Circuits and Magnetism Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 2

Question 21Question →

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 22Question →

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

25\ \Omega

6\ \Omega

12.5\ \Omega

5\ \Omega

Show answer & explanation

Answer: $6\ \Omega$

Answer

The total equivalent resistance of the parallel circuit is

6\ \Omega

.

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

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

. Substituting

10\ \Omega

and

15\ \Omega

gives

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

.

Step-by-Step Solution

1

Identify the formula for equivalent resistance of two parallel resistors

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

or

\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}

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

2

Substitute the given resistances (

R_1 = 10\ \Omega

and

R_2 = 15\ \Omega

) into the product-over-sum formula

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

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

3

Calculate the final equivalent resistance

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

Dividing

150

by

25

yields the total resistance of

6\ \Omega

.

Key Concept

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

Estimated Time:45s

Question 23Question →

An industrial automation control loop is powered by a $48\text{ V}$ direct current (DC) source connected in series with three resistors: $R_1 = 20\ \Omega$ , $R_2 = 50\ \Omega$ , and $R_3 = 90\ \Omega$ . What is the voltage drop across resistor $R_2$ ?

48\text{ V}

15\text{ V}

0.3\text{ V}

16\text{ V}

Show answer & explanation

Answer: $15\text{ V}$

Answer

The voltage drop across the

50\ \Omega

resistor (

R_2

) is

15\text{ V}

.

The correct answer is

15\text{ V}

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

20\ \Omega + 50\ \Omega + 90\ \Omega = 160\ \Omega

. The circuit current is therefore

48\text{ V} / 160\ \Omega = 0.3\text{ A}

. Applying Ohm's Law across

R_2

gives a voltage drop of

0.3\text{ A} \times 50\ \Omega = 15\text{ V}

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) by summing the three series resistances.

R_T = R_1 + R_2 + R_3 = 20\ \Omega + 50\ \Omega + 90\ \Omega = 160\ \Omega

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

2

Determine the total circuit current (

I_T

) using Ohm's Law.

I_T = \frac{V_S}{R_T} = \frac{48\text{ V}}{160\ \Omega} = 0.3\text{ A}

Current equals total voltage divided by total resistance, and current remains constant through every component in a single series loop.

3

Calculate the specific voltage drop (

V_2

) across resistor

R_2

.

V_2 = I_T \times R_2 = 0.3\text{ A} \times 50\ \Omega = 15\text{ V}

Applying Ohm's Law (

V = I \times R

) to an individual component determines the potential difference dropped across that load.

Key Concept

Series circuit total resistance, current constancy, and individual component voltage drop calculation using Ohm's Law.

Estimated Time:1m 0s

Question 24Question →

An electronics maintenance panel is powered by a $120\text{ V}$ DC bus. The circuit features a primary $40\ \Omega$ load resistor ( $R_1$ ) connected in series with a parallel junction containing two secondary resistors: $R_2$ ( $60\ \Omega$ ) and $R_3$ ( $30\ \Omega$ ). During a routine inspection, a digital multimeter indicates a $48\text{ V}$ drop across $R_1$ . Assuming the source voltage is nominal, which specific component failure is causing this reading?

The

30\ \Omega

secondary resistor (

R_3

) has failed open.

The

60\ \Omega

secondary resistor (

R_2

) has failed open.

The primary resistor (

R_1

) has degraded and reduced in value to

24\ \Omega

.

The parallel junction has short-circuited to ground.

Show answer & explanation

Answer: The $30\ \Omega$ secondary resistor ( $R_3$ ) has failed open.

Answer

The

30\ \Omega

secondary resistor (

R_3

) has failed open.

Based on the

48\text{ V}

drop across the

40\ \Omega

series resistor, the actual circuit current is

1.2\text{ A}

. To draw exactly

1.2\text{ A}

from a

120\text{ V}

source, the total circuit resistance must be

100\ \Omega

. Since the series resistor accounts for

40\ \Omega

, the parallel junction must be contributing exactly

60\ \Omega

. The only way the parallel junction (normally containing a

60\ \Omega

and a

30\ \Omega

resistor) equals

60\ \Omega

is if the

30\ \Omega

resistor has failed open, leaving only the

60\ \Omega

resistor in the path.

