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

Question 21Question →

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 22Question →

A DC power supply is connected across a series circuit containing two resistive components. Fill in the blanks with the correct numerical values to complete the circuit analysis.

Fill in the blanks below

A

50\text{ V}

DC source is connected in series with a

15\ \Omega

resistor and a second resistor

R_2

. If the circuit draws a current of

2\text{ A}

, the total circuit resistance is

\Omega

, and the resistance of

R_2

is

\Omega

.

Show answer & explanation

Answer

The total circuit resistance is

25\ \Omega

, and the resistance of

R_2

is

10\ \Omega

.

Applying Ohm's Law to the total circuit yields a total resistance of

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

. In a series circuit, total resistance equals the sum of the individual component resistances (

R_T = R_1 + R_2

). Subtracting the known resistor value (

15\ \Omega

) from the total resistance (

25\ \Omega

) results in

R_2 = 10\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law with source voltage and total current.

R_T = \frac{V_T}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

In any complete circuit, the total equivalent resistance equals total source voltage divided by the total circuit current.

2

Calculate the resistance of

R_2

using the series circuit resistance addition formula (

R_T = R_1 + R_2

).

R_2 = R_T - R_1 = 25\ \Omega - 15\ \Omega = 10\ \Omega

In a series circuit, total resistance is the direct sum of individual component resistances.

Key Concept

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

R_T = R_1 + R_2 + \dots

), and the circuit current is constant through all components as governed by Ohm's Law (

R_T = V_T / I

).

Estimated Time:1m 0s

Question 23Question →

A direct current (DC) circuit consists of a $48\text{ V}$ power supply connected in series with three resistors: $R_1 = 12\ \Omega$ , $R_2 = 24\ \Omega$ , and $R_3 = 24\ \Omega$ . In its baseline operating state, the total circuit resistance is $R_T = 60\ \Omega$ and the circuit loop current is $I = 0.8\text{ A}$ .

Match each circuit alteration on the left to its corresponding electrical outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair: (1) shorting resistor

R_3

with total resistance dropping to

36\ \Omega

, current rising to

1.33\text{ A}

, and

0\text{ V}

across

R_3

; (2) an open at

R_2

with loop current dropping to

0\text{ A}

and

48\text{ V}

appearing across

R_2

; (3) adding a

20\ \Omega

series resistor with total resistance rising to

80\ \Omega

, current dropping to

0.6\text{ A}

, and

7.2\text{ V}

across

R_1

; and (4) raising the source to

72\text{ V}

with total resistance remaining

60\ \Omega

, current rising to

1.2\text{ A}

, and

28.8\text{ V}

across

R_2

.

Each modification directly follows the governing principles of series DC circuits. In series circuits, total resistance equals the direct sum of all individual resistances (

R_T = R_1 + R_2 + \dots + R_n

), and current is constant throughout the single loop (

I_T = \frac{V_T}{R_T}

). Short circuits eliminate individual component resistance and drop component voltage to zero while raising loop current. Open circuits eliminate current entirely and place full source voltage across the open terminals. Adding series resistance reduces current throughout the loop, and increasing source voltage increases current and all individual voltage drops proportionally.

Step-by-Step Solution

1

Analyze the short circuit condition across

R_3

R_T = 12 + 24 = 36\ \Omega

;

I = \frac{48}{36} \approx 1.33\text{ A}

;

V_3 = 0\text{ V}

A short circuit bypasses the component with zero resistance, eliminating its resistance from the total series sum and producing zero potential difference across itself.

2

Analyze the open circuit condition at

R_2

I = 0\text{ A}

;

V_1 = 0\text{ V}

;

V_3 = 0\text{ V}

;

V_2 = 48\text{ V}

An open in a series circuit interrupts the sole conductive loop, causing current to cease everywhere; infinite resistance across the open component causes the full source potential to appear across the gap.

3

Analyze the addition of a fourth series resistor (

R_4 = 20\ \Omega

)

R_T = 60 + 20 = 80\ \Omega

;

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

;

V_1 = 0.6 \times 12 = 7.2\text{ V}

In series circuits, equivalent resistance is additive (

R_T = \sum R

). Increasing total resistance reduces total current according to Ohm's Law (

I = \frac{V}{R_T}

), which proportionally reduces individual component voltage drops.

