Circuits and Magnetism

163 questions

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 mA150\text{ mA} when connected to a 24 V24\text{ V} direct current source. The internal resistance of the lamp is Ω\Omega, and if the supply voltage is increased to 36 V36\text{ V}, the current drawn by the lamp will increase to mA\text{mA}.
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Answer

The internal resistance of the lamp is 160 Ω, and the new current at 36 V is 225 mA.
Converting 150 mA150\text{ mA} to standard amperes yields 0.15 A0.15\text{ A}. Using Ohm's law (R=V/IR = V / I), the internal resistance is 24 V/0.15 A=160 Ω24\text{ V} / 0.15\text{ A} = 160\ \Omega. When the voltage rises to 36 V36\text{ V}, applying I=V/RI = V / R produces 36 V/160 Ω=0.225 A36\text{ V} / 160\ \Omega = 0.225\text{ A}, which corresponds to 225 mA225\text{ mA}.

Step-by-Step Solution

1
Convert the nominal current from milliamperes to base amperes
I1=150 mA=0.15 AI_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=V1I1=24 V0.15 A=160 ΩR = \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
I2=V2R=36 V160 Ω=0.225 A=225 mAI_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/RI = V / R).

Key Concept

Ohm's Law (V=I×RV = 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.

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Items

Solenoid
Transformer
Relay
Inductor

Matches

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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 V12\text{ V} power source draws a current of 2 A2\text{ A}. What is the total electrical power consumed by the lamp?

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Answer: 24 W24\text{ W}

Answer

24 W24\text{ W}
According to Watt's Law, the electrical power (PP) consumed by a circuit load is equal to the voltage (VV) across the component multiplied by the current (II) flowing through it: P=V×IP = V \times I. Substituting the given values yields P=12 V×2 A=24 WP = 12\text{ V} \times 2\text{ A} = 24\text{ W}.

Step-by-Step Solution

1
Identify the given circuit parameters
Voltage (VV) = 12 V12\text{ V}, Current (II) = 2 A2\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×IP = 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 V×2 A=24 WP = 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×IP = 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 V12\text{ V} DC power supply. If the internal electrical resistance of the pump motor is 6 Ω6\ \Omega, what is the steady-state current drawn by the pump during operation?

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Answer: 2 A2\text{ A}

Answer

The current drawn by the pump is 2 A2\text{ A}.
According to Ohm's Law, current (II) equals voltage (VV) divided by resistance (RR). Dividing the 12 V12\text{ V} supply by the 6 Ω6\ \Omega load resistance yields exactly 2 A2\text{ A}.

Step-by-Step Solution

1
Identify the given circuit parameters and the formula to solve for unknown current.
Given: Voltage V=12 VV = 12\text{ V}, Resistance R=6 ΩR = 6\ \Omega. Ohm's Law states V=I×RV = I \times R, which rearranges to I=VRI = \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=12 V6 Ω=2 AI = \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=VRI = \frac{V}{R}
Estimated Time:40s
Question 5Question →

An electrical test bench delivers a constant 80 V80\text{ V} direct-current supply to four separate parallel circuit arrangements. Match each circuit configuration on the left with its corresponding equivalent resistance (ReqR_{eq}) and total supply current (ITI_T) on the right.

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Items

Two resistors of 40 Ω40\ \Omega and 10 Ω10\ \Omega connected in parallel across the supply
Four identical 64 Ω64\ \Omega resistive heating elements connected in parallel
Three parallel branches containing 120 Ω120\ \Omega, 60 Ω60\ \Omega, and 40 Ω40\ \Omega resistors
Two parallel branches containing a 150 Ω150\ \Omega sensor and a 75 Ω75\ \Omega resistor