Step-by-Step Solution

1

Calculate the actual current flowing through the primary resistor (

R_1

) under fault conditions.

I_{total} = \frac{48\text{ V}}{40\ \Omega} = 1.2\text{ A}

By Ohm's Law, the measured voltage drop across a known resistance reveals the actual current flowing through that component.

2

Determine the new total circuit resistance (

R_{total}

) required to restrict the current to

1.2\text{ A}

.

R_{total} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

The total resistance determines how much current is drawn from the main voltage bus.

3

Calculate the equivalent resistance of the parallel network (

R_p

) under the current faulted condition.

R_p = R_{total} - R_1 = 100\ \Omega - 40\ \Omega = 60\ \Omega

Subtracting the series resistance from the total circuit resistance isolates the effective resistance of the parallel branch.

4

Identify the component failure that changes the parallel network's resistance from its normal

20\ \Omega

to exactly

60\ \Omega

.

The

30\ \Omega

resistor (

R_3

) must be an open circuit.

If

R_3

is open (infinite resistance), no current flows through it. The only path through the junction is via

R_2

(

60\ \Omega

), perfectly matching the required

60\ \Omega

equivalent resistance.

Key Concept

Troubleshooting Series-Parallel Circuits using Voltage Drops and Ohm's Law

Estimated Time:3m 0s

Question 25Question →

Which electrical principle or formula defines each parameter in a standard parallel circuit? Match each circuit parameter on the left with its corresponding characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Voltage across each branch matches 'Remains equal across every branch and matches the source voltage'; Total circuit current matches 'Equals the sum of all individual branch currents'; Total equivalent resistance matches 'Is always less than the resistance of the smallest branch resistor'; Current through an individual branch matches 'Calculated by dividing source voltage by the branch resistance (

I_n = \frac{V}{R_n}

)'.

In parallel circuits, voltage is identical across every branch. Total current is the sum of all individual branch currents. Total equivalent resistance decreases as more branches are added and is always less than the smallest individual resistor. Each branch current is calculated independently using Ohm's Law with the common voltage.

Step-by-Step Solution

1

Determine the voltage rule for parallel connections

Branch voltage is identical to source voltage (

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

)

All parallel branches share common input and return connection points.

2

Determine the total current rule for parallel connections

Total current is the sum of branch currents (

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

)

Current divides among all available parallel conducting branches according to Kirchhoff's Current Law.

3

Determine the equivalent resistance behavior

Equivalent resistance is less than the smallest branch resistance (

R_{\text{eq}} < R_{\text{smallest}}

)

Each additional parallel path increases overall circuit conductance.

4

Apply Ohm's law to calculate individual branch current

Branch current is

I_n = \frac{V}{R_n}

Ohm's law applies directly across each branch using the common source voltage.

Key Concept

Fundamental voltage, current, and equivalent resistance relationships in parallel circuits

Question 26Question →

In direct current (DC) electronics, circuit modifications and faults alter overall circuit parameters. Match each series circuit condition on the left with its correct electrical effect on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding a series resistor increases total resistance and decreases loop current; an open circuit drops current to zero; shorting a resistor reduces total resistance and increases current and remaining voltage drops; doubling source voltage doubles both total current and individual voltage drops.

In a series circuit, current flows through a single continuous path. Total resistance is the sum of all individual resistances (

R_T = R_1 + R_2 + \dots

). Adding a resistor increases total resistance, reducing current. Shorting a resistor bypasses that element, lowering total resistance and increasing loop current, which increases the voltage drop across the remaining resistors (

V = I \times R

). An open circuit breaks the loop, reducing current to zero. Doubling the source voltage across a fixed total resistance doubles the current and doubles each individual voltage drop proportionally.