4

Analyze the voltage source increase to

72\text{ V}

R_T = 60\ \Omega

;

I = \frac{72}{60} = 1.2\text{ A}

;

V_2 = 1.2 \times 24 = 28.8\text{ V}

Resistor values are independent of source voltage. Applying Ohm's Law with higher voltage across constant resistance yields directly proportional increases in loop current and individual voltage drops.

Key Concept

Series Circuit Governing Rules: Additive Resistance, Current Continuity, Ohm's Law, and Fault Dynamics

Estimated Time:2m 0s

Question 24Question →

A naval vessel alarm panel is powered by a $72\text{ V}$ DC supply connected in series with three resistive loads: an alert strobe ( $R_1 = 8\ \Omega$ ), an auxiliary buzzer ( $R_2 = 10\ \Omega$ ), and a line-balancing resistor ( $R_3 = 6\ \Omega$ ). What is the voltage drop across the auxiliary buzzer ( $R_2$ )?

30\text{ V}

24\text{ V}

18\text{ V}

3\text{ V}

Show answer & explanation

Answer: $30\text{ V}$

Answer

The voltage drop across the auxiliary buzzer (

R_2

) is

30\text{ V}

.

The correct answer is

30\text{ V}

. First, sum the resistances in series to obtain

R_T = 8\ \Omega + 10\ \Omega + 6\ \Omega = 24\ \Omega

. Next, determine the loop current using Ohm's Law:

I = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

. Finally, multiply the current by the resistance of the auxiliary buzzer (

10\ \Omega

) to get

V_2 = 3\text{ A} \times 10\ \Omega = 30\text{ V}

.

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_T

) of the series circuit.

R_T = R_1 + R_2 + R_3 = 8\ \Omega + 10\ \Omega + 6\ \Omega = 24\ \Omega

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

2

Determine the common loop current (

I

) flowing through the circuit.

I = \frac{V_T}{R_T} = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

According to Ohm's Law, current equals total voltage divided by total resistance, and current remains constant throughout a series path.

3

Calculate the voltage drop across the target component (

R_2

).

V_2 = I \times R_2 = 3\text{ A} \times 10\ \Omega = 30\text{ V}

The voltage drop across a specific resistor in series is the product of the circuit current and that resistor's value.

Key Concept

In a series DC circuit, total resistance is the sum of all individual resistances, current is identical through every component, and the voltage drop across any individual resistor is proportional to its resistance (

V = I \times R

).

Estimated Time:1m 0s

Question 25Question →

An industrial security monitoring loop is powered by a $36\text{ V}$ DC power supply connected in series with a current-limiting resistor $R_1 = 150\ \Omega$ , an optical sensor module $R_2$ , and a line termination resistor $R_3 = 250\ \Omega$ . When the circuit is active, a voltmeter measures a $12\text{ V}$ drop across the optical sensor module $R_2$ . What is the electrical resistance of the optical sensor module $R_2$ ?

133\ \Omega

200\ \Omega

400\ \Omega

600\ \Omega

Show answer & explanation

Answer: $200\ \Omega$

Answer

The electrical resistance of the optical sensor module

R_2

is

200\ \Omega

.

The correct answer is determined by first finding the remaining voltage across the known resistors (

36\text{ V} - 12\text{ V} = 24\text{ V}

). Dividing this by their combined resistance (

150\ \Omega + 250\ \Omega = 400\ \Omega

) yields a constant loop current of

0.06\text{ A}

. Dividing the sensor's

12\text{ V}

drop by

0.06\text{ A}

gives

200\ \Omega

.

Step-by-Step Solution

1

Calculate the combined voltage drop across the known series resistors

R_1

and

R_3

.

V_{1,3} = V_{\text{total}} - V_2 = 36\text{ V} - 12\text{ V} = 24\text{ V}

In a series circuit, Kirchhoff's Voltage Law dictates that the sum of all individual voltage drops equals the total supply voltage.

2

Find the sum of the known series resistances.

R_{1,3} = R_1 + R_3 = 150\ \Omega + 250\ \Omega = 400\ \Omega

Resistances in a series path are directly additive.

3

Determine the common loop current flowing through the circuit.

I = \frac{V_{1,3}}{R_{1,3}} = \frac{24\text{ V}}{400\ \Omega} = 0.06\text{ A}

Current is constant at all points in a series circuit according to Ohm's Law.

4

Calculate the resistance of

R_2

using its measured voltage drop and the loop current.