Matches

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Answer

The configuration with 40 Ω40\ \Omega and 10 Ω10\ \Omega matches Req=8 Ω,IT=10 AR_{eq} = 8\ \Omega, I_T = 10\text{ A}; four 64 Ω64\ \Omega elements match Req=16 Ω,IT=5 AR_{eq} = 16\ \Omega, I_T = 5\text{ A}; three branches of 120 Ω,60 Ω,40 Ω120\ \Omega, 60\ \Omega, 40\ \Omega match Req=20 Ω,IT=4 AR_{eq} = 20\ \Omega, I_T = 4\text{ A}; and two branches of 150 Ω,75 Ω150\ \Omega, 75\ \Omega match Req=50 Ω,IT=1.6 AR_{eq} = 50\ \Omega, I_T = 1.6\text{ A}.
In any direct-current parallel network, the total equivalent resistance (ReqR_{eq}) is calculated using the reciprocal formula 1Req=∑1Rn\frac{1}{R_{eq}} = \sum \frac{1}{R_n} (or R/NR/N for identical branches). Total current drawn from the supply is determined by Ohm's Law: IT=VSReqI_T = \frac{V_S}{R_{eq}}. Evaluating each configuration against the constant 80 V80\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 Ω40\ \Omega and 10 Ω10\ \Omega parallel branch pair.
Req=40×1040+10=8 ΩR_{eq} = \frac{40 \times 10}{40 + 10} = 8\ \Omega; IT=80 V8 Ω=10 AI_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/RI = V/R).
2
Calculate the equivalent resistance and current for the four parallel 64 Ω64\ \Omega elements.
Req=644=16 ΩR_{eq} = \frac{64}{4} = 16\ \Omega; IT=80 V16 Ω=5 AI_T = \frac{80\text{ V}}{16\ \Omega} = 5\text{ A}.
Identical parallel resistors simplify to branch resistance divided by the number of branches (R/NR/N).
3
Calculate the equivalent resistance and current for the three parallel branches (120 Ω,60 Ω,40 Ω120\ \Omega, 60\ \Omega, 40\ \Omega).
1Req=1120+2120+3120=6120=120  ⟹  Req=20 Ω\frac{1}{R_{eq}} = \frac{1}{120} + \frac{2}{120} + \frac{3}{120} = \frac{6}{120} = \frac{1}{20} \implies R_{eq} = 20\ \Omega; IT=80 V20 Ω=4 AI_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 Ω150\ \Omega and 75 Ω75\ \Omega parallel branches.
Req=150×75150+75=11250225=50 ΩR_{eq} = \frac{150 \times 75}{150 + 75} = \frac{11250}{225} = 50\ \Omega; IT=80 V50 Ω=1.6 AI_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 Ω10\ \Omega, 15 Ω15\ \Omega, and 25 Ω25\ \Omega are connected in series across a direct current (DC) voltage source. What is the total equivalent resistance of the circuit?

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Answer: 50 Ω50\ \Omega

Answer

The total equivalent resistance is 50 Ω50\ \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: RT=10 Ω+15 Ω+25 Ω=50 ΩR_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.
RT=R1+R2+R3R_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.
RT=10 Ω+15 Ω+25 ΩR_T = 10\ \Omega + 15\ \Omega + 25\ \Omega
Each component contributes its full resistive value directly to the total opposition.
3
Calculate the arithmetic sum.
RT=50 ΩR_T = 50\ \Omega
Simple addition yields the equivalent circuit resistance.

Key Concept

In a series circuit, total equivalent resistance (RTR_T) equals the direct sum of all individual resistances (RT=R1+R2+⋯+RnR_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.
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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 (Itotal=I1=I2I_{\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 (Itotal=I1=I2=…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 V24\text{ V} direct current (DC) power supply is connected to a circuit with two resistors in series: a 4 Ω4\ \Omega resistor and an 8 Ω8\ \Omega resistor. What is the current flowing through the 8 Ω8\ \Omega resistor?

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Answer: 2 A2\text{ A}

Answer

2 A2\text{ A}
The total resistance of the series circuit is the direct sum of the resistors (4 Ω+8 Ω=12 Ω4\ \Omega + 8\ \Omega = 12\ \Omega). According to Ohm's Law (I=V/RI = V / R), dividing the 24 V24\text{ V} source by 12 Ω12\ \Omega yields a total current of 2 A2\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 A2\text{ A} flows through the 8 Ω8\ \Omega resistor.

