Circuits and Magnetism Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 7

Question 121Question →

A $24\text{ V}$ direct-current (DC) power source is connected to a circuit where resistor $R_1$ ( $4\ \Omega$ ) is in series with a parallel pair consisting of $R_2$ ( $12\ \Omega$ ) and $R_3$ ( $6\ \Omega$ ). Under normal operation, the total current is $3.0\text{ A}$ and the voltage drop across the parallel node is $12\text{ V}$ . An electronics technician records abnormal multimeter readings indicating different single-component faults. Match each set of diagnostic symptoms to the corresponding circuit fault.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The symptoms map to the faults as follows: 2.4 A and 14.4 V indicates R2 is open; 1.5 A and 18.0 V indicates R3 is open; 6.0 A and 0 V indicates a short in the parallel section (R2 or R3); 0 A and 24 V across R1 indicates R1 is open.

Each diagnostic reading mathematically corresponds to a specific component failure. When a component fails open, its resistance becomes infinite, altering the equivalent circuit resistance and redistributing voltage drops. When a component short-circuits, it bypasses that section of the circuit with near-zero resistance, causing a current spike and eliminating the voltage drop across that section. Placing a voltmeter across an open component in a series path will read the full source voltage.

Step-by-Step Solution

1

Analyze the normal circuit parameters to establish a baseline for comparison.

Normal equivalent parallel resistance (

R_p

) is

4\ \Omega

. Total resistance (

R_t

) is

8\ \Omega

. Normal current is

3.0\text{ A}

.

Baseline parameters are necessary to evaluate how mathematical deviations correlate to specific component failures.

2

Evaluate the symptom with

I = 2.4\text{ A}

and

V_p = 14.4\text{ V}

.

Total resistance increased to

10\ \Omega

(

24\text{ V} / 2.4\text{ A}

). Since

R_1

is

4\ \Omega

, the parallel section must now be

6\ \Omega

. This matches the value of

R_3

alone.

If

R_2

fails open, its branch has infinite resistance, leaving only

R_3

(

6\ \Omega

) to carry current in the parallel node.

3

Evaluate the symptom with

I = 1.5\text{ A}

and

V_p = 18.0\text{ V}

.

Total resistance increased to

16\ \Omega

(

24\text{ V} / 1.5\text{ A}

). The parallel section must now be

12\ \Omega

, which matches the value of

R_2

alone.

If

R_3

fails open, its branch has infinite resistance, leaving only

R_2

to carry current in the parallel section.

4

Evaluate the symptom with

I = 6.0\text{ A}

and

V_p = 0\text{ V}

.

Total resistance dropped to

4\ \Omega

, exactly equal to

R_1

. The parallel section resistance is

0\ \Omega

.

A short circuit in either

R_2

or

R_3

bypasses the resistance of the parallel node entirely, dropping its voltage to zero and causing a current spike.

5

Evaluate the symptom with

I = 0\text{ A}

and

24\text{ V}

measured directly across

R_1

.

An open circuit exists at

R_1

.

When a component fails open, the entire source voltage drops across the break because there is no voltage drop across the rest of the intact circuit components (since current is zero).

Key Concept

Circuit Fault Diagnostics using Ohm's Law and Series-Parallel Analysis

Estimated Time:3m 0s

Question 122Question →

An aircraft DC power distribution bus supplies $36\text{ V}$ across two parallel branch circuits. The first branch contains a resistive load of $12\ \Omega$ , and an ammeter at the main bus measures a total circuit current of $5\text{ A}$ . What is the resistance of the load in the second branch?

7.2\ \Omega

18\ \Omega

12\ \Omega

72\ \Omega

Show answer & explanation

Answer: $18\ \Omega$

Answer

The resistance of the second branch load is

18\ \Omega

.

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

36\text{ V}

. Using Ohm's Law, the first branch draws

I_1 = 36\text{ V} / 12\ \Omega = 3\text{ A}

. Subtracting this from the total current of

5\text{ A}

gives a second branch current of

I_2 = 2\text{ A}

. Dividing the

36\text{ V}

bus voltage by

2\text{ A}

yields a resistance of

18\ \Omega

for the second branch.

