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

Question 1Question →

Complete the statement regarding the electrical parameters of the DC circuit by calculating the missing resistance and current values.

Fill in the blanks below

An indicator lamp with a constant resistance draws

150\text{ mA}

when connected to a

24\text{ V}

direct current source. The internal resistance of the lamp is

\Omega

, and if the supply voltage is increased to

36\text{ V}

, the current drawn by the lamp will increase to

\text{mA}

.

Show answer & explanation

Answer

The internal resistance of the lamp is 160 Ω, and the new current at 36 V is 225 mA.

Converting

150\text{ mA}

to standard amperes yields

0.15\text{ A}

. Using Ohm's law (

R = V / I

), the internal resistance is

24\text{ V} / 0.15\text{ A} = 160\ \Omega

. When the voltage rises to

36\text{ V}

, applying

I = V / R

produces

36\text{ V} / 160\ \Omega = 0.225\text{ A}

, which corresponds to

225\text{ mA}

.

Step-by-Step Solution

1

Convert the nominal current from milliamperes to base amperes

I_1 = 150\text{ mA} = 0.15\text{ A}

Ohm's law requires base SI units (amperes, volts, ohms) for direct calculation.

2

Calculate the constant internal resistance using Ohm's law

R = \frac{V_1}{I_1} = \frac{24\text{ V}}{0.15\text{ A}} = 160\ \Omega

Resistance equals voltage divided by current.

3

Calculate the new current at the elevated voltage level

I_2 = \frac{V_2}{R} = \frac{36\text{ V}}{160\ \Omega} = 0.225\text{ A} = 225\text{ mA}

Current across a fixed resistance is directly proportional to applied voltage (

I = V / R

).

Key Concept

Ohm's Law (

V = I \times R

) and Direct Proportionality in DC Circuits

Estimated Time:45s

Question 2Question →

Match each electromagnetic component to its primary operating principle or function.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Solenoid pairs with converting electrical energy into linear mechanical motion; Transformer pairs with transferring electrical energy via mutual electromagnetic induction; Relay pairs with using an electromagnet to actuate switch contacts; Inductor pairs with storing energy in a magnetic field to oppose current changes.

Each electromagnetic component performs a unique function based on electromagnetic principles: solenoids produce direct mechanical stroke; transformers transfer power across isolated windings via alternating magnetic flux; relays actuate circuit contacts remotely via electromagnetic force; and inductors provide inductive reactance and self-induced counter-EMF.

Step-by-Step Solution

1

Identify the primary mechanism of a solenoid.

A solenoid converts magnetic force into linear mechanical movement.

When energized, the coil's magnetic field draws an iron core inward to actuate a mechanical load.

2

Identify the operating principle of a transformer.

A transformer utilizes mutual induction across coupled coils.

Changing current in the primary coil induces an alternating voltage in the secondary coil without direct electrical connection.

3

Identify the primary function of an electromechanical relay.

A relay operates switch contacts using electromagnetic attraction.

An electromagnet pulls an armature to switch a separate circuit on or off.

4

Identify the foundational circuit property of an inductor.

An inductor stores magnetic energy and opposes changes in current flow.

By Lenz's and Faraday's laws of induction, self-inductance generates counter-electromotive force when current changes.

Key Concept

Electromagnetic Components and Operating Principles

Estimated Time:45s

Question 3Question →

A direct current (DC) automotive lamp connected to a $12\text{ V}$ power source draws a current of $2\text{ A}$ . What is the total electrical power consumed by the lamp?

6\text{ W}

14\text{ W}

24\text{ W}

48\text{ W}

Show answer & explanation

Answer: $24\text{ W}$

Answer

24\text{ W}

According to Watt's Law, the electrical power (

P

) consumed by a circuit load is equal to the voltage (

V

) across the component multiplied by the current (

I

) flowing through it:

P = V \times I

. Substituting the given values yields

P = 12\text{ V} \times 2\text{ A} = 24\text{ W}

.

Step-by-Step Solution

1

Identify the given circuit parameters

Voltage (

V

) =

12\text{ V}

, Current (

I

) =

2\text{ A}

These are the two electrical quantities necessary to calculate power directly.

2

Select the appropriate electrical power formula (Watt's Law)

P = V \times I

When voltage and current are known, power is equal to their product.

3

Substitute the given values into the formula and calculate

P = 12\text{ V} \times 2\text{ A} = 24\text{ W}

Multiplying electromotive force by current gives total power in watts.

