrange the Ohm's law equation to solve for resistance (

R

).

R = \frac{V}{I}

Resistance is equal to the voltage divided by the current flowing through the component.

3

Substitute the measured values into the formula and solve.

R = \frac{10\text{ V}}{0.5\text{ A}} = 20\text{ }\Omega

Dividing

10\text{ V}

by

0.5\text{ A}

yields the total resistance of

20\text{ }\Omega

.

Key Concept

Ohm's Law defines the relationship between voltage, current, and resistance in an electrical circuit:

V = I \times R

, which rearranges to

R = \frac{V}{I}

.

Estimated Time:45s

Question 122Question →

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 123Question →

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 124Question →

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 125Question →

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 126Question →

A medical detachment is tracking a patient's resting heart rate over a six-day period. The mean heart rate over the six days is $69$ beats per minute (bpm). The readings for five of those days are $66$ , $72$ , $68$ , $78$ , and $65$ . What is the median heart rate for the entire six-day period?

Show answer & explanation

Answer: 67

Answer

67

The correct answer is derived in two phases. First, the missing sixth value is found by subtracting the sum of the five known values (349) from the total sum of all six values (69 * 6 = 414), which yields 65. Second, all six values are sorted from least to greatest: 65, 65, 66, 68, 72, 78. Since there are 6 values, the median is the exact average of the two middle numbers (66 and 68). (66 + 68) / 2 = 67 bpm.

Step-by-Step Solution

1

Calculate the total sum of the heart rates for all six days.

414

The total sum is required to find the value of the missing sixth day. Multiply the given mean (

69

) by the total number of days (

6

).

2

Calculate the sum of the five known heart rates.

349

Adding the given values (

66 + 72 + 68 + 78 + 65

) allows us to see how much of the total sum is accounted for.

3

Determine the heart rate for the sixth day.

65

Subtracting the sum of the known days (

349

) from the total sum (

414

) reveals the missing data point.

4

Arrange all six heart rates in ascending order.

65, 65, 66, 68, 72, 78

The median is a measure of position, which strictly requires the entire dataset to be ordered from least to greatest.

5

Identify the two middle values and calculate their average.

67

Because there is an even number of values (

n=6

), the median is the mean of the 3rd and 4th values:

(66 + 68) / 2

.

Key Concept

Solving for a missing data value using the arithmetic mean, followed by determining the median of an even-numbered dataset.

Question 127Question →

A copper feed wire in a naval radar subsystem has a length of $50\text{ m}$ , a circular cross-sectional diameter of $2.0\text{ mm}$ , and a measured DC resistance of $0.27\ \Omega$ . During maintenance, this line is replaced with an alloy wire that has a resistivity $4.0$ times that of copper, a total length of $75\text{ m}$ , and a diameter of $3.0\text{ mm}$ . What is the electrical resistance of the replacement wire in ohms?

Show answer & explanation

Answer: 0.72

Answer

The electrical resistance of the replacement wire is 0.72 ohms.

Electrical resistance is given by the formula

R = \rho \frac{L}{A} = \frac{4\rho L}{\pi d^2}

. Comparing the replacement wire to the original: resistivity increases by a factor of

4.0

, length increases by a factor of

\frac{75}{50} = 1.5

, and cross-sectional area increases by a factor of

\left(\frac{3.0}{2.0}\right)^2 = 2.25

. Combining these factors gives an overall resistance change factor of

\frac{4.0 \times 1.5}{2.25} = \frac{8}{3}

. Multiplying the baseline resistance of

0.27\ \Omega

by

\frac{8}{3}

gives exactly

0.72\ \Omega

.

Step-by-Step Solution

1

Relate electrical resistance to material resistivity and conductor dimensions.

R = \rho \frac{L}{A} = \frac{4\rho L}{\pi d^2}

Resistance is directly proportional to resistivity and length, and inversely proportional to cross-sectional area (which depends on the square of the diameter).

2

Determine the relative scaling factors for resistivity, length, and diameter.

Resistivity factor

= 4.0

; Length factor

= \frac{75}{50} = 1.5

; Diameter factor

= \frac{3.0}{2.0} = 1.5

Proportional scaling allows direct calculation without calculating intermediate microscopic values.

3

Calculate the combined scaling factor for resistance.