Step-by-Step Solution

1

Calculate the current flowing through the first branch using Ohm's Law (

I_1 = V / R_1

).

I_1 = 36\text{ V} / 12\ \Omega = 3\text{ A}

In a parallel circuit, the full supply voltage of

36\text{ V}

is present across every parallel branch.

2

Determine the current flowing through the second branch using Kirchhoff's Current Law (

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

).

I_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}

Total circuit current equals the sum of all individual branch currents in a parallel circuit.

3

Calculate the resistance of the second branch using Ohm's Law (

R_2 = V / I_2

).

R_2 = 36\text{ V} / 2\text{ A} = 18\ \Omega

Dividing the common branch voltage by the branch current yields the branch resistance.

Key Concept

Parallel branch current relationships and individual component resistance calculations using Ohm's Law and Kirchhoff's Current Law.

Alternative Method

Find the total circuit equivalent resistance first:

R_{\text{eq}} = 36\text{ V} / 5\text{ A} = 7.2\ \Omega

. Then apply the parallel resistance formula:

1/R_2 = 1/R_{\text{eq}} - 1/R_1 = 1/7.2 - 1/12 = 5/36 - 3/36 = 2/36 = 1/18

, which gives

R_2 = 18\ \Omega

.

Estimated Time:1m 0s

Question 123Question →

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 124Question →

A backup telemetry transmitter operates from a regulated $60\text{ V}$ direct-current (DC) power source connected across two parallel branch circuits. The primary transmitter branch has a resistance of $15\ \Omega$ , and the secondary cooling fan branch has a resistance of $30\ \Omega$ . What is the total current drawn from the power source?

8\text{ A}

6\text{ A}

1.33\text{ A}

10\text{ A}

Show answer & explanation

Answer: $6\text{ A}$

Answer

The total current drawn from the power source is

6\text{ A}

.

The correct answer is

6\text{ A}

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

60\text{ V}

. Applying Ohm's Law to each branch yields

I_1 = \frac{60\text{ V}}{15\ \Omega} = 4\text{ A}

and

I_2 = \frac{60\text{ V}}{30\ \Omega} = 2\text{ A}

. Summing these branch currents gives a total current of

4\text{ A} + 2\text{ A} = 6\text{ A}

. Alternatively, finding the equivalent resistance (

R_{\text{eq}} = \frac{15 \times 30}{15 + 30} = 10\ \Omega

) and applying Ohm's Law (

I_{\text{total}} = \frac{60\text{ V}}{10\ \Omega}

) confirms

6\text{ A}

.

Step-by-Step Solution

1

Determine the voltage across each parallel branch

V_1 = V_2 = 60\text{ V}

In a parallel circuit, the full supply voltage drops equally across all parallel branches.

2

Calculate the individual branch currents using Ohm's Law (

I = \frac{V}{R}

)

I_1 = \frac{60\text{ V}}{15\ \Omega} = 4\text{ A}

,

I_2 = \frac{60\text{ V}}{30\ \Omega} = 2\text{ A}

Each branch current is determined independently by the branch resistance and the common voltage.

3

Sum the branch currents to find the total circuit current (

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

)

I_{\text{total}} = 4\text{ A} + 2\text{ A} = 6\text{ A}

According to Kirchhoff's Current Law, total current entering a parallel junction equals the sum of currents leaving through the branches.

Key Concept

In a DC parallel circuit, voltage remains constant across all branches, and the total current supplied is the sum of the individual branch currents (

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

).

Alternative Method

Find equivalent resistance first:

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{15 \times 30}{15 + 30} = \frac{450}{45} = 10\ \Omega

. Then calculate total current:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{60\text{ V}}{10\ \Omega} = 6\text{ A}

.

Estimated Time:1m 0s

Question 125Question →

In an auxiliary radio circuit, a direct-current source applies $24\text{ V}$ across a fixed resistor, producing a measured current flow of $8\text{ mA}$ . What is the resistance of the component?