Key Concept

Electrical power in a DC circuit is calculated using Watt's Law (

P = V \times I

), where power in watts is the product of voltage in volts and current in amperes.

Estimated Time:45s

Question 4Question →

An auxiliary electric bilge pump operates on a $12\text{ V}$ DC power supply. If the internal electrical resistance of the pump motor is $6\ \Omega$ , what is the steady-state current drawn by the pump during operation?

0.5\text{ A}

2\text{ A}

6\text{ A}

72\text{ A}

Show answer & explanation

Answer: $2\text{ A}$

Answer

The current drawn by the pump is

2\text{ A}

.

According to Ohm's Law, current (

I

) equals voltage (

V

) divided by resistance (

R

). Dividing the

12\text{ V}

supply by the

6\ \Omega

load resistance yields exactly

2\text{ A}

.

Step-by-Step Solution

1

Identify the given circuit parameters and the formula to solve for unknown current.

Given: Voltage

V = 12\text{ V}

, Resistance

R = 6\ \Omega

. Ohm's Law states

V = I \times R

, which rearranges to

I = \frac{V}{R}

.

Ohm's Law defines the fundamental mathematical relationship between voltage, current, and resistance in a direct current circuit.

2

Substitute the given values into the rearranged formula and compute the current.

I = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}

.

Carrying out the division provides the magnitude of electron flow (amperage) through the load.

Key Concept

Ohm's Law formula for calculating current:

I = \frac{V}{R}

Estimated Time:40s

Question 5Question →

An electrical test bench delivers a constant $80\text{ V}$ direct-current supply to four separate parallel circuit arrangements. Match each circuit configuration on the left with its corresponding equivalent resistance ( $R_{eq}$ ) and total supply current ( $I_T$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The configuration with

40\ \Omega

and

10\ \Omega

matches

R_{eq} = 8\ \Omega, I_T = 10\text{ A}

; four

64\ \Omega

elements match

R_{eq} = 16\ \Omega, I_T = 5\text{ A}

; three branches of

120\ \Omega, 60\ \Omega, 40\ \Omega

match

R_{eq} = 20\ \Omega, I_T = 4\text{ A}

; and two branches of

150\ \Omega, 75\ \Omega

match

R_{eq} = 50\ \Omega, I_T = 1.6\text{ A}

.

In any direct-current parallel network, the total equivalent resistance (

R_{eq}

) is calculated using the reciprocal formula

\frac{1}{R_{eq}} = \sum \frac{1}{R_n}

(or

R/N

for identical branches). Total current drawn from the supply is determined by Ohm's Law:

I_T = \frac{V_S}{R_{eq}}

. Evaluating each configuration against the constant

80\text{ V}

bus correctly establishes all four resistance and current pairs.

Step-by-Step Solution

1

Calculate the equivalent resistance and total current for the

40\ \Omega

and

10\ \Omega

parallel branch pair.

R_{eq} = \frac{40 \times 10}{40 + 10} = 8\ \Omega

;

I_T = \frac{80\text{ V}}{8\ \Omega} = 10\text{ A}

.

Two unequal parallel resistors follow the product-over-sum rule, and total current is determined by Ohm's Law (

I = V/R

).

2

Calculate the equivalent resistance and current for the four parallel

64\ \Omega

elements.

R_{eq} = \frac{64}{4} = 16\ \Omega

;

I_T = \frac{80\text{ V}}{16\ \Omega} = 5\text{ A}

.

Identical parallel resistors simplify to branch resistance divided by the number of branches (

R/N

).

3

Calculate the equivalent resistance and current for the three parallel branches (

120\ \Omega, 60\ \Omega, 40\ \Omega

).

\frac{1}{R_{eq}} = \frac{1}{120} + \frac{2}{120} + \frac{3}{120} = \frac{6}{120} = \frac{1}{20} \implies R_{eq} = 20\ \Omega

;

I_T = \frac{80\text{ V}}{20\ \Omega} = 4\text{ A}

.

Three or more unequal parallel branches require finding the reciprocal sum over a common denominator.

4

Calculate the equivalent resistance and current for the

150\ \Omega

and

75\ \Omega

parallel branches.

R_{eq} = \frac{150 \times 75}{150 + 75} = \frac{11250}{225} = 50\ \Omega

;

I_T = \frac{80\text{ V}}{50\ \Omega} = 1.6\text{ A}

.