R_2 = \frac{V_2}{I} = \frac{12\text{ V}}{0.06\text{ A}} = 200\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) to the individual component provides its exact resistance.

Key Concept

Series circuit current constancy and voltage division using Ohm's Law and Kirchhoff's Voltage Law

Alternative Method

Using the voltage divider formula:

\frac{V_2}{V_{\text{total}}} = \frac{R_2}{R_1 + R_2 + R_3} \implies \frac{12}{36} = \frac{R_2}{400 + R_2} \implies \frac{1}{3} = \frac{R_2}{400 + R_2} \implies 400 + R_2 = 3R_2 \implies 2R_2 = 400 \implies R_2 = 200\ \Omega

.

Estimated Time:1m 30s

Question 26Question →

During the calibration of a naval radar control system, engineers measure a continuous current of $1.5\text{ A}$ flowing through a series safety loop. The loop is energized by a $60\text{ V}$ direct current source and contains three elements: a $15\ \Omega$ current-limiting resistor ( $R_1$ ), a calibration rheostat ( $R_2$ ), and a primary sensor coil ( $R_3$ ). If the primary sensor coil dissipates exactly $36\text{ W}$ of power, what is the required resistance of the calibration rheostat ( $R_2$ )?

1\ \Omega

40\ \Omega

9\ \Omega

59\ \Omega

Show answer & explanation

Answer: $9\ \Omega$

Answer

The required resistance of the calibration rheostat is

9\ \Omega

.

In this series circuit, the total resistance must equal the source voltage divided by the total current (

60\text{ V} / 1.5\text{ A} = 40\ \Omega

). Since components are in series, they share the same

1.5\text{ A}

current. The sensor coil dissipates

36\text{ W}

, meaning its resistance is

P / I^2

(

36 / 2.25 = 16\ \Omega

). Subtracting the

15\ \Omega

resistor and the

16\ \Omega

coil from the

40\ \Omega

total leaves

9\ \Omega

for the calibration rheostat.

Step-by-Step Solution

1

Calculate the total equivalent resistance of the series circuit using Ohm's Law.

R_T = \frac{V_T}{I_T} = \frac{60\text{ V}}{1.5\text{ A}} = 40\ \Omega

Total resistance is needed to find the missing component value in a series string.

2

Determine the resistance of the primary sensor coil (

R_3

) using the power formula

P = I^2R

.

R_3 = \frac{P}{I^2} = \frac{36\text{ W}}{(1.5\text{ A})^2} = \frac{36}{2.25} = 16\ \Omega

Since the circuit is in series, the full

1.5\text{ A}

current flows through all components, allowing the coil's resistance to be calculated from its power dissipation.

3

Subtract the known series resistances from the total resistance to find the rheostat's value (

R_2

).

R_2 = R_T - R_1 - R_3 = 40\ \Omega - 15\ \Omega - 16\ \Omega = 9\ \Omega

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

R_T = R_1 + R_2 + R_3

).

Key Concept

Series Circuit Resistance and Power Dissipation

Question 27Question →

An electronics technician is analyzing the operational principles of a DC series circuit. Match each series circuit characteristic on the left with its correct electrical rule or behavioral description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Total circuit resistance matches with the sum of all individual load values; Current flow matches with remaining constant and identical at every point; Individual component voltage drop matches with being directly proportional to resistance (

I \times R_n

); Open-circuit fault matches with reducing total loop current to zero and disabling all components.

In a series circuit, all components are connected sequentially in a single loop. Therefore, total resistance is the direct sum of all resistor values, the current is uniform throughout, the individual voltage drops are proportional to component resistance via Ohm's Law, and an open circuit completely eliminates current flow across all components.

Step-by-Step Solution

1

Identify the relationship for total resistance in a series string

R_T = R_1 + R_2 + \dots + R_n

Series components add their oppositions consecutively to the flow of current.

2

Determine the path and behavior of electrical current in a series loop

I_T = I_1 = I_2 = \dots = I_n

A single conducting path allows no branching, ensuring the rate of charge flow is uniform everywhere.

3

Apply Ohm's Law to find the voltage drop across an individual resistor

V_n = I \times R_n

With current constant, the voltage drop across each component depends directly on its resistance value.

4

Analyze the impact of a discontinuity (open circuit)

I = 0\text{ A}

throughout the entire loop

Breaking any point in a single closed loop stops all current from circulating.

Key Concept

Core governing laws of DC series circuits: additive resistance, uniform loop current, proportional voltage division, and single-path continuity dependency.