Step-by-Step Solution

1

Analyze the resistance relationship for series circuits.

Total resistance is the direct sum of all individual resistors:

R_T = R_1 + R_2 + \dots + R_n

.

Adding a resistor increases

R_T

, while shorting out a resistor removes its resistance from the sum, lowering

R_T

.

2

Apply Ohm's Law (

I = \frac{V_T}{R_T}

) to determine current behavior under each modification.

Higher resistance reduces current, lower resistance increases current, and doubled voltage with fixed resistance doubles current.

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

3

Evaluate the single-path property of a series circuit under open fault conditions.

An open circuit provides infinite resistance (

R = \infty

), causing

I = 0\text{ A}

.

A continuous closed path is mandatory for current flow in any series circuit.

Key Concept

Series Circuit Governing Rules and Fault Characteristics

Estimated Time:1m 15s

Question 27Question →

A direct current (DC) relay coil with a fixed resistance draws $30\text{ mA}$ when connected across an $18\text{ V}$ power supply. When the same relay coil is connected to a different DC power supply, an ammeter measures a current flow of $75\text{ mA}$ through the coil. What is the voltage of the second power supply?

7.2\text{ V}

45\text{ V}

4.5\text{ V}

75\text{ V}

Show answer & explanation

Answer: $45\text{ V}$

Answer

45\text{ V}

The correct answer is determined by using Ohm's law (

R = \frac{V}{I}

) to find the coil's fixed resistance from the initial state (

R = \frac{18\text{ V}}{0.030\text{ A}} = 600\,\Omega

) and then determining the new voltage (

V = 0.075\text{ A} \times 600\,\Omega = 45\text{ V}

).

Step-by-Step Solution

1

Convert the initial current from milliamperes to amperes and calculate the resistance of the relay coil using Ohm's law (

R = \frac{V}{I}

).

I_1 = 30\text{ mA} = 0.030\text{ A}

;

R = \frac{18\text{ V}}{0.030\text{ A}} = 600\,\Omega

The resistance of the coil remains constant regardless of the applied supply voltage.

2

Convert the second current reading to amperes and calculate the new supply voltage using Ohm's law (

V = I \times R

).

I_2 = 75\text{ mA} = 0.075\text{ A}

;

V_2 = 0.075\text{ A} \times 600\,\Omega = 45\text{ V}

Ohm's law states that the potential difference across a resistor equals current multiplied by resistance.

Key Concept

Direct proportionality between voltage and current across a fixed resistance (

V = I \times R

) and metric unit conversion.

Estimated Time:45s

Question 28Question →

An emergency communications console is powered by a $120\text{ V}$ DC source connected in series with three resistive elements: $R_1 = 15\ \Omega$ , $R_2 = 25\ \Omega$ , and an unknown load resistor $R_3$ . If a voltmeter placed across $R_1$ reads $18\text{ V}$ , what is the resistance value of $R_3$ ?

85\ \Omega

60\ \Omega

100\ \Omega

104\ \Omega

Show answer & explanation

Answer: $60\ \Omega$

Answer

The resistance value of

R_3

is

60\ \Omega

.

In a series circuit, the current is identical at all points. Using Ohm's Law on the first resistor gives a loop current of

I = \frac{18\text{ V}}{15\ \Omega} = 1.2\text{ A}

. The total circuit resistance is

R_T = \frac{V_T}{I} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

. Since series resistances are additive (

R_T = R_1 + R_2 + R_3

), the unknown resistance is

R_3 = 100\ \Omega - (15\ \Omega + 25\ \Omega) = 60\ \Omega

.

Step-by-Step Solution

1

Calculate the common series circuit current using the known voltage drop and resistance of the first resistor.

I = \frac{V_1}{R_1} = \frac{18\text{ V}}{15\ \Omega} = 1.2\text{ A}

In a series circuit, current remains constant through every component.

2

Calculate the total equivalent resistance of the entire circuit using Ohm's Law with the total source voltage.