Applying product-over-sum and dividing bus voltage by equivalent resistance yields the total circuit current.

Key Concept

Parallel circuit rules require that full bus voltage appears across each branch, equivalent resistance is always lower than the smallest branch resistance, and total current equals the sum of all individual branch currents.

Estimated Time:2m 0s

Question 6Question →

Three resistors with values of $10\ \Omega$ , $15\ \Omega$ , and $25\ \Omega$ are connected in series across a direct current (DC) voltage source. What is the total equivalent resistance of the circuit?

4.8\ \Omega

50\ \Omega

25\ \Omega

50\text{ k}\Omega

Show answer & explanation

Answer: $50\ \Omega$

Answer

The total equivalent resistance is

50\ \Omega

.

In a series circuit, electrical current flows through each component sequentially along a single path. Therefore, the total opposition to current flow is the sum of the individual resistor values:

R_T = 10\ \Omega + 15\ \Omega + 25\ \Omega = 50\ \Omega

.

Step-by-Step Solution

1

Identify the circuit configuration and the formula for total series resistance.

R_T = R_1 + R_2 + R_3

In a series circuit, there is only one continuous path for electric current, meaning the opposition to current flow is cumulative.

2

Substitute the given resistor values into the series formula.

R_T = 10\ \Omega + 15\ \Omega + 25\ \Omega

Each component contributes its full resistive value directly to the total opposition.

3

Calculate the arithmetic sum.

R_T = 50\ \Omega

Simple addition yields the equivalent circuit resistance.

Key Concept

In a series circuit, total equivalent resistance (

R_T

) equals the direct sum of all individual resistances (

R_T = R_1 + R_2 + \dots + R_n

).

Estimated Time:45s

Question 7Question →

Complete the statement regarding the fundamental electrical behavior of a series circuit.

Fill in the blanks below

In a series circuit, because all electrical components are connected end-to-end in a single continuous path, the electric is the same at every point throughout the circuit.

Show answer & explanation

Answer

current (or amperage)

In any series circuit, electrical components share a single continuous conductive loop. Under the principle of conservation of electric charge, the rate at which charge passes through any cross-section of the conductor (current) must be identical at all points (

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

).

Step-by-Step Solution

1

Analyze the structural path of a series circuit.

A series circuit consists of a single closed loop with no branch points or alternate paths.

By definition, components in series are connected sequentially in one continuous line.

2

Apply the law of conservation of charge to determine which parameter remains constant across all components.

The rate of electron flow must be equal across all parts of the circuit (

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

).

Because charge cannot accumulate or disappear at any junction in a single loop, current is uniform everywhere.

Key Concept

Electric current is identical at all points in a series circuit because there is only one pathway for charge to flow.

Estimated Time:30s

Question 8Question →

A $24\text{ V}$ direct current (DC) power supply is connected to a circuit with two resistors in series: a $4\ \Omega$ resistor and an $8\ \Omega$ resistor. What is the current flowing through the $8\ \Omega$ resistor?

3\text{ A}

2\text{ A}

6\text{ A}

9\text{ A}

Show answer & explanation

Answer: $2\text{ A}$

Answer

2\text{ A}

The total resistance of the series circuit is the direct sum of the resistors (

4\ \Omega + 8\ \Omega = 12\ \Omega

). According to Ohm's Law (

I = V / R

), dividing the

24\text{ V}

source by

12\ \Omega

yields a total current of

2\text{ A}

. Because a series circuit provides only a single continuous path for electron flow, the current is identical through every component in the loop, meaning

2\text{ A}

flows through the

8\ \Omega

resistor.

Step-by-Step Solution

1

Calculate the total equivalent resistance of the series circuit.

R_{\text{total}} = R_1 + R_2 = 4\ \Omega + 8\ \Omega = 12\ \Omega

In any series circuit, individual resistances add directly together to form the total circuit resistance.

2

Apply Ohm's Law to determine the total circuit current.

I_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}

Current equals total voltage divided by total resistance.

3

Determine the current flowing specifically through the

8\ \Omega

resistor.

I_{R2} = I_{\text{total}} = 2\text{ A}

In a single-loop series circuit, current is uniform at all points and through all components.