Estimated Time:1m 0s

Question 28Question →

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

Show answer & explanation

Answer: 60

Answer

The resistance of the humidity sensor

R_3

is

60\ \Omega

.

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

R_2

(

21.6\text{ V}

) and its known resistance (

27\ \Omega

), the loop current is

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

. The total equivalent resistance connected across the

96\text{ V}

source is

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

. Because series resistances are directly additive (

R_T = R_1 + R_2 + R_3 + R_4

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

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

.

Step-by-Step Solution

1

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

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

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

2

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

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

Total resistance equals source voltage divided by total circuit current.

3

Calculate the sum of all known resistances.

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

Resistances in a series circuit are directly additive.

4

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

R_3

.

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

R_T = R_1 + R_2 + R_3 + R_4

, so

R_3 = R_T - (R_1 + R_2 + R_4)

.

Key Concept

Series Circuit Loop Current and Total Resistance Analysis

Alternative Method

Calculate individual voltage drops using loop current

I = 0.8\text{ A}

:

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

and

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

. By Kirchhoff's Voltage Law, the voltage across

R_3

is

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

. Applying Ohm's Law directly to

R_3

yields

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

.

Estimated Time:1m 30s

Question 29Question →

A technician is testing a direct-current (DC) control circuit connected to a $20\text{ V}$ power supply. An ammeter in the loop indicates a steady current of $0.5\text{ A}$ . The circuit contains three resistors wired in series: two have fixed values of $12\ \Omega$ and $16\ \Omega$ , while the third is an unlabelled component. What is the resistance of the unlabelled resistor?

28\ \Omega

12\ \Omega

10\ \Omega

40\ \Omega

Show answer & explanation

Answer: $12\ \Omega$

Answer

12\ \Omega

The correct answer of

12\ \Omega

is found by first calculating the total circuit resistance using Ohm's Law (

R_T = \frac{20\text{ V}}{0.5\text{ A}} = 40\ \Omega

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

R_T = R_1 + R_2 + R_3

). Subtracting the two known values (

12\ \Omega + 16\ \Omega = 28\ \Omega

) from

40\ \Omega

leaves exactly

12\ \Omega

for the unlabelled resistor.

Step-by-Step Solution

1

Calculate the total resistance (

R_T

) of the series circuit using Ohm's Law.

R_T = \frac{V_{\text{total}}}{I} = \frac{20\text{ V}}{0.5\text{ A}} = 40\ \Omega

The total resistance of a DC circuit is equal to the total applied voltage divided by the total circuit current.

2

Sum the resistances of the two known series resistors.

R_{\text{known}} = 12\ \Omega + 16\ \Omega = 28\ \Omega

In a series circuit, individual resistances add directly to form the total equivalent resistance.

3

Subtract the sum of the known resistors from the total circuit resistance to find the unknown resistance (

R_3

).

R_3 = R_T - R_{\text{known}} = 40\ \Omega - 28\ \Omega = 12\ \Omega

Because

R_T = R_1 + R_2 + R_3

, the unlabelled resistor's value is the remaining resistance needed to reach

40\ \Omega

.

Key Concept

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

R_T = R_1 + R_2 + \dots + R_n

), and the total current is determined by Ohm's Law (

I = \frac{V_T}{R_T}

).

Estimated Time:1m 0s

Question 30Question →

Analyze the series circuit parameters under normal operation and after a short-circuit fault by filling in the correct numerical values (use decimal format without units, e.g., '0.5' or '10').

Fill in the blanks below

An aviation warning beacon is powered by a

120\text{ V}

DC supply connected in series with three resistive components:

R_1 = 100\ \Omega

,

R_2 = 150\ \Omega

, and

R_3 = 50\ \Omega

. Under normal operation, the total current flowing through the circuit is A and the voltage drop across

R_1

is V. If component

R_2

experiences a catastrophic failure and completely shorts out (its internal resistance drops to

0\ \Omega

), the equivalent resistance of the entire circuit drops to

\Omega

, causing the voltage drop across

R_3

to become V.

Show answer & explanation

Answer

Under normal operation, the current is 0.4 A, and the voltage drop across R1 is 40 V. After R2 shorts out, the total resistance drops to 150 ohms, and the increased current causes the voltage drop across R3 to become 40 V.