R_T = \frac{V_T}{I} = \frac{120\text{ V}}{1.2\text{ A}} = 100\ \Omega

Total resistance equals total supply voltage divided by the total circuit current.

3

Determine the resistance of the unknown resistor by subtracting the known resistors from the total series resistance.

R_3 = R_T - R_1 - R_2 = 100\ \Omega - 15\ \Omega - 25\ \Omega = 60\ \Omega

The total resistance in a series circuit is the linear sum of all individual resistances:

R_T = R_1 + R_2 + R_3

.

Key Concept

Series Circuit Analysis and Ohm's Law Multi-Step Synthesis

Estimated Time:1m 30s

Question 29Question →

An electrical technician is analyzing a perfectly ohmic heating element in a circuit. During initial operation, when an unknown voltage $V$ is applied, the element draws a current of $45 \text{ mA}$ . The technician notes that if the applied voltage is increased by exactly $500 \text{ mV}$ , the current drawn by the element increases by $2.5 \text{ mA}$ . Assuming the element's resistance remains completely constant throughout the test, what was the original applied voltage $V$ ?

9.5 V

9 V

225 V

0.9 V

Show answer & explanation

Answer: 9 V

Answer

9 V

For an ohmic component, resistance (

R

) is constant. This means the change in voltage is directly proportional to the change in current. The resistance can be found using the changes:

R = \Delta V / \Delta I

. After converting to base units,

R = 0.5 \text{ V} / 0.0025 \text{ A} = 200 \, \Omega

. Using Ohm's Law (

V = I \times R

), the original voltage is the original current (

0.045 \text{ A}

) multiplied by this resistance, resulting in

0.045 \text{ A} \times 200 \, \Omega = 9 \text{ V}

.

Step-by-Step Solution

1

Identify the relationship between the change in voltage (ΔV) and the change in current (ΔI).

R = \frac{\Delta V}{\Delta I}

For a perfectly ohmic component, the resistance is constant. This means the slope of the voltage-current relationship defines the resistance.

2

Convert the change values into standard base units (Volts and Amperes) and calculate the resistance.

\Delta V = 500 \text{ mV} = 0.5 \text{ V}

, and

\Delta I = 2.5 \text{ mA} = 0.0025 \text{ A}

.

R = \frac{0.5 \text{ V}}{0.0025 \text{ A}} = 200 \, \Omega

Standardizing metric prefixes prevents decimal shift errors before applying Ohm's Law.

3

Apply Ohm's Law to the original state to find the initial voltage.

V = I_{original} \times R = 0.045 \text{ A} \times 200 \, \Omega = 9 \text{ V}

Multiplying the originally measured current by the established component resistance yields the original voltage applied.

Key Concept

Applying Ohm's Law using proportional changes (ΔV and ΔI) while accurately managing metric prefix conversions.

Question 30Question →

Four identical $40\ \Omega$ heating elements are connected in parallel across a $120\text{ V}$ power line. What is the total equivalent resistance of this circuit?

160\ \Omega

10\ \Omega

40\ \Omega

20\ \Omega

Show answer & explanation

Answer: $10\ \Omega$

Answer

The total equivalent resistance of the circuit is

10\ \Omega

.

For

N

identical resistors of equal resistance

R

connected in parallel, the total equivalent resistance is simply the resistance of one branch divided by the total number of branches:

R_{\text{eq}} = \frac{R}{N} = \frac{40\ \Omega}{4} = 10\ \Omega

.

Step-by-Step Solution

1

Identify the number of parallel branches (

N

) and individual branch resistance (

R

).

N = 4

identical branches, each with

R = 40\ \Omega

.

Recognizing identical parallel branches allows the use of the equal-resistor shortcut formula.

2

Apply the parallel equivalent resistance formula for identical resistors:

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

.

R_{\text{eq}} = \frac{40\ \Omega}{4} = 10\ \Omega

.