Key Concept

Series Circuit Current and Total Resistance

Estimated Time:45s

Question 9Question →

Match each series circuit electrical property or condition on the left with its correct fundamental rule on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Circuit Current matches 'Remains identical and continuous through every connected component'; Total Resistance matches 'Equals the direct sum of all individual component resistances'; Source Voltage matches 'Divides across components such that individual drops sum to the source total'; Open Component Fault matches 'Completely interrupts electron flow throughout the entire single loop'.

Each relationship reflects the core principles of series circuit analysis: a single pathway guarantees uniform current throughout, sequential resistive elements add directly to total opposition, the sum of individual voltage drops balances the source voltage, and any break disrupts the entire closed conductive loop.

Step-by-Step Solution

1

Analyze current flow in a single-loop circuit

Identify that charge has only one path to follow, making current uniform throughout all series components.

Conservation of charge dictates current cannot accumulate or branch without alternative parallel paths.

2

Determine total resistance formula for series elements

Establish that each resistor adds directly to the total opposition (

R_T = R_1 + R_2 + \dots

).

Electrons must push through each resistive element sequentially.

3

Analyze voltage division across series loads

Confirm that source voltage is distributed proportionally across loads, summing to

V_T

.

Energy provided by the power supply equals total energy dropped across the circuit loads.

4

Evaluate the effect of a break or open circuit

Conclude that an open element halts all current in the entire circuit.

Breaking the single loop prevents complete continuity back to the power source.

Key Concept

Fundamental Series Circuit Laws (Current Uniformity, Resistance Additivity, Voltage Division, and Loop Continuity)

Estimated Time:45s

Question 10Question →

A direct current ( $12\text{ V}$ DC) circuit contains two indicator lamps connected in series. If a voltmeter measures a $7\text{ V}$ drop across the first lamp, what is the voltage drop across the second lamp?

12\text{ V}

5\text{ V}

7\text{ V}

19\text{ V}

Show answer & explanation

Answer: $5\text{ V}$

Answer

The voltage drop across the second lamp is

5\text{ V}

.

In a series circuit, total voltage supplied by the source is distributed across all series components such that the sum of the individual voltage drops equals the source voltage (

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

). Subtracting the

7\text{ V}

drop across the first lamp from the

12\text{ V}

total supply gives

5\text{ V}

across the second lamp.

Step-by-Step Solution

1

Identify the fundamental rule for voltage distribution in a series circuit.

The total supply voltage equals the sum of the individual voltage drops across each series load:

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

.

According to Kirchhoff's Voltage Law, the algebraic sum of all voltages around any closed loop in a circuit must equal zero.

2

Substitute the known values into the equation and solve for the unknown voltage drop (

V_2

).

V_2 = V_{\text{total}} - V_1 = 12\text{ V} - 7\text{ V} = 5\text{ V}

.

Subtracting the known component voltage drop from the source voltage yields the remaining voltage dropped across the second component.

Key Concept

Series Voltage Distribution (Kirchhoff's Voltage Law)

Estimated Time:45s

Question 11Question →

In a series-connected aircraft instrument lighting circuit, three resistive elements have values of $15\ \Omega$ , $35\ \Omega$ , and $50\ \Omega$ . What is the total equivalent resistance of the circuit in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 100

Answer

100 ohms (

\Omega

)

In a series circuit, total resistance is calculated by adding the individual resistance values together (

R_T = R_1 + R_2 + R_3

). Summing

15\ \Omega

,

35\ \Omega

, and

50\ \Omega

gives exactly

100\ \Omega

.

Step-by-Step Solution

1

Identify the formula for total resistance in a series circuit

R_T = R_1 + R_2 + R_3

In a series circuit, there is only one continuous path for current flow, making total opposition to current equal to the sum of all individual resistances.

2

Substitute the individual component values and calculate the sum

R_T = 15\ \Omega + 35\ \Omega + 50\ \Omega = 100\ \Omega

Adding the three series resistances yields the overall equivalent resistance.

Key Concept

Total resistance in a series circuit is equal to the direct sum of all individual resistances.

Estimated Time:45s

Question 12Question →

In direct-current (DC) power distribution, parallel circuit configurations govern how voltage, current, and resistance interact under various operating states. Match each parallel circuit modification on the left with its corresponding electrical outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding an additional resistive branch matches with decreasing total equivalent resistance and increasing total source current; an open-circuit fault matches with decreasing total circuit current while remaining branches operate normally; a short-circuit fault matches with dropping total resistance near zero and triggering circuit protection; and increasing main supply voltage matches with increasing the voltage drop and current across all branches simultaneously.