Step-by-Step Solution

1
Calculate the total equivalent resistance of the series circuit.
Rtotal=R1+R2=4 Ω+8 Ω=12 ΩR_{\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.
Itotal=VtotalRtotal=24 V12 Ω=2 AI_{\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 Ω8\ \Omega resistor.
IR2=Itotal=2 AI_{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

Items

Circuit Current (II)
Total Resistance (RTR_T)
Source Voltage (VTV_T)
Open Component Fault

Matches

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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 (RT=R1+R2+…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 VTV_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 V12\text{ V} DC) circuit contains two indicator lamps connected in series. If a voltmeter measures a 7 V7\text{ V} drop across the first lamp, what is the voltage drop across the second lamp?

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Answer: 5 V5\text{ V}

Answer

The voltage drop across the second lamp is 5 V5\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 (Vtotal=V1+V2V_{\text{total}} = V_1 + V_2). Subtracting the 7 V7\text{ V} drop across the first lamp from the 12 V12\text{ V} total supply gives 5 V5\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: Vtotal=V1+V2V_{\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 (V2V_2).
V2=Vtotal−V1=12 V−7 V=5 VV_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 Ω15\ \Omega, 35 Ω35\ \Omega, and 50 Ω50\ \Omega. What is the total equivalent resistance of the circuit in ohms (Ω\Omega)?

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Answer: 100

Answer

100 ohms (Ω\Omega)
In a series circuit, total resistance is calculated by adding the individual resistance values together (RT=R1+R2+R3R_T = R_1 + R_2 + R_3). Summing 15 Ω15\ \Omega, 35 Ω35\ \Omega, and 50 Ω50\ \Omega gives exactly 100 Ω100\ \Omega.

Step-by-Step Solution

1
Identify the formula for total resistance in a series circuit
RT=R1+R2+R3R_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
RT=15 Ω+35 Ω+50 Ω=100 ΩR_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

Items

Adding an additional resistive branch in parallel
Developing an open-circuit fault in one parallel branch
Developing a short-circuit fault across one branch
Increasing the main DC supply voltage

Matches

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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/RI = 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?

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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=∞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 AI = 0\text{ A}).
According to Ohm's Law (I=VRI = \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 .
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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 (RT=R1+R2+R3R_T = R_1 + R_2 + R_3). Inserting an additional resistor increases the total circuit resistance. By Ohm's Law (I=VRTI = \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 (RT=R1+R2+⋯+RnR_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=VRTI = \frac{V}{R_T}), when source voltage VV remains constant and total resistance RTR_T increases, the total current II must decrease.

Key Concept

Series Resistance and Current Relationship

Alternative Method

Consider a numerical example: a 12 V12\text{ V} source with a single 6 Ω6\ \Omega resistor yields I=12 V6 Ω=2 AI = \frac{12\text{ V}}{6\ \Omega} = 2\text{ A}. Adding another 6 Ω6\ \Omega resistor in series makes RT=12 ΩR_T = 12\ \Omega, which drops current to I=12 V12 Ω=1 AI = \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 Ω30\ \Omega and 60 Ω60\ \Omega are connected in series. What is the total equivalent resistance of this circuit?

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Answer: 90 Ω90\ \Omega

Answer

90 Ω90\ \Omega
The total resistance of a series circuit is the sum of the individual component resistances (RT=R1+R2R_T = R_1 + R_2). Adding 30 Ω30\ \Omega and 60 Ω60\ \Omega gives 90 Ω90\ \Omega.

Step-by-Step Solution

1
Identify the formula for total resistance in a series circuit.
RT=R1+R2R_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.
RT=30 Ω+60 Ω=90 ΩR_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 (RT=R1+R2+⋯+RnR_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

Items

Total Circuit Resistance (RTR_T)
Circuit Current (II)
Total Voltage (VTV_T)
Open Circuit Condition

Matches

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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 (RT=R1+R2+…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 (IT=I1=I2I_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 (VT=V1+V2+…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 V12\ \text{V} direct current (DC) power source in series with two resistors valued at 2 Ω2\ \Omega and 4 Ω4\ \Omega. What is the total current flowing through the circuit?