Connecting resistors in parallel creates multiple paths for electron flow, reducing the total circuit resistance below that of any individual branch.

Key Concept

Total Equivalent Resistance of Equal Parallel Resistors

Estimated Time:45s

Question 31Question →

A surveillance camera heating assembly is powered by a $120\text{ V}$ DC power supply connected in series with three heating elements: $R_1 = 18\ \Omega$ , $R_2 = 22\ \Omega$ , and $R_3 = 40\ \Omega$ . What is the voltage drop across the $22\ \Omega$ element ( $R_2$ )?

120\text{ V}

33\text{ V}

40\text{ V}

1.5\text{ V}

Show answer & explanation

Answer: $33\text{ V}$

Answer

33\text{ V}

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

18\ \Omega + 22\ \Omega + 40\ \Omega = 80\ \Omega

). Applying Ohm's Law with a

120\text{ V}

supply gives a uniform loop current of

1.5\text{ A}

(

120\text{ V} / 80\ \Omega

). Multiplying this current by the resistance of the specified component yields

1.5\text{ A} \times 22\ \Omega = 33\text{ V}

.

Step-by-Step Solution

1

Calculate the total series resistance (

R_T

) by summing individual resistances.

R_T = R_1 + R_2 + R_3 = 18\ \Omega + 22\ \Omega + 40\ \Omega = 80\ \Omega

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

2

Determine the total circuit current (

I_T

) using Ohm's Law (

I = V / R

).

I_T = \frac{V_T}{R_T} = \frac{120\text{ V}}{80\ \Omega} = 1.5\text{ A}

Current remains constant through every component in a single continuous series loop.

3

Calculate the voltage drop (

V_2

) across the target resistor (

R_2 = 22\ \Omega

) using

V = I \times R

.

V_2 = I_T \times R_2 = 1.5\text{ A} \times 22\ \Omega = 33\text{ V}

The voltage drop across any individual series resistor is proportional to its resistance and the loop current.

Key Concept

Series Circuit Resistance and Voltage Division

Estimated Time:1m 0s

Question 32Question →

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 33Question →

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 34Question →

Two resistors with values of $12\ \Omega$ and $6\ \Omega$ are wired in parallel across a circuit board. What is the total equivalent resistance of this combination?

18\ \Omega

4\ \Omega

9\ \Omega

72\ \Omega

Show answer & explanation

Answer: $4\ \Omega$

Answer

4\ \Omega

For two resistors in parallel, the equivalent resistance is calculated by dividing their product by their sum:

\frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega

. In parallel circuits, the equivalent resistance is always lower than any single branch resistance.

Step-by-Step Solution

1

Identify the given branch resistances and select the parallel equivalent resistance formula.

R_1 = 12\ \Omega

,

R_2 = 6\ \Omega

, and

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

The product-over-sum formula determines the total equivalent resistance of two parallel branches.

2

Calculate the product and sum of the two resistors, then divide.

R_{eq} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega

Completing the arithmetic yields the equivalent resistance, which must be less than the smallest individual branch.

Key Concept

Total equivalent resistance in a parallel circuit is always less than the lowest individual branch resistance and can be calculated using the reciprocal or product-over-sum formula.

Estimated Time:45s

Question 35Question →

An aircraft avionics sensor circuit is powered by a $28\text{ V}$ direct current (DC) power source connected in series with three resistors: $R_1 = 30\ \Omega$ , $R_2 = 45\ \Omega$ , and an unknown resistor $R_3$ . A technician measures the voltage drop across $R_2$ and finds it to be $9\text{ V}$ . Based on these circuit measurements, what are the total loop current and the resistance value of $R_3$ ?

Fill in the blanks below

The total current flowing through the circuit is A, and the resistance value of

R_3

is

\Omega

.

Show answer & explanation

Answer

The circuit current is 0.2 A and the resistance of resistor R3 is 65 ohms.