In parallel networks, the total equivalent resistance decreases when branches are added and increases when branches are removed. Full line voltage is maintained across every branch independently, meaning faults in one branch (such as an open) do not disrupt the operation of other branches, whereas a direct short bridges the common voltage bus and triggers circuit protection.

Step-by-Step Solution

1

Analyze the impact of adding a parallel branch

Additional current paths reduce equivalent resistance and raise total current without altering existing branch conditions.

Parallel branches operate independently across a constant voltage source.

2

Analyze an open circuit in a parallel branch

Only the opened branch stops conducting current; other branches retain full voltage and normal current.

An open in one parallel leg does not break the continuity of the remaining parallel paths.

3

Analyze a short circuit across a parallel network

A zero-ohm bridge shorts out all parallel loads and draws extreme current.

Current follows the path of least resistance, bypassing the loads entirely.

4

Analyze a change in primary DC source voltage

Every branch experiences the higher voltage and draws proportionally more current via Ohm's law (

I = V/R

).

The full source voltage is applied directly across every individual branch in a parallel circuit.

Key Concept

Parallel Circuit Governing Rules and Fault Characteristics

Estimated Time:1m 15s

Question 13Question →

An electrical technician is inspecting a circuit containing three resistive heating elements wired in series with a direct current power source. If one of the heating elements burns out and creates an open circuit, what happens to the current in the circuit?

Current continues to flow at the same level through the other two heating elements.

Current flow stops completely throughout the entire circuit.

Current increases through the remaining heating elements due to reduced total resistance.

Current bypasses the broken heating element and creates a short circuit back to the source.

Show answer & explanation

Answer: Current flow stops completely throughout the entire circuit.

Answer

Current flow stops completely throughout the entire circuit.

In a series circuit, there is only one continuous path for electrical current to travel from the power source through all connected loads and back. When one component burns out, it creates a physical break in that single conductive path (an open circuit). Because current requires an uninterrupted closed loop, an open circuit anywhere in the series loop causes total current to drop to zero across all components.

Step-by-Step Solution

1

Identify the circuit topology and the path rule for series circuits.

In a series circuit, all electrical components are connected end-to-end to form a single continuous loop.

Current must flow sequentially through every component in the loop because there are no alternative branches.

2

Determine the electrical impact of a burned-out component.

A burned-out element creates an open circuit, creating a physical break with infinite electrical resistance (

R = \infty

).

Electrons require an uninterrupted conductive path to travel from one terminal of the power source to the other.

3

Apply Ohm's Law to calculate the resulting loop current.

The total circuit current drops to zero (

I = 0\text{ A}

).

According to Ohm's Law (

I = \frac{V}{R}

), an infinite resistance in the loop causes the current throughout the entire series circuit to drop to zero.

Key Concept

Single continuous conductive path and open circuit fault behavior in series circuits

Estimated Time:45s

Question 14Question →

Complete the statement describing how adding components affects a series circuit connected to a constant voltage source.

Fill in the blanks below

When an additional resistor is added in series to a circuit connected to a constant voltage source, the total circuit resistance will and the total current in the circuit will .

Show answer & explanation

Answer

increase (for total resistance) and decrease (for total current)

In a series circuit, all components share a single continuous path. The total equivalent resistance is the arithmetic sum of each individual resistance (

R_T = R_1 + R_2 + R_3

). Inserting an additional resistor increases the total circuit resistance. By Ohm's Law (

I = \frac{V}{R_T}

), an increase in total resistance with an unchanging source voltage results in a decrease in total current.

Step-by-Step Solution

1

Determine the effect of adding a resistor on total series resistance.

Total resistance increases.

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

R_T = R_1 + R_2 + \dots + R_n

), so adding another resistor always increases the total resistance.

2

Apply Ohm's Law to find the effect on total loop current with a fixed voltage source.

Total circuit current decreases.

According to Ohm's Law (

I = \frac{V}{R_T}

), when source voltage

V

remains constant and total resistance

R_T

increases, the total current

I

must decrease.

Key Concept

Series Resistance and Current Relationship

Alternative Method

Consider a numerical example: a

12\text{ V}

source with a single

6\ \Omega

resistor yields

I = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}

. Adding another

6\ \Omega

resistor in series makes

R_T = 12\ \Omega

, which drops current to

I = \frac{12\text{ V}}{12\ \Omega} = 1\text{ A}

.