In a series circuit, total resistance is the linear sum of all resistive components. Current remains uniform throughout the entire loop and is determined by the total voltage divided by the total resistance. When a component fails via a short circuit, it effectively acts as a direct wire (zero ohms). This absence of resistance lowers the total circuit resistance, which consequently forces a higher current to flow from the constant voltage source. The increased current then results in proportionately higher voltage drops across the surviving resistors.

Step-by-Step Solution

1

Calculate the total equivalent resistance under normal operating conditions.

R_{total} = R_1 + R_2 + R_3 = 100 + 150 + 50 = 300\ \Omega

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

2

Use Ohm's Law to find the total current flowing through the normal circuit.

I_{total} = \frac{V_{total}}{R_{total}} = \frac{120\text{ V}}{300\ \Omega} = 0.4\text{ A}

(This is blank 1).

Total current is equal to the source voltage divided by the total equivalent resistance.

3

Calculate the voltage drop across

R_1

under normal conditions.

V_{R1} = I_{total} \times R_1 = 0.4\text{ A} \times 100\ \Omega = 40\text{ V}

(This is blank 2).

The voltage drop across any single component in a series circuit is the total current multiplied by that component's resistance.

4

Determine the new total resistance after

R_2

shorts out.

R_{new\_total} = 100 + 0 + 50 = 150\ \Omega

(This is blank 3).

A shorted component has

0\ \Omega

resistance. The circuit remains closed, but the total resistance decreases.

5

Calculate the new total current and the resulting new voltage drop across

R_3

.

I_{new} = \frac{120\text{ V}}{150\ \Omega} = 0.8\text{ A}

. Then,

V_{R3} = 0.8\text{ A} \times 50\ \Omega = 40\text{ V}

(This is blank 4).

The lowered total resistance causes a higher total current to flow from the same voltage source, which proportionally increases the voltage drop across the remaining functional resistors.

Key Concept

Series Circuit Analysis and Component Fault (Short Circuit) Behavior

Question 31Question →

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

Show answer & explanation

Answer: 2.1

Answer

The resistance of a single glow plug is

2.1\ \Omega

.

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

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

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

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

). Subtracting the ballast resistor (

1.8\ \Omega

) leaves

4.2\ \Omega

across the two identical glow plugs. Dividing

4.2\ \Omega

equally between them gives

2.1\ \Omega

per glow plug.

Step-by-Step Solution

1

Calculate the total circuit resistance using Ohm's Law.

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

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

2

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

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

Resistances in a series circuit are directly additive (

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

).

3

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

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

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

Key Concept

Series Total Resistance and Ohm's Law

Estimated Time:1m 0s

Question 32Question →

An emergency backup lighting circuit is powered by a $48\text{ V}$ DC supply connected in series with a $30\ \Omega$ current-limiting resistor and two identical LED lamp modules. If a steady current of $0.4\text{ A}$ flows through the circuit during operation, what is the voltage drop across each individual LED lamp module?

12\text{ V}

36\text{ V}

18\text{ V}

24\text{ V}

Show answer & explanation

Answer: $18\text{ V}$

Answer

18\text{ V}

In a series circuit, the total voltage of

48\text{ V}

is distributed across all components according to Ohm's Law. With

0.4\text{ A}

flowing through the

30\ \Omega

resistor, that resistor drops

12\text{ V}

. Subtracting

12\text{ V}

from the total

48\text{ V}

source leaves

36\text{ V}

dropped across the two identical LED modules. Because the modules are identical and in series, they share this remaining voltage equally, resulting in

18\text{ V}

across each module.

Step-by-Step Solution

1

Calculate the voltage drop across the

30\ \Omega

series resistor using Ohm's Law (

V = I \times R

).

V_{R} = 0.4\text{ A} \times 30\ \Omega = 12\text{ V}

Current is identical through every component in a series circuit.

2

Determine the remaining voltage available for the two LED modules by applying Kirchhoff's Voltage Law (

V_{\text{total}} - V_R

).

V_{\text{remaining}} = 48\text{ V} - 12\text{ V} = 36\text{ V}

The sum of all voltage drops in a series loop must equal the total supply voltage.

3

Divide the remaining voltage equally between the two identical LED modules.

V_{\text{module}} = \frac{36\text{ V}}{2} = 18\text{ V}

Identical series components carry the same current and therefore drop equal portions of the remaining voltage.

Key Concept

In a series circuit, current is constant throughout the single loop, and the total supply voltage equals the sum of the individual component voltage drops.