3\ \Omega

192\ \Omega

300\ \Omega

3{}000\ \Omega

Show answer & explanation

Answer: $3{}000\ \Omega$

Answer

3{}000\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) with the current converted to amperes (

8\text{ mA} = 0.008\text{ A}

) gives

\frac{24\text{ V}}{0.008\text{ A}} = 3{}000\ \Omega

.

Step-by-Step Solution

1

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

8\text{ mA} = 8 \times 10^{-3}\text{ A} = 0.008\text{ A}

Ohm's Law calculations require current in base units of amperes.

2

Apply Ohm's Law to solve for resistance (

R

)

R = \frac{V}{I} = \frac{24\text{ V}}{0.008\text{ A}} = 3{}000\ \Omega

Resistance equals voltage divided by current.

Key Concept

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

R = \frac{V}{I}

).

Estimated Time:45s

Question 126Question →

A direct-current (DC) power supply delivers a constant $90\text{ V}$ to two resistors connected in parallel. An ammeter measures the total circuit current leaving the source as $5\text{ A}$ . If the resistance of the first resistor ( $R_1$ ) is $30\ \Omega$ , what is the resistance of the second resistor ( $R_2$ )?

12\ \Omega

18\ \Omega

36\ \Omega

45\ \Omega

Show answer & explanation

Answer: $45\ \Omega$

Answer

The resistance of the second resistor (

R_2

) is

45\ \Omega

.

In any parallel circuit, the voltage across each branch is identical to the source voltage (

90\text{ V}

). The current in the first branch is

I_1 = 90\text{ V} / 30\ \Omega = 3\text{ A}

. According to Kirchhoff's Current Law, the total current equals the sum of branch currents (

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

). Therefore, the current in the second branch is

I_2 = 5\text{ A} - 3\text{ A} = 2\text{ A}

. Applying Ohm's Law to the second branch gives

R_2 = 90\text{ V} / 2\text{ A} = 45\ \Omega

.

Step-by-Step Solution

1

Determine the current flowing through the first parallel branch (

I_1

)

I_1 = \frac{V}{R_1} = \frac{90\text{ V}}{30\ \Omega} = 3\text{ A}

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

90\text{ V}

).

2

Calculate the current flowing through the second branch (

I_2

) using Kirchhoff's Current Law

I_2 = I_{\text{total}} - I_1 = 5\text{ A} - 3\text{ A} = 2\text{ A}

The sum of individual branch currents equals the total current supplied by the source.

3

Calculate the resistance of the second resistor (

R_2

) using Ohm's Law

R_2 = \frac{V}{I_2} = \frac{90\text{ V}}{2\text{ A}} = 45\ \Omega

Ohm's Law defines resistance as the voltage drop across the component divided by the current through it.

Key Concept

Parallel branch voltage equality and Kirchhoff's Current Law

Estimated Time:1m 15s

Question 127Question →

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 128Question →

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 129Question →

An off-grid solar installation supplies direct-current (DC) power to two ventilation fan circuits connected in parallel across the main distribution bus. The primary intake fan circuit presents a resistance of $40\ \Omega$ , while the secondary exhaust fan circuit presents a resistance of $60\ \Omega$ . What is the total equivalent resistance of these two parallel circuits combined?

100\ \Omega

24\ \Omega

50\ \Omega

20\ \Omega

Show answer & explanation

Answer: $24\ \Omega$

Answer

24\ \Omega

For two resistors connected in parallel, the total equivalent resistance (

R_{\text{eq}}

) is calculated using the product-over-sum formula:

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

. Substituting the circuit values yields

R_{\text{eq}} = \frac{40 \times 60}{40 + 60} = \frac{2400}{100} = 24\ \Omega

. In any parallel circuit, the total equivalent resistance must always be less than the resistance of the smallest branch (

24\ \Omega < 40\ \Omega

).

Step-by-Step Solution

1

Identify the parallel resistance formula for two branch loads

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

The equivalent resistance of two parallel resistors is given by their product divided by their sum.