Estimated Time:45s

Question 15Question →

In a direct current (DC) electronic device, two resistors with values of $30\ \Omega$ and $60\ \Omega$ are connected in series. What is the total equivalent resistance of this circuit?

90\ \Omega

20\ \Omega

30\ \Omega

1800\ \Omega

Show answer & explanation

Answer: $90\ \Omega$

Answer

90\ \Omega

The total resistance of a series circuit is the sum of the individual component resistances (

R_T = R_1 + R_2

). Adding

30\ \Omega

and

60\ \Omega

gives

90\ \Omega

.

Step-by-Step Solution

1

Identify the formula for total resistance in a series circuit.

R_T = R_1 + R_2

In a series circuit, current flows through each resistor sequentially, meaning total opposition to current is the sum of all individual resistances.

2

Substitute the given resistor values into the series formula.

R_T = 30\ \Omega + 60\ \Omega = 90\ \Omega

Adding the individual resistor values produces the total equivalent resistance.

Key Concept

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

R_T = R_1 + R_2 + \dots + R_n

).

Estimated Time:30s

Question 16Question →

Match each series circuit parameter or condition to its corresponding electrical rule.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Total Circuit Resistance pairs with 'Equal to the sum of the individual component resistances'; Circuit Current pairs with 'Remains constant and identical throughout all parts of the circuit'; Total Voltage pairs with 'Equal to the sum of all individual component voltage drops'; Open Circuit Condition pairs with 'Completely interrupts electron flow throughout the entire circuit loop'.

In a series circuit, electric charge has only one continuous loop to travel. Consequently, current is identical at every point, total resistance is the direct sum of all individual resistances, total voltage equals the sum of the individual component voltage drops, and an open circuit completely halts current flow.

Step-by-Step Solution

1

Identify the resistance relationship in a single-path series circuit.

Total resistance is additive (

R_T = R_1 + R_2 + \dots

).

Electrons must travel through each resistive element sequentially, adding together their opposition to current.

2

Identify the current behavior in a series circuit.

Current is uniform everywhere (

I_T = I_1 = I_2

).

There are no branch splits or junctions, so the rate of charge flow is constant throughout the single loop.

3

Identify how voltage is distributed across a series circuit.

Source voltage equals the sum of individual component voltage drops (

V_T = V_1 + V_2 + \dots

).

The total electrical potential provided by the power supply is completely dropped across the resistive elements.

4

Determine the effect of an open fault in a series circuit.

Current flow stops entirely throughout the circuit.

A break at any point opens the single continuous conductive pathway required for current to flow.

Key Concept

Series Circuit Fundamental Rules

Estimated Time:45s

Question 17Question →

An auxiliary automotive circuit connects a $12\ \text{V}$ direct current (DC) power source in series with two resistors valued at $2\ \Omega$ and $4\ \Omega$ . What is the total current flowing through the circuit?

2\ \text{A}

9\ \text{A}

0.5\ \text{A}

72\ \text{A}

Show answer & explanation

Answer: $2\ \text{A}$

Answer

The total current flowing through the circuit is

2\ \text{A}

.

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

R_T = 2\ \Omega + 4\ \Omega = 6\ \Omega

. Using Ohm's law (

I = V / R_T

), dividing the source voltage of

12\ \text{V}

by the total resistance of

6\ \Omega

yields a total current of

2\ \text{A}

.

Step-by-Step Solution

1

Calculate the total equivalent resistance (

R_T

) of the series circuit.

R_T = R_1 + R_2 = 2\ \Omega + 4\ \Omega = 6\ \Omega

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

2

Calculate the total circuit current (

I

) using Ohm's law.

I = \frac{V_T}{R_T} = \frac{12\ \text{V}}{6\ \Omega} = 2\ \text{A}

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

Key Concept

Series Circuit Resistance and Ohm's Law

Estimated Time:45s

Question 18Question →

A $36\text{ V}$ DC power supply is connected across a series circuit containing a fixed $16\ \Omega$ resistor and a variable rheostat. If an in-line ammeter measures a circuit current of $1.5\text{ A}$ , what is the resistance setting of the rheostat?