Estimated Time:1m 0s

Question 33Question →

An industrial direct-current (DC) sensor loop is powered by a $50\text{ V}$ power source. The loop consists of two series-connected components: a fixed resistor $R_1 = 10\ \Omega$ and an unknown resistor $R_2$ . An ammeter placed in the circuit measures a continuous current of $2\text{ A}$ . What is the resistance of $R_2$ , and what is the voltage drop across $R_2$ ?

Fill in the blanks below

The resistance of

R_2

is

\Omega

, and the voltage drop across

R_2

is

\text{V}

.

Show answer & explanation

Answer

The resistance of

R_2

is

15\ \Omega

, and the voltage drop across

R_2

is

30\text{ V}

.

In a series circuit, the total resistance is calculated as

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

. Since

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

, the unknown resistance is

R_2 = 25\ \Omega - 10\ \Omega = 15\ \Omega

. The voltage drop across

R_2

is found by multiplying the common series current by

R_2

, yielding

V_2 = 2\text{ A} \times 15\ \Omega = 30\text{ V}

(which also satisfies Kirchhoff's Voltage Law:

50\text{ V} - 20\text{ V} = 30\text{ V}

).

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_{\text{total}}

) of the series circuit using Ohm's Law.

R_{\text{total}} = \frac{V_{\text{total}}}{I} = \frac{50\text{ V}}{2\text{ A}} = 25\ \Omega

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

2

Determine the value of the unknown resistor

R_2

using the series resistance rule.

R_2 = R_{\text{total}} - R_1 = 25\ \Omega - 10\ \Omega = 15\ \Omega

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

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

).

3

Calculate the individual voltage drop across

R_2

(

V_2

).

V_2 = I \times R_2 = 2\text{ A} \times 15\ \Omega = 30\text{ V}

The same current flows through all series components; voltage drop across any component equals current multiplied by its resistance (or

V_{\text{total}} - V_1 = 50\text{ V} - 20\text{ V} = 30\text{ V}

).

Key Concept

Series Circuit Resistance and Voltage Division

Estimated Time:1m 0s

Question 34Question →

In direct-current (DC) series circuit analysis, distinct electrical principles govern how current, voltage, resistance, and faults behave across a single closed pathway. Match each series circuit parameter or condition on the left with its correct governing rule on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Total equivalent resistance matches the direct sum of all individual resistive values; Circuit current matches remaining constant at every point; Total source voltage matches the sum of the individual voltage drops across all loads; An open-circuit fault matches immediately halting all current flow.

Series circuits are defined by a single pathway for current. Consequently, current (

I

) is identical at all points, total resistance (

R_T

) is the arithmetic sum of the component resistances, total voltage (

V_T

) equals the sum of the individual component voltage drops, and an interruption anywhere breaks the complete circuit, dropping total current to zero.

Step-by-Step Solution

1

Identify the resistance relationship for series circuits.

Total resistance increases with each added resistor and is found by adding resistances directly (

R_T = R_1 + R_2 + \dots

).

Electrons must pass sequentially through every resistive element along the single pathway.

2

Determine the current distribution rule in a single-loop circuit.

Current magnitude is uniform everywhere (

I_T = I_1 = I_2

).

Charge cannot accumulate or branch off in an unbranched series loop.

3

Apply Kirchhoff's Voltage Law to the series loop.

Total applied voltage equals the sum of the series voltage drops (

V_T = V_1 + V_2 + \dots

).

The electrical potential energy supplied by the source must be completely dissipated across the series loads.

4

Evaluate the effect of an open-circuit fault.

An open condition creates infinite resistance, stopping all loop current (

I = 0\text{ A}

).

A continuous physical path is required for DC electron flow.

Key Concept

Fundamental series circuit characteristics: additive resistance, constant current, additive voltage drops (Kirchhoff's Voltage Law), and single-point failure dependency.

Estimated Time:1m 0s

Question 35Question →

An aircraft avionics system is powered by a $28\text{ V}$ DC supply connected in series with three components: a $40\ \Omega$ indicator lamp, a $70\ \Omega$ current-limiting resistor, and a $30\ \Omega$ monitoring resistor. What is the voltage drop across the $70\ \Omega$ current-limiting resistor?