\frac{R_2}{R_1} = \frac{\rho_2}{\rho_1} \times \frac{L_2}{L_1} \times \left(\frac{d_1}{d_2}\right)^2 = 4.0 \times 1.5 \times \left(\frac{1}{1.5}\right)^2 = \frac{4.0}{1.5} = \frac{8}{3}

Area scales quadratically with diameter, so diameter increase reduces resistance by a factor of

1.5^2 = 2.25

.

4

Calculate the final resistance of the replacement wire.

R_2 = 0.27\ \Omega \times \frac{8}{3} = 0.72\ \Omega

Multiplying the original resistance by the overall scaling factor yields the new total resistance.

Key Concept

Dependence of electrical resistance on resistivity, conductor length, and cross-sectional area

Alternative Method

Calculate the resistivity of copper first using

\rho_{\text{Cu}} = \frac{R_1 A_1}{L_1} = \frac{0.27 \times \pi (0.001)^2}{50} \approx 1.696 \times 10^{-8}\ \Omega\cdot\text{m}

. Then calculate

\rho_{\text{alloy}} = 4.0 \times \rho_{\text{Cu}} \approx 6.786 \times 10^{-8}\ \Omega\cdot\text{m}

, and compute

R_2 = \frac{\rho_{\text{alloy}} \times 75}{\pi (0.0015)^2} = 0.72\ \Omega

.

Estimated Time:1m 30s

Question 128Question →

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

Show answer & explanation

Answer: 4.8

Answer

4.8

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

Step-by-Step Solution

1

Convert the cross-sectional area from

mm^2

to

m^2

.

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

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

2

Calculate the resistance at

20^\circ C

(

R_{20}

) using Pouillet's law (

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

).

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

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

3

Calculate the change in temperature (

\Delta T

).

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

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

4

Calculate the new resistance at

70^\circ C

using the temperature coefficient formula:

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

.

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

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

Key Concept

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

Estimated Time:3m 0s

Question 129Question →

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

Show answer & explanation

Answer: 10

Answer

The total equivalent capacitance of the series circuit is

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

.

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

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

. Substituting

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

,

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

, and

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

gives

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

. Inverting this fraction yields the correct total capacitance of

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

, which is less than the smallest individual capacitor (

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

).

Step-by-Step Solution

1

State the series capacitance relationship

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

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

2

Calculate the sum of the reciprocals using a common denominator

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

Finding a least common multiple of

60

allows direct fractional addition.

3

Invert the result to find

C_{\text{total}}

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

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

Key Concept

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

Estimated Time:1m 0s

Question 130Question →

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

Show answer & explanation

Answer: 80

Answer

80 turns

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

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

. Since the secondary voltage (

24\text{ V}

) is one-tenth of the primary input voltage (

240\text{ V}

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

800 \times 0.10 = 80

turns.

Step-by-Step Solution

1

Identify known electrical parameters from the problem statement.

V_p = 240\text{ V}

,

V_s = 24\text{ V}

, and

N_p = 800\text{ turns}

.

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

2

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

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

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

3

Rearrange the equation to isolate the unknown secondary winding turns (

N_s

).

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

Algebraic isolation allows direct computation of secondary turns.

4

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

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

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

Key Concept

Transformer Turns Ratio and Voltage Transformation

Alternative Method

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

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

. Then multiply by the desired secondary voltage:

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

.

Estimated Time:45s

Question 131Question →

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

Show answer & explanation

Answer: 2

Answer

The current flowing through the solenoid coil is

2\text{ A}

.

According to Ohm's law, electrical current (

I

) equals voltage (

V

) divided by resistance (

R

). Substituting the given values yields

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

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage

V = 30\text{ V}

, Resistance

R = 15\text{ }\Omega

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

2

State Ohm's law formula for current

I = \frac{V}{R}

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

3

Substitute the values and evaluate

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

Dividing

30

by

15

yields the circuit current in amperes.

Key Concept

Ohm's Law (

I = \frac{V}{R}

)

Estimated Time:45s

Question 132Question →

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

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

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

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

Show answer & explanation

Answer: 250

Answer

250 turns

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

V_p/V_s = N_p/N_s

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

1{}200\text{ V}

input to

240\text{ V}

. Transformer 2 receives this

240\text{ V}

and must further step it down to

48\text{ V}

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

50

secondary turns by this ratio yields

250

primary turns.

Step-by-Step Solution

1

Calculate the secondary voltage of Transformer 1.

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

Using the turns ratio formula

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

, we find

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

.