Because the circuit components are connected in series, the current flowing through

R_2

(

I = 9\text{ V} / 45\ \Omega = 0.2\text{ A}

) is the exact same current passing through the entire loop. Applying Ohm's Law to the full circuit gives a total resistance of

R_T = 28\text{ V} / 0.2\text{ A} = 140\ \Omega

. Subtracting the known resistances (

30\ \Omega + 45\ \Omega = 75\ \Omega

) leaves

65\ \Omega

for

R_3

.

Step-by-Step Solution

1

Calculate the constant loop current using the known values for resistor

R_2

I = \frac{V_2}{R_2} = \frac{9\text{ V}}{45\ \Omega} = 0.2\text{ A}

In a series circuit, current is identical through every component.

2

Calculate the total equivalent resistance of the entire series circuit using Ohm's Law and the source voltage

R_T = \frac{V_{\text{source}}}{I} = \frac{28\text{ V}}{0.2\text{ A}} = 140\ \Omega

Total circuit resistance equals the total applied voltage divided by the common series current.

3

Determine the resistance of

R_3

by subtracting known series resistances from total resistance

R_3 = R_T - (R_1 + R_2) = 140\ \Omega - (30\ \Omega + 45\ \Omega) = 140\ \Omega - 75\ \Omega = 65\ \Omega

Total series resistance is the direct algebraic sum of all individual resistances in the loop.

Key Concept

In a single-loop series circuit, current is constant at all points (

I_T = I_1 = I_2 = I_3

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

R_T = R_1 + R_2 + R_3

).

Estimated Time:1m 15s

Question 36Question →

An electronics technician is testing a DC bias circuit containing a fixed resistor connected to an adjustable DC power supply. When the supply voltage is set to $28\text{ V}$ , the circuit ammeter measures a current of $3.5\text{ mA}$ . To increase the current flow through the resistor to $11\text{ mA}$ , what must the power supply voltage be set to?

60\text{ V}

8.8\text{ V}

88\text{ V}

22\text{ V}

Show answer & explanation

Answer: $88\text{ V}$

Answer

The power supply voltage must be set to

88\text{ V}

.

The power supply must be set to

88\text{ V}

. First, determine the component's constant resistance:

R = \frac{28\text{ V}}{3.5\text{ mA}} = 8\text{ k}\Omega

. Next, apply Ohm's Law to find the voltage required for the target current:

V = 11\text{ mA} \times 8\text{ k}\Omega = 88\text{ V}

. Alternatively, using direct proportionality:

V_2 = 28\text{ V} \times \left(\frac{11\text{ mA}}{3.5\text{ mA}}\right) = 88\text{ V}

.

Step-by-Step Solution

1

Calculate the fixed resistance using the initial voltage and current measurements.

R = \frac{V_1}{I_1} = \frac{28\text{ V}}{3.5\text{ mA}} = \frac{28\text{ V}}{0.0035\text{ A}} = 8{}000\,\Omega = 8\text{ k}\Omega

Ohm's Law defines resistance as the ratio of voltage to current (

R = \frac{V}{I}

).

2

Calculate the new voltage required to produce

11\text{ mA}

through the

8\text{ k}\Omega

resistor.

V_2 = I_2 \times R = 0.011\text{ A} \times 8{}000\,\Omega = 88\text{ V}

Applying Ohm's Law in the form

V = I \times R

determines the potential difference needed for the desired current.

Key Concept

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

V = I \times R

). In a circuit with constant resistance, voltage is directly proportional to current.

Alternative Method

Because resistance is constant, voltage is directly proportional to current:

V_2 = V_1 \times \frac{I_2}{I_1} = 28\text{ V} \times \frac{11\text{ mA}}{3.5\text{ mA}} = 28 \times \frac{22}{7} = 4 \times 22 = 88\text{ V}

.