24\ \Omega

8\ \Omega

54\ \Omega

9.6\ \Omega

Show answer & explanation

Answer: $8\ \Omega$

Answer

8\ \Omega

In a series circuit, the current is identical at all points. Applying Ohm's Law gives the total resistance:

R_T = \frac{V_T}{I} = \frac{36\text{ V}}{1.5\text{ A}} = 24\ \Omega

. Since resistances in series add together linearly (

R_T = R_{\text{fixed}} + R_{\text{rheostat}}

), the rheostat resistance is

24\ \Omega - 16\ \Omega = 8\ \Omega

.

Step-by-Step Solution

1

Calculate the total circuit resistance (

R_T

) using Ohm's Law.

R_T = \frac{V_T}{I} = \frac{36\text{ V}}{1.5\text{ A}} = 24\ \Omega

The current in a series circuit is uniform throughout the entire loop, allowing Ohm's Law to determine total loop resistance from total voltage and current.

2

Subtract the fixed resistance from the total series resistance to solve for the unknown rheostat resistance (

R_{\text{rheostat}}

).

R_{\text{rheostat}} = R_T - R_{\text{fixed}} = 24\ \Omega - 16\ \Omega = 8\ \Omega

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

R_T = R_1 + R_2

).

Key Concept

Series Total Resistance and Ohm's Law

Estimated Time:1m 0s

Question 19Question →

A direct current circuit contains two resistors connected in parallel across a power source. The first resistor has a resistance of $30\ \Omega$ and the second resistor has a resistance of $60\ \Omega$ . What is the total equivalent resistance of the circuit in ohms ( $\Omega$ )?

Show answer & explanation

Answer: 20

Answer

The total equivalent resistance of the circuit is

20\ \Omega

.

For two resistors in parallel, total equivalent resistance is determined by

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

. Substituting

30\ \Omega

and

60\ \Omega

yields

\frac{1800}{90} = 20\ \Omega

.

Step-by-Step Solution

1

Select the appropriate equivalent resistance formula for parallel branches.

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

When two resistors are connected in parallel, the product-over-sum formula directly provides the total equivalent resistance.

2

Substitute the resistor values into the formula and solve.

R_{eq} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\ \Omega

Evaluate the arithmetic expression to calculate the equivalent circuit resistance.

Key Concept

Total equivalent resistance of parallel resistors

Alternative Method

Using the standard reciprocal formula:

\frac{1}{R_{eq}} = \frac{1}{30} + \frac{1}{60} = \frac{2}{60} + \frac{1}{60} = \frac{3}{60} = \frac{1}{20}

. Inverting both sides gives

R_{eq} = 20\ \Omega

.

Estimated Time:45s

Question 20Question →

A marine navigation console contains three indicator loads connected in series across a $75\text{ V}$ direct current (DC) power supply. The individual resistances of the loads are $8\ \Omega$ , $12\ \Omega$ , and $5\ \Omega$ . What is the total equivalent resistance of the circuit, and what is the current flowing through each individual load? Complete the statements below.

Fill in the blanks below

The total equivalent resistance of the series circuit is

\Omega

, and the current flowing through each load is

\text{A}

.

Show answer & explanation

Answer

The total equivalent resistance is

25\ \Omega

and the current through each load is

3\ \text{A}

.

In a series circuit, the total resistance is found by directly summing all individual component resistances (

8\ \Omega + 12\ \Omega + 5\ \Omega = 25\ \Omega

). Using Ohm's Law (

I = V / R_T

), the total circuit current is

75\text{ V} / 25\ \Omega = 3\text{ A}

. In any series circuit, the current is constant throughout the entire loop, meaning every connected component experiences the exact same

3\text{ A}

of current.

Step-by-Step Solution

1

Calculate the total equivalent resistance of the series circuit by adding the individual resistance values.

R_T = R_1 + R_2 + R_3 = 8\ \Omega + 12\ \Omega + 5\ \Omega = 25\ \Omega

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

2

Calculate the total circuit current using Ohm's Law (

I = V_T / R_T

).

I = \frac{75\text{ V}}{25\ \Omega} = 3\text{ A}

Ohm's Law defines current as total applied electromotive force divided by total series resistance.

3

Determine the current flowing through each individual component.

I_1 = I_2 = I_3 = I_T = 3\text{ A}

Because there is only one continuous path for charge flow in a series circuit, current is identical at all points.

Key Concept

Series Circuit Resistance and Current Characteristics

Estimated Time:1m 0s

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