2

Identify the primary voltage of Transformer 2.

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

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

3

Set up the turns ratio equation for Transformer 2.

N_{p2} / 50 = 240 / 48

Applying the turns ratio formula

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

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

4

Solve for the primary turns of Transformer 2.

N_{p2} = 250\text{ turns}

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

.

Key Concept

Cascaded Transformer Turns Ratios

Question 133Question →

A $36\text{ V}$ direct-current power supply is connected across a parallel circuit containing two branches. If the first branch contains an $18\ \Omega$ resistor and the second branch contains a $9\ \Omega$ resistor, what is the current flowing through the second branch in amperes?

Show answer & explanation

Answer: 4

Answer

4 A

In a parallel circuit, the voltage across all individual branches is identical to the supply voltage (

36\text{ V}

). Using Ohm's law for the second branch, the current is calculated as

I = \frac{V}{R} = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}

.

Step-by-Step Solution

1

Determine the voltage drop across the second branch.

The voltage across the second branch is

36\text{ V}

.

In a parallel circuit, each individual branch experiences the full source voltage (

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

).

2

Calculate the branch current using Ohm's law.

I_2 = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}

.

Current through an individual resistor branch is found by dividing the branch voltage by the branch resistance (

I = \frac{V}{R}

).

Key Concept

In a parallel circuit, the voltage across every parallel branch is equal to the source voltage, and the current through any branch depends solely on that branch's resistance via Ohm's law (

I = \frac{V}{R}

).

Estimated Time:45s

Question 134Question →

An electronics technician tests a DC audio indicator module connected to a $24\text{ V}$ power supply. An in-line meter measures a steady current flow of $80\text{ mA}$ through the component. According to Ohm's law, what is the electrical resistance of the module in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 300

Answer

The total resistance of the module is

300\text{ }\Omega

.

According to Ohm's law (

V = I \times R

), resistance equals voltage divided by current (

R = \frac{V}{I}

). Converting

80\text{ mA}

to standard amperes gives

0.08\text{ A}

. Dividing the

24\text{ V}

source by

0.08\text{ A}

yields a resistance of

300\text{ }\Omega

.

Step-by-Step Solution

1

Convert current from milliamperes (mA) to base amperes (A)

80\text{ mA} = 80 \times 10^{-3}\text{ A} = 0.08\text{ A}

Ohm's law requires current to be expressed in standard SI units (amperes) when working with volts and ohms.

2

Rearrange Ohm's law (

V = I \times R

) to isolate resistance (

R

)

R = \frac{V}{I}

Dividing both sides of the standard Ohm's law equation by current isolates the unknown resistance.

3

Substitute the known voltage (

24\text{ V}

) and current (

0.08\text{ A}

) into the formula

R = \frac{24\text{ V}}{0.08\text{ A}} = 300\text{ }\Omega

Performing the division calculates the exact resistance value in ohms.

Key Concept

Ohm's law resistance calculation with metric prefix conversion (

R = \frac{V}{I}

)

Estimated Time:1m 0s

Question 135Question →

An industrial controller module requires a total electromotive force (EMF) of $0.12\text{ kV}$ for its backup power rail. If standard $1.5\text{ V}$ alkaline cells are connected in series-aiding to supply the rail, how many cells are required?

Show answer & explanation

Answer: 80

Answer

80 cells

The total requirement is

0.12\text{ kV}

, which equals

120\text{ V}

. Since series-aiding cells add their electromotive forces directly, dividing the total required voltage of

120\text{ V}

by

1.5\text{ V}

per cell gives exactly

80

cells.

Step-by-Step Solution

1

Convert the required electromotive force from kilovolts (kV) to standard volts (V)

0.12\text{ kV} \times 1000 = 120\text{ V}

The prefix kilo- denotes a multiplier of

1000

, so the potential difference must be converted to base units to match individual cell ratings.

2

Calculate the number of series cells needed by dividing total EMF by the EMF of a single cell

n = \frac{120\text{ V}}{1.5\text{ V/cell}} = 80\text{ cells}

When electrical cells are connected in series-aiding configuration, their individual voltages add directly (

V_{\text{total}} = n \times V_{\text{cell}}

).