28\text{ V}

14\text{ V}

8\text{ V}

5.6\text{ V}

Show answer & explanation

Answer: $14\text{ V}$

Answer

14\text{ V}

The correct answer is

14\text{ V}

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

40\ \Omega + 70\ \Omega + 30\ \Omega = 140\ \Omega

. The circuit current is

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

. Multiplying this current by the

70\ \Omega

resistance gives a voltage drop of

0.2\text{ A} \times 70\ \Omega = 14\text{ V}

.

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_{\text{total}}

) of the series circuit.

R_{\text{total}} = 40\ \Omega + 70\ \Omega + 30\ \Omega = 140\ \Omega

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

2

Determine the common circuit current (

I

) using Ohm's Law.

I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{28\text{ V}}{140\ \Omega} = 0.2\text{ A}

Current remains constant through every component along a single series path.

3

Calculate the voltage drop across the

70\ \Omega

resistor (

V_{70\Omega}

).

V_{70\Omega} = I \times R = 0.2\text{ A} \times 70\ \Omega = 14\text{ V}

Applying Ohm's Law (

V = I \times R

) to the specific component yields its proportional voltage drop.

Key Concept

Series Circuit Resistance and Voltage Division

Alternative Method

Use the voltage divider formula directly:

V_{70\Omega} = 28\text{ V} \times \left(\frac{70\ \Omega}{40\ \Omega + 70\ \Omega + 30\ \Omega}\right) = 28\text{ V} \times \frac{70}{140} = 28\text{ V} \times 0.5 = 14\text{ V}

.

Estimated Time:1m 0s

Question 36Question →

A marine communication console operates on a $36\text{ V}$ DC power bus. The circuit consists of three components connected in series: a status indicator resistor $R_1 = 18\ \Omega$ , an internal filter resistor $R_2 = 24\ \Omega$ , and an unknown current-limiting resistor $R_3$ . An inline ammeter measures the total circuit current as $0.5\text{ A}$ . What is the resistance of resistor $R_3$ ?

18\ \Omega

30\ \Omega

24\ \Omega

72\ \Omega

Show answer & explanation

Answer: $30\ \Omega$

Answer

The resistance of resistor

R_3

is

30\ \Omega

.

Using Ohm's Law, the total series resistance is

R_T = \frac{V_T}{I} = \frac{36\text{ V}}{0.5\text{ A}} = 72\ \Omega

. Since resistances in a series circuit are additive (

R_T = R_1 + R_2 + R_3

), the unknown resistance is found by subtracting the sum of the known resistors (

18\ \Omega + 24\ \Omega = 42\ \Omega

) from the total:

R_3 = 72\ \Omega - 42\ \Omega = 30\ \Omega

.

Step-by-Step Solution

1

Calculate the total series circuit resistance (

R_T

) using Ohm's Law.

R_T = \frac{V_T}{I_T} = \frac{36\text{ V}}{0.5\text{ A}} = 72\ \Omega

In a series circuit, total resistance equals total applied voltage divided by the common loop current.

2

Sum the resistances of the known series components.

R_{1+2} = R_1 + R_2 = 18\ \Omega + 24\ \Omega = 42\ \Omega

Resistances add directly in a series configuration.

3

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

R_3

.

R_3 = R_T - (R_1 + R_2) = 72\ \Omega - 42\ \Omega = 30\ \Omega

The total resistance is the sum of all individual resistors:

R_T = R_1 + R_2 + R_3

.

Key Concept

In a series circuit, current remains constant throughout the entire loop, and total resistance equals the sum of all individual resistances (

R_T = R_1 + R_2 + R_3

). Ohm's Law (

R = \frac{V}{I}

) relates total voltage to total resistance and current.

Estimated Time:1m 0s

Question 37Question →

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

Show answer & explanation

Answer: 60

Answer

The resistance of the protective resistor

R_x

is

60\ \Omega

.

Applying Ohm's Law (

R_T = \frac{V_T}{I}

) gives a total series resistance of

180\ \Omega

. Since resistances in series add directly (

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

), solving for the unknown yields

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

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law.

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

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

2

Determine the combined resistance of the known series components.

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

Resistances in a series circuit are directly additive.

3

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

R_x

.

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

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

R_T = R_1 + R_2 + R_x

).

Key Concept

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

I_T = I_1 = I_2 = I_3

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

R_T = R_1 + R_2 + \dots + R_n

).

Alternative Method

Find the voltage drops across the known components first:

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

and

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

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

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

. Then find

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

.

Estimated Time:1m 0s

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