2

Calculate the product of the two branch resistances

40\ \Omega \times 60\ \Omega = 2400\ \Omega^2

Multiplying the individual branch values forms the numerator of the formula.

3

Calculate the sum of the two branch resistances

40\ \Omega + 60\ \Omega = 100\ \Omega

Adding the branch values forms the denominator of the formula.

4

Divide the product by the sum to determine equivalent resistance

R_{\text{eq}} = \frac{2400}{100} = 24\ \Omega

Evaluating the quotient provides the total opposition to current flow across the parallel network.

Key Concept

Parallel Equivalent Resistance

Estimated Time:1m 0s

Question 130Question →

A telemetry sensor on an unmanned aerial vehicle is connected to a $15\text{ V}$ direct-current (DC) power supply. If the sensor circuit has an effective resistance of $2.5\text{ k}\Omega$ , what is the steady current flowing through the circuit in milliamperes ( $\text{mA}$ )?

37.5\text{ mA}

6\text{ mA}

60\text{ mA}

167\text{ mA}

Show answer & explanation

Answer: $6\text{ mA}$

Answer

The steady current flowing through the sensor circuit is

6\text{ mA}

.

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

I = \frac{V}{R}

). Converting

2.5\text{ k}\Omega

to

2{}500\ \Omega

gives

I = \frac{15\text{ V}}{2{}500\ \Omega} = 0.006\text{ A}

, which equals

6\text{ mA}

.

Step-by-Step Solution

1

Convert the resistance value from kilohms (

\text{k}\Omega

) to base ohms (

\Omega

).

R = 2.5\text{ k}\Omega = 2.5 \times 10^3\ \Omega = 2{}500\ \Omega

Ohm's Law calculations require base electrical units (volts, amperes, and ohms).

2

Apply Ohm's Law to solve for current in amperes (

I = \frac{V}{R}

).

I = \frac{15\text{ V}}{2{}500\ \Omega} = 0.006\text{ A}

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

3

Convert the resulting current from amperes (

\text{A}

) to milliamperes (

\text{mA}

).

I = 0.006\text{ A} \times 1{}000\text{ mA/A} = 6\text{ mA}

The question asks for the final answer in milliamperes.

Key Concept

Ohm's Law defines the relationship between voltage, current, and resistance (

I = \frac{V}{R}

).

Estimated Time:45s

Question 131Question →

In direct-current (DC) parallel circuit troubleshooting and analysis, changes in circuit configuration directly affect voltage, current, and equivalent resistance. Match each circuit condition or modification with its corresponding electrical effect.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding a parallel branch matches with decreasing overall equivalent resistance and increasing total current; opening one branch matches with reducing total current while remaining branches stay unchanged; connecting unequal branches matches with lower resistance branches drawing higher current; measuring voltage drops matches with identical potential difference readings across every branch.

In parallel DC circuits, adding a branch reduces total resistance and increases total current, opening a branch isolates only that path while others continue operating normally, current divides inversely with branch resistance under constant voltage, and voltage is identical across every parallel branch.

Step-by-Step Solution

1

Analyze the effect of adding a parallel branch to an active circuit

Equivalent resistance decreases and total current increases

Every additional parallel path offers a new route for electron flow, reducing the total opposition to current.

2

Analyze the effect of an open fault in a single branch of a parallel network

Total current drops while other branches continue normal operation

Parallel branches are electrically independent across the supply rails, so opening one branch subtracts only its individual branch current from the total.

3

Determine how current distributes across branches with different resistances

Current divides inversely with resistance

With uniform branch voltages, Ohm's law (

I = \frac{V}{R}

) ensures that smaller resistances allow larger branch currents.

4

Verify voltage characteristics across parallel circuit elements

Voltage drop across every branch equals the source voltage

All parallel branches connect directly across the common positive and negative power distribution rails.