Key Concept

Series electromotive force summation and metric prefix voltage conversion

Estimated Time:1m 0s

Question 136Question →

An aircraft cockpit lighting circuit is connected to a $24\text{ V}$ direct current (DC) power source. The circuit consists of three loads connected in series: a $15\ \Omega$ panel lamp, a $25\ \Omega$ dimming rheostat, and a $20\ \Omega$ ballast resistor. What is the voltage drop across the $25\ \Omega$ dimming rheostat in volts?

Show answer & explanation

Answer: 10

Answer

The voltage drop across the

25\ \Omega

dimming rheostat is

10\text{ V}

.

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

15\ \Omega + 25\ \Omega + 20\ \Omega = 60\ \Omega

. Applying Ohm's Law to the entire circuit gives a current of

24\text{ V} / 60\ \Omega = 0.4\text{ A}

. Because current is constant throughout a series circuit, the voltage drop across the

25\ \Omega

rheostat is

0.4\text{ A} \times 25\ \Omega = 10\text{ V}

.

Step-by-Step Solution

1

Calculate the total series resistance (

R_T

) of the circuit.

R_T = 15\ \Omega + 25\ \Omega + 20\ \Omega = 60\ \Omega

In a series circuit, total resistance equals the algebraic sum of all individual series resistances.

2

Calculate the common series current (

I

) flowing through the circuit.

I = \frac{V_T}{R_T} = \frac{24\text{ V}}{60\ \Omega} = 0.4\text{ A}

Ohm's Law (

I = \frac{V}{R}

) relates the total supply voltage to the total circuit resistance.

3

Calculate the voltage drop (

V_2

) across the

25\ \Omega

resistor.

V_2 = I \times R_2 = 0.4\text{ A} \times 25\ \Omega = 10\text{ V}

Current is constant throughout all components in a series circuit, so the individual voltage drop is the product of loop current and component resistance.

Key Concept

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

R_T = R_1 + R_2 + \dots + R_n

), current is identical through all components (

I_T = I_1 = I_2 = \dots = I_n

), and individual voltage drops are directly proportional to component resistance (

V_n = I \times R_n

).

Estimated Time:1m 0s

Question 137Question →

A direct-current (DC) ventilation blower operates on a $28\text{ V}$ aircraft electrical bus and draws a steady current of $5\text{ A}$ . How many watts ( $\text{W}$ ) of electrical power are consumed by the blower?

Show answer & explanation

Answer: 140

Answer

The power consumed by the blower is

140\text{ W}

.

According to Watt's Law, electrical power (

P

) dissipated by a direct-current device equals the product of the applied voltage (

V

) and the current (

I

). Multiplying

28\text{ V}

by

5\text{ A}

gives

140\text{ W}

.

Step-by-Step Solution

1

Identify the known electrical quantities given in the scenario.

Voltage

V = 28\text{ V}

and Current

I = 5\text{ A}

.

Identifying the known parameters determines which form of Watt's Law to use.

2

Apply the fundamental Watt's Law formula for electrical power (

P = V \times I

).

P = 28\text{ V} \times 5\text{ A} = 140\text{ W}

.

Electrical power in a direct-current circuit is calculated by multiplying the potential difference across the load by the current flowing through it.

Key Concept

Watt's Law for DC Power (

P = V \times I

)

Estimated Time:45s

Question 138Question →

A piece of copper hookup wire with a length of $15\text{ m}$ has an electrical resistance of $0.6\ \Omega$ . If the wire is replaced with a $45\text{ m}$ piece of the exact same wire type and gauge at the same temperature, what is the resistance of the new wire in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 1.8

Answer

1.8 \Omega

According to the fundamental resistance relationship

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

, resistance is directly proportional to the length

L

of the conductor when resistivity

\rho

and cross-sectional area

A

remain unchanged. Tripling the wire length from

15\text{ m}

to

45\text{ m}

multiplies the resistance by a factor of 3, resulting in

0.6\ \Omega \times 3 = 1.8\ \Omega

.

Step-by-Step Solution

1

Calculate the proportional change in conductor length.

The new length is 3 times the original length (

45\text{ m} \div 15\text{ m} = 3

).

Identifying the scaling factor allows direct application of the resistance formula.

2

Apply the direct proportionality between conductor length and electrical resistance (

R \propto L

).

R_{\text{new}} = R_{\text{old}} \times 3 = 0.6\ \Omega \times 3 = 1.8\ \Omega

.

Since material resistivity (

\rho

) and cross-sectional area (

A

) are held constant, tripling the length triples the total resistance.