Estimated Time:45s

Question 37Question →

A technician is evaluating the electrical characteristics of a direct-current parallel circuit network. Match each circuit parameter on the left with its corresponding behavior or rule on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Branch Voltage matches 'Remains constant and equal across all connected branches'; Total Current matches 'Equals the sum of the individual currents flowing through all branches'; Equivalent Resistance matches 'Is always lower than the resistance of the smallest branch resistor'; Branch Current matches 'Varies inversely with the branch resistance according to

I = \frac{V}{R}

'.

In parallel circuits, all components share common connection points, maintaining equal voltage across each branch. Total current equals the sum of all individual branch currents. Total equivalent resistance decreases as more branches are added (remaining less than the smallest branch resistance). Individual branch current is calculated via Ohm's Law as voltage divided by that branch's resistance.

Step-by-Step Solution

1

Determine the voltage relationship across parallel branches.

Branch Voltage (

V_{\text{branch}}

) is equal across all branches (

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

).

All branches connect across the same two common nodes to the power supply.

2

Determine the total current relationship for a parallel network.

Total Current (

I_{\text{total}}

) equals the sum of branch currents (

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

).

Current splits at junction nodes according to Kirchhoff's Current Law.

3

Determine how total equivalent resistance behaves in a parallel circuit.

Equivalent Resistance (

R_{\text{eq}}

) is always smaller than the smallest branch resistance.

Adding parallel branches adds additional conductive paths for electron flow.

4

Determine how current through a specific branch is calculated.

Branch Current (

I_{\text{branch}}

) is calculated using Ohm's Law (

I_{\text{branch}} = \frac{V}{R_{\text{branch}}}

).

Branch current depends on the common branch voltage divided by the individual branch resistance.

Key Concept

Fundamental rules governing voltage, current, and resistance in DC parallel circuits.

Estimated Time:45s

Question 38Question →

A military vehicle communication module is powered by a $60\text{ V}$ DC supply connected in series with three resistive components: a fixed input resistor $R_1 = 40\ \Omega$ , an unknown matching resistor $R_2$ , and a load resistor $R_3 = 80\ \Omega$ . If a technician measures a voltage drop of $12\text{ V}$ across resistor $R_1$ , what is the resistance value of $R_2$ ?

40\ \Omega

160\ \Omega

80\ \Omega

200\ \Omega

Show answer & explanation

Answer: $80\ \Omega$

Answer

80\ \Omega

Because current is uniform throughout a single series path, the current through the entire circuit is determined by the known component

R_1

:

I = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}

. With a

60\text{ V}

supply, the total circuit resistance is

R_T = \frac{60\text{ V}}{0.3\text{ A}} = 200\ \Omega

. Since

R_T = R_1 + R_2 + R_3

, subtracting the known values gives

R_2 = 200\ \Omega - 40\ \Omega - 80\ \Omega = 80\ \Omega

.

Step-by-Step Solution

1

Calculate the common series loop current using Ohm's Law across the first known resistor.

I = \frac{V_1}{R_1} = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}

In a series circuit, the current is identical at all points along the loop.

2

Calculate the total equivalent resistance of the circuit using the total supply voltage and loop current.

R_T = \frac{V_{\text{total}}}{I} = \frac{60\text{ V}}{0.3\text{ A}} = 200\ \Omega

Ohm's Law relates the source voltage to the total circuit resistance and total current.

3

Subtract the known series resistances from the total circuit resistance to isolate the unknown resistor.

R_2 = R_T - R_1 - R_3 = 200\ \Omega - 40\ \Omega - 80\ \Omega = 80\ \Omega

Total resistance in a series circuit is the algebraic sum of all individual resistances (

R_T = R_1 + R_2 + R_3

).

Key Concept

In a series circuit, current remains constant through every component, and total circuit resistance is the sum of individual resistances (

R_T = R_1 + R_2 + R_3

).

Alternative Method

Find the loop current

I = \frac{12\text{ V}}{40\ \Omega} = 0.3\text{ A}

. Then find the voltage drop across

R_3

:

V_3 = 0.3\text{ A} \times 80\ \Omega = 24\text{ V}

. Using Kirchhoff's Voltage Law, find the voltage drop across

R_2

:

V_2 = 60\text{ V} - 12\text{ V} - 24\text{ V} = 24\text{ V}

. Finally, calculate

R_2 = \frac{V_2}{I} = \frac{24\text{ V}}{0.3\text{ A}} = 80\ \Omega

.