Key Concept

Fundamental operational principles of parallel circuits: equal branch voltage, independent branch currents, and reciprocal equivalent resistance

Estimated Time:1m 0s

Question 132Question →

An avionics technician is analyzing a direct-current (DC) subassembly with an applied voltage of $36\text{ V}$ across a load resistor rated at $1.8\text{ k}\Omega$ . What is the resistance in ohms and the resulting current in milliamperes? Fill in the blanks below to complete the circuit analysis.

Fill in the blanks below

The total resistance of the load is

\Omega

, and the steady current flowing through the circuit is

\text{mA}

.

Show answer & explanation

Answer

The resistance is 1800 Ω and the current is 20 mA.

Converting

1.8\text{ k}\Omega

to base units gives

1800\ \Omega

. Dividing the applied voltage of

36\text{ V}

by

1800\ \Omega

yields

0.02\text{ A}

, which corresponds to

20\text{ mA}

.

Step-by-Step Solution

1

Convert the load resistance from kilohms (kΩ) to ohms (Ω)

R = 1.8\text{ k}\Omega = 1.8 \times 1000\ \Omega = 1800\ \Omega

Standard electrical calculations using Ohm's Law require base units of ohms, volts, and amperes.

2

Apply Ohm's Law (

I = \frac{V}{R}

) to calculate the circuit current in amperes

I = \frac{36\text{ V}}{1800\ \Omega} = 0.02\text{ A}

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

3

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

I = 0.02\text{ A} \times 1000\text{ mA/A} = 20\text{ mA}

The required unit for the current blank is milliamperes.

Key Concept

Ohm's Law (

I = \frac{V}{R}

) and metric prefix conversions (kilo- and milli-).

Estimated Time:45s

Question 133Question →

A mobile radar tracking unit contains two heating elements connected in parallel across a common direct-current voltage rail. The primary heating element has a resistance of $18\ \Omega$ , and the secondary heating element has a resistance of $9\ \Omega$ . What is the total equivalent resistance of this circuit?

27\ \Omega

6\ \Omega

13.5\ \Omega

2\ \Omega

Show answer & explanation

Answer: $6\ \Omega$

Answer

The total equivalent resistance of the parallel circuit is

6\ \Omega

.

The equivalent resistance of parallel resistors is found using the formula

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

or the shortcut

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

. Substituting

18\ \Omega

and

9\ \Omega

yields

\frac{18 \times 9}{18 + 9} = \frac{162}{27} = 6\ \Omega

, which is properly less than the smallest branch resistance (

9\ \Omega

).

Step-by-Step Solution

1

Identify the standard equivalent resistance relationship for two unequal parallel branches

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

or

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

Parallel branches provide multiple paths for current, resulting in a total equivalent resistance that is always smaller than the lowest individual branch resistance.

2

Substitute the given resistances (

R_1 = 18\ \Omega

and

R_2 = 9\ \Omega

) into the product-over-sum formula

R_{eq} = \frac{18 \times 9}{18 + 9} = \frac{162}{27}

Using the product-over-sum method allows direct calculation without converting to fractions with a common denominator.

3

Divide the product by the sum to determine the equivalent resistance

R_{eq} = 6\ \Omega

Simplifying

\frac{162}{27}

yields the total opposition to current offered by the two parallel elements.

Key Concept

Total Equivalent Resistance of Unequal Parallel Branches

Estimated Time:1m 0s

Question 134Question →

A technician is testing a direct-current (DC) indicator lamp assembly in an armored vehicle. The lamp is connected to a $24\text{ V}$ power supply and draws a steady current of $75\text{ mA}$ . What is the total electrical resistance of the lamp assembly?

0.32\ \Omega

1.8\ \Omega

320\ \Omega

1,800\ \Omega

Show answer & explanation

Answer: $320\ \Omega$

Answer

The total resistance of the lamp assembly is

320\ \Omega

.

According to Ohm's Law (

V = I \times R

), resistance equals voltage divided by current (

R = \frac{V}{I}

). Converting

75\text{ mA}

to

0.075\text{ A}

and dividing

24\text{ V}

by

0.075\text{ A}

yields

320\ \Omega

.

Step-by-Step Solution

1

Convert the measured current from milliamperes (

\text{mA}

) to base amperes (

\text{A}

).