Key Concept

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

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

).

Estimated Time:45s

Question 139Question →

A technician is troubleshooting a test circuit connected to a $32\text{V}$ DC power supply. The circuit consists of a fixed resistor $R_1$ in series with a parallel block. The parallel block contains a $12\text{ k}\Omega$ fixed resistor and a variable resistor (rheostat) $R_v$ . The resistor $R_1$ has four color bands: Green, Blue, Red, and Gold. The technician adjusts the rheostat $R_v$ until the total current drawn from the power supply is exactly $4\text{ mA}$ .

What is the resistance setting of the rheostat $R_v$ in kilohms (k $\Omega$ )?

Show answer & explanation

Answer: 3

Answer

3

The total resistance required by the circuit to draw

4\text{ mA}

from a

32\text{V}

source is

8\text{ k}\Omega

. The series resistor

R_1

(Green, Blue, Red) contributes

5.6\text{ k}\Omega

. This leaves exactly

2.4\text{ k}\Omega

required for the parallel section. For a

12\text{ k}\Omega

resistor in parallel with an unknown

R_v

to yield an equivalent resistance of

2.4\text{ k}\Omega

,

R_v

must be set to

3\text{ k}\Omega

.

Step-by-Step Solution

1

Decode the color code of resistor

R_1

.

R_1 = 5.6\text{ k}\Omega

The color bands are Green (5), Blue (6), and Red (multiplier of

10^2

or 100). Thus,

56 \times 100 = 5,600\ \Omega

, which is

5.6\text{ k}\Omega

. The Gold band indicates a 5% tolerance, which applies to real-world variance but does not alter our nominal calculation.

2

Calculate the total equivalent resistance (

R_{eq}

) of the circuit.

R_{eq} = 8\text{ k}\Omega

Using Ohm's Law (

R = \frac{V}{I}

), divide the source voltage (

32\text{V}

) by the total current (

4\text{ mA}

or

0.004\text{ A}

) to find the total resistance required to draw that current.

3

Determine the required equivalent resistance of the parallel block.

R_{parallel} = 2.4\text{ k}\Omega

Since

R_1

is in series with the parallel block, subtract

R_1

from the total circuit resistance to find the resistance of the parallel segment:

8\text{ k}\Omega - 5.6\text{ k}\Omega = 2.4\text{ k}\Omega

.

4

Solve for the rheostat's resistance (

R_v

).

R_v = 3\text{ k}\Omega

The parallel block contains a

12\text{ k}\Omega

resistor and

R_v

. Using the parallel resistance formula

\frac{1}{R_p} = \frac{1}{R_a} + \frac{1}{R_b}

, we get

\frac{1}{2.4} = \frac{1}{12} + \frac{1}{R_v}

. Rearranging gives

\frac{1}{R_v} = \frac{1}{2.4} - \frac{1}{12} = \frac{10}{24} - \frac{2}{24} = \frac{8}{24} = \frac{1}{3}

, meaning

R_v = 3\text{ k}\Omega

.

Key Concept

Integrating Ohm's law with series-parallel equivalent resistance and resistor color code decoding.

Estimated Time:3m 0s

Question 140Question →

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

Show answer & explanation

Answer: 8

Answer

8 Ω

For two resistors in parallel with values of

10\ \Omega

and

40\ \Omega

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

R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 40}{10 + 40} = \frac{400}{50} = 8\ \Omega

. In any parallel circuit, the total equivalent resistance is always smaller than the smallest branch resistor.

Step-by-Step Solution

1

Select the appropriate formula for calculating the equivalent resistance of two parallel resistors.

R_eq = (R_1 * R_2) / (R_1 + R_2) or 1/R_eq = 1/R_1 + 1/R_2

In parallel circuits, total resistance is found by taking the reciprocal of the sum of the reciprocals, which simplifies to the product-over-sum rule for two branches.

2

Substitute the given resistor values (10 Ω and 40 Ω) into the product-over-sum formula.

R_eq = (10 * 40) / (10 + 40) = 400 / 50

Multiplying the branch values gives the numerator, and adding them gives the denominator.

3

Divide the numerator by the denominator to find the equivalent resistance.

R_eq = 8 Ω

400 divided by 50 equals 8, which is less than the smallest branch resistance (10 Ω), confirming the validity of the parallel calculation.

Key Concept

Parallel Circuit Equivalent Resistance

Estimated Time:45s

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