Estimated Time:1m 30s

Question 39Question →

An electric vehicle telemetry subsystem is powered by a $45\text{ V}$ direct current (DC) source connected in series with three resistive sensors: $R_1 = 7\ \Omega$ , $R_2 = 11\ \Omega$ , and $R_3 = 12\ \Omega$ . What is the voltage drop across the $11\ \Omega$ sensor ( $R_2$ )?

15.0\text{ V}

16.5\text{ V}

45.0\text{ V}

1.5\text{ V}

Show answer & explanation

Answer: $16.5\text{ V}$

Answer

16.5\text{ V}

(the voltage drop across the

11\ \Omega

resistor)

The correct answer is

16.5\text{ V}

. In a series circuit, total resistance is found by adding all individual resistors (

7\ \Omega + 11\ \Omega + 12\ \Omega = 30\ \Omega

). Dividing the

45\text{ V}

source by the

30\ \Omega

total resistance yields a circuit current of

1.5\text{ A}

. Multiplying this

1.5\text{ A}

current by the

11\ \Omega

resistance of the specified resistor gives a voltage drop of

16.5\text{ V}

.

Step-by-Step Solution

1

Calculate the total series resistance (

R_T

) by summing the three individual resistances.

R_T = R_1 + R_2 + R_3 = 7\ \Omega + 11\ \Omega + 12\ \Omega = 30\ \Omega

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

2

Calculate the total loop current (

I

) using Ohm's Law.

I = \frac{V_T}{R_T} = \frac{45\text{ V}}{30\ \Omega} = 1.5\text{ A}

Current remains constant and identical through all components in a series loop.

3

Calculate the voltage drop across

R_2

using Ohm's Law.

V_2 = I \times R_2 = 1.5\text{ A} \times 11\ \Omega = 16.5\text{ V}

The voltage drop across any individual resistor is the product of the common loop current and that resistor's resistance.

Key Concept

In a series circuit, current is identical through every component, and the source voltage divides across each resistor in direct proportion to its resistance value (

V_x = I \times R_x

).

Alternative Method

Use the voltage divider formula:

V_2 = V_T \times \left(\frac{R_2}{R_T}\right) = 45\text{ V} \times \left(\frac{11\ \Omega}{30\ \Omega}\right) = 16.5\text{ V}

.

Estimated Time:1m 0s

Question 40Question →

What is required to induce an electric current in a stationary coil of wire using a magnet?

A constant, stationary magnetic field held next to the coil

Relative motion between the magnetic field and the coil

An electrostatic charge stored inside an insulating dielectric

A fixed potential difference that prevents magnetic line movement

Show answer & explanation

Answer: Relative motion between the magnetic field and the coil

Answer

Relative motion between the magnetic field and the coil is required to induce an electric current.

According to the principle of electromagnetic induction, an electric current is generated in a conductor when magnetic lines of force cut across the wire. This requires relative motion between the magnetic field and the coil, creating a changing magnetic flux that produces an electromotive force (voltage).

Step-by-Step Solution

1

Identify the fundamental principle of electromagnetic induction defined by Faraday's Law.

An electromotive force (EMF) is induced in a wire loop only when the loop experiences a changing magnetic flux.

Stationary magnetic fields do not cut across stationary conductors, producing zero induced voltage.

2

Determine the physical condition necessary to produce this change in magnetic flux.

There must be relative motion between the magnet and the coil so that magnetic lines of force cut across the turns of wire.

Moving the magnet relative to the coil causes the magnetic flux density through the coil to vary with time, driving electron movement.

Key Concept

Faraday's Law of Electromagnetic Induction

Estimated Time:45s

Circuits and Magnetism

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