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Answer: 2 A2\ \text{A}

Answer

The total current flowing through the circuit is 2 A2\ \text{A}.
In a series circuit, total resistance is the direct sum of the individual resistors: RT=2 Ω+4 Ω=6 ΩR_T = 2\ \Omega + 4\ \Omega = 6\ \Omega. Using Ohm's law (I=V/RTI = V / R_T), dividing the source voltage of 12 V12\ \text{V} by the total resistance of 6 Ω6\ \Omega yields a total current of 2 A2\ \text{A}.

Step-by-Step Solution

1
Calculate the total equivalent resistance (RTR_T) of the series circuit.
RT=R1+R2=2 Ω+4 Ω=6 ΩR_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 (II) using Ohm's law.
I=VTRT=12 V6 Ω=2 AI = \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 V36\text{ V} DC power supply is connected across a series circuit containing a fixed 16 Ω16\ \Omega resistor and a variable rheostat. If an in-line ammeter measures a circuit current of 1.5 A1.5\text{ A}, what is the resistance setting of the rheostat?

Show answer & explanation

Answer: 8 Ω8\ \Omega

Answer

8 Ω8\ \Omega
In a series circuit, the current is identical at all points. Applying Ohm's Law gives the total resistance: RT=VTI=36 V1.5 A=24 ΩR_T = \frac{V_T}{I} = \frac{36\text{ V}}{1.5\text{ A}} = 24\ \Omega. Since resistances in series add together linearly (RT=Rfixed+RrheostatR_T = R_{\text{fixed}} + R_{\text{rheostat}}), the rheostat resistance is 24 Ω−16 Ω=8 Ω24\ \Omega - 16\ \Omega = 8\ \Omega.

Step-by-Step Solution

1
Calculate the total circuit resistance (RTR_T) using Ohm's Law.
RT=VTI=36 V1.5 A=24 ΩR_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 (RrheostatR_{\text{rheostat}}).
Rrheostat=RT−Rfixed=24 Ω−16 Ω=8 ΩR_{\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 (RT=R1+R2R_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 Ω30\ \Omega and the second resistor has a resistance of 60 Ω60\ \Omega. What is the total equivalent resistance of the circuit in ohms (Ω\Omega)?

Show answer & explanation

Answer: 20

Answer

The total equivalent resistance of the circuit is 20 Ω20\ \Omega.
For two resistors in parallel, total equivalent resistance is determined by Req=R1×R2R1+R2R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2}. Substituting 30 Ω30\ \Omega and 60 Ω60\ \Omega yields 180090=20 Ω\frac{1800}{90} = 20\ \Omega.

Step-by-Step Solution

1
Select the appropriate equivalent resistance formula for parallel branches.
Req=R1×R2R1+R2R_{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.
Req=30×6030+60=180090=20 ΩR_{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: 1Req=130+160=260+160=360=120\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 Req=20 ΩR_{eq} = 20\ \Omega.
Estimated Time:45s
Question 20Question →

A marine navigation console contains three indicator loads connected in series across a 75 V75\text{ V} direct current (DC) power supply. The individual resistances of the loads are 8 Ω8\ \Omega, 12 Ω12\ \Omega, and 5 Ω5\ \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 Ω\Omega, and the current flowing through each load is A\text{A}.
Show answer & explanation

Answer

The total equivalent resistance is 25 Ω25\ \Omega and the current through each load is 3 A3\ \text{A}.
In a series circuit, the total resistance is found by directly summing all individual component resistances (8 Ω+12 Ω+5 Ω=25 Ω8\ \Omega + 12\ \Omega + 5\ \Omega = 25\ \Omega). Using Ohm's Law (I=V/RTI = V / R_T), the total circuit current is 75 V/25 Ω=3 A75\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 A3\text{ A} of current.

Step-by-Step Solution

1
Calculate the total equivalent resistance of the series circuit by adding the individual resistance values.
RT=R1+R2+R3=8 Ω+12 Ω+5 Ω=25 ΩR_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=VT/RTI = V_T / R_T).
I=75 V25 Ω=3 AI = \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.
I1=I2=I3=IT=3 AI_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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