I = 75\text{ mA} = \frac{75}{1000}\text{ A} = 0.075\text{ A}

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

2

Rearrange Ohm's Law (

V = I \times R

) to solve for resistance (

R = \frac{V}{I}

) and substitute the known values.

R = \frac{24\text{ V}}{0.075\text{ A}} = 320\ \Omega

Dividing the applied voltage by the circuit current gives the total resistance.

Key Concept

Ohm's Law defines the relationship between voltage, current, and resistance (

V = I \times R

, rearranged as

R = \frac{V}{I}

). Current must always be converted to base amperes before computing resistance.

Estimated Time:45s

Question 135Question →

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 136Question →

In direct-current (DC) circuit analysis, modifications to individual parallel paths directly impact total resistance, total current, and branch behavior. Match each parallel circuit modification on the left with its resulting electrical effect on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding an additional branch matches with decreased total resistance, increased total current, and unaffected existing branches; Increasing one branch resistance matches with decreased branch current, decreased total current, and increased total resistance; Opening one branch matches with increased total resistance, decreased total current, and normal operation of remaining branches; Connecting a short circuit matches with total resistance dropping to zero, bypassing parallel loads, and causing an excessive current surge.

In parallel circuits, each branch receives full source voltage independently. Adding branches decreases total resistance and increases total current without changing existing branch currents. Changing one branch resistance alters only that branch's current and total current. An open circuit disables only the affected branch, whereas a short circuit drops total resistance to zero across all branches.

Step-by-Step Solution

1

Analyze the impact of adding a parallel branch

R_{\text{eq}}

decreases and

I_{\text{total}}

increases

Each additional parallel path increases overall circuit conductance (

G_{\text{total}} = G_1 + G_2 + \dots

), lowering equivalent resistance.

2

Apply Ohm's law to evaluate increasing a single branch's resistance

I_{\text{branch}}

drops,

I_{\text{total}}

drops, and

R_{\text{eq}}

rises

With branch voltage fixed by the power source, current through that branch varies inversely with its resistance.

3

Determine the effect of an open-circuit fault in one parallel branch

Defective branch current goes to zero while intact branches continue functioning

Parallel branches are electrically independent because each branch connects directly across the full supply voltage.

4

Evaluate a short-circuit condition across any parallel branch

R_{\text{eq}} \approx 0\ \Omega

and total current surges drastically

A parallel short bridges the supply rails directly, shunting current away from all resistive paths.

Key Concept

Independent branch operation and equivalent resistance behavior in parallel DC circuits

Estimated Time:1m 0s

Question 137Question →

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 138Question →

A $120\text{ V}$ direct-current power source energizes three parallel branch circuits in an automated conveyor control box. Branch 1 has a resistance of $60\ \Omega$ , while Branch 2 and Branch 3 each have a resistance of $30\ \Omega$ . What is the total current flowing from the power source through the entire circuit?

1\text{ A}

6\text{ A}

10\text{ A}

12\text{ A}

Show answer & explanation

Answer: $10\text{ A}$

Answer

10\text{ A}

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

120\text{ V}

). The current in each branch is determined by dividing voltage by resistance:

I_1 = 120\text{ V} / 60\ \Omega = 2\text{ A}

,

I_2 = 120\text{ V} / 30\ \Omega = 4\text{ A}

, and

I_3 = 120\text{ V} / 30\ \Omega = 4\text{ A}

. Summing these branch currents yields a total current of

10\text{ A}

(

2\text{ A} + 4\text{ A} + 4\text{ A} = 10\text{ A}

). Alternatively, calculating equivalent resistance gives

\frac{1}{R_{\text{eq}}} = \frac{1}{60} + \frac{1}{30} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}

, so

R_{\text{eq}} = 12\ \Omega

, yielding

I_{\text{total}} = \frac{120\text{ V}}{12\ \Omega} = 10\text{ A}

.

Step-by-Step Solution

1

Calculate the individual branch currents using Ohm's Law (

I = \frac{V}{R}

).

I_1 = \frac{120\text{ V}}{60\ \Omega} = 2\text{ A}

,

I_2 = \frac{120\text{ V}}{30\ \Omega} = 4\text{ A}

, and

I_3 = \frac{120\text{ V}}{30\ \Omega} = 4\text{ A}

.

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

120\text{ V}

.

2

Sum all individual branch currents to find the total circuit current (

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

).

I_{\text{total}} = 2\text{ A} + 4\text{ A} + 4\text{ A} = 10\text{ A}

.

By Kirchhoff's Current Law, total current supplied to a parallel network equals the sum of currents through all branches.

Key Concept

Parallel circuit branch current calculation and Kirchhoff's Current Law

Estimated Time:1m 0s

Question 139Question →

A technician is servicing a communications shelter ventilation fan controller that operates on a $48\text{ V}$ direct-current (DC) power supply. A diagnostic test shows that the control circuit's primary sensing resistor has a resistance of $1.2\text{ k}\Omega$ . What is the current flowing through this resistor?

25\text{ mA}

40\text{ mA}

400\text{ mA}

57.6\text{ mA}

Show answer & explanation

Answer: $40\text{ mA}$

Answer

40\text{ mA}

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

I = \frac{V}{R}

). First converting

1.2\text{ k}\Omega

to

1200\ \Omega

gives

I = \frac{48\text{ V}}{1200\ \Omega} = 0.04\text{ A}

, which is equivalent to

40\text{ mA}

.

Step-by-Step Solution

1

Convert the resistance value from kilohms (

\text{k}\Omega

) to base ohms (

\Omega

).

R = 1.2\text{ k}\Omega = 1.2 \times 1000\ \Omega = 1200\ \Omega

Standard electrical calculations using Ohm's Law require base units of volts, amperes, and ohms.

2

Apply Ohm's Law (

I = \frac{V}{R}

) to solve for the circuit current.

I = \frac{48\text{ V}}{1200\ \Omega} = 0.04\text{ A}

Current equals voltage divided by resistance.

3

Convert the current in amperes to milliamperes (

\text{mA}

).

0.04\text{ A} \times 1000\text{ mA/A} = 40\text{ mA}

The answer choices are expressed in milliamperes.

Key Concept

Ohm's Law (

I = \frac{V}{R}

) and metric prefix conversions

Estimated Time:45s

Question 140Question →

A field technician is testing a direct-current (DC) sensor module. What is the current flowing through the circuit when a potential difference of $90\text{ V}$ is applied across a total circuit resistance of $15\text{ k}\Omega$ ?

Fill in the blanks below

Under an applied potential difference of

90\text{ V}

, a circuit having a resistance of

15\text{ k}\Omega

conducts a current of milliamperes.

Show answer & explanation

Answer

6 milliamperes (6 mA)

According to Ohm's law (

I = \frac{V}{R}

), dividing the

90\text{ V}

source by the

15,000\ \Omega

(

15\text{ k}\Omega

) total resistance yields

0.006\text{ A}

, which converts directly to

6\text{ mA}

.

Step-by-Step Solution

1

Identify known circuit quantities and convert resistance from kilohms to ohms.

V = 90\text{ V}

and

R = 15\text{ k}\Omega = 15 \times 1,000\ \Omega = 15,000\ \Omega

.

Standard electrical formulas require resistance in base units of ohms (

\Omega

) to calculate current in amperes (

\text{A}

).

2

Apply Ohm's law (

I = \frac{V}{R}

) to solve for the current in amperes.

I = \frac{90\text{ V}}{15,000\ \Omega} = 0.006\text{ A}

.

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

3

Convert current from amperes to milliamperes as requested by the statement.

0.006\text{ A} \times 1,000\text{ mA/A} = 6\text{ mA}

.

One milliampere equals one thousandth of an ampere (

1\text{ mA} = 10^{-3}\text{ A}

).

Key Concept

Ohm's Law (

I = \frac{V}{R}

) defines the proportional relationship between potential difference, current, and resistance in a direct-current circuit.

Estimated Time:45s

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