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

Question 21Question →

A field communication station is powered by a $72\text{ V}$ direct-current supply connected across three parallel load branches. The resistance of the first branch is $18\ \Omega$ , the resistance of the second branch is $36\ \Omega$ , and the resistance of the third branch is $72\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), supplied to the circuit?

Show answer & explanation

Answer: 7

Answer

7 A

Each of the three branches is connected directly across the

72\text{ V}

source. Using Ohm's Law (

I = V/R

), the branch currents are

4\text{ A}

,

2\text{ A}

, and

1\text{ A}

. Adding these currents together yields the total current of

7\text{ A}

.

Step-by-Step Solution

1

Calculate the current in the first parallel branch using

I_1 = \frac{V}{R_1}

.

I_1 = \frac{72\text{ V}}{18\ \Omega} = 4\text{ A}

In a parallel DC circuit, the full supply voltage of

72\text{ V}

appears across every parallel branch.

2

Calculate the current in the second parallel branch using

I_2 = \frac{V}{R_2}

.

I_2 = \frac{72\text{ V}}{36\ \Omega} = 2\text{ A}

Applying Ohm's Law to the second branch with its specific resistance.

3

Calculate the current in the third parallel branch using

I_3 = \frac{V}{R_3}

.

I_3 = \frac{72\text{ V}}{72\ \Omega} = 1\text{ A}

Applying Ohm's Law to the third branch with its specific resistance.

4

Add all three individual branch currents together to determine the total current.

I_{\text{total}} = 4\text{ A} + 2\text{ A} + 1\text{ A} = 7\text{ A}

By Kirchhoff's Current Law, the total current leaving the power supply equals the sum of the currents through all parallel branches.

Key Concept

In a parallel circuit, the voltage across each parallel branch is identical to the source voltage, and the total circuit current is the sum of the individual branch currents (

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

).

Alternative Method

Find the total equivalent resistance (

R_{\text{eq}}

) first:

\frac{1}{R_{\text{eq}}} = \frac{1}{18} + \frac{1}{36} + \frac{1}{72} = \frac{4}{72} + \frac{2}{72} + \frac{1}{72} = \frac{7}{72}\ \Omega^{-1}

, so

R_{\text{eq}} = \frac{72}{7}\ \Omega

. Then find total current using

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{72}{72/7} = 7\text{ A}

.

Estimated Time:1m 0s

Question 22Question →

A security alarm control panel operates on a $12\text{ V}$ direct-current power supply connected across two parallel load branches. The first branch has a resistance of $20\ \Omega$ , and the second branch has a resistance of $30\ \Omega$ . What is the total current drawn by the alarm panel from the power supply?

0.24\text{ A}

0.60\text{ A}

1.0\text{ A}

144\text{ A}

Show answer & explanation

Answer: $1.0\text{ A}$

Answer

The total current drawn by the alarm panel is

1.0\text{ A}

.

The total current of

1.0\text{ A}

is correct because each parallel branch experiences the full source voltage of

12\text{ V}

. Dividing the voltage by each branch resistance yields branch currents of

0.60\text{ A}

and

0.40\text{ A}

. Adding these together gives the total circuit current of

1.0\text{ A}

.

Step-by-Step Solution

1

Calculate the current flowing through the first parallel branch (

I_1

)

I_1 = \frac{V}{R_1} = \frac{12\text{ V}}{20\ \Omega} = 0.60\text{ A}

In a parallel circuit, the full supply voltage of

12\text{ V}

appears across each individual branch.

2

Calculate the current flowing through the second parallel branch (

I_2

)

I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{30\ \Omega} = 0.40\text{ A}

Apply Ohm's Law to the second branch using the constant supply voltage.

3

Sum the branch currents to determine total circuit current (

I_{\text{total}}

)

I_{\text{total}} = I_1 + I_2 = 0.60\text{ A} + 0.40\text{ A} = 1.0\text{ A}

According to Kirchhoff's Current Law, total current entering a parallel network equals the sum of all individual branch currents.

Key Concept

In a parallel DC circuit, voltage remains equal across all branches, and total source current equals the sum of the individual branch currents (

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

).

Alternative Method

Find total equivalent resistance using the product-over-sum formula:

R_{\text{total}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{20 \times 30}{20 + 30} = \frac{600}{50} = 12\ \Omega

. Then calculate total current directly:

I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}

.

Estimated Time:1m 0s

Question 23Question →

In direct-current (DC) parallel circuit analysis, modifying circuit components or introducing faults alters electrical parameters according to fundamental circuit laws. Match each parallel circuit condition or parameter on the left with its correct electrical behavior or outcome on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Adding a branch matches decreasing total equivalent resistance; an open branch matches dropping current to zero only in that branch; branch voltage drop matches remaining identical across all branches; unequal current division matches directing larger current through the smaller resistor.

In parallel circuits: (1) Adding branches decreases equivalent resistance and raises total current; (2) An open in one branch affects only that path; (3) Voltage is equal across all branches; and (4) Branch current is inversely proportional to resistance.

Step-by-Step Solution

1

Analyze equivalent resistance behavior when adding parallel paths.

R_{\text{eq}}

decreases as branches are added because

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

.

Each added branch provides an additional path for electron flow, reducing overall opposition to current.

2

Evaluate the effect of an open circuit in an isolated parallel branch.

Current ceases in the open branch, while voltage and current across other branches remain steady.

Independent branch connections across common rails preserve full line voltage across unaffected branches.

3

Determine voltage distribution across parallel loads.

Branch voltages are identical (

V_{\text{source}} = V_1 = V_2

).

All parallel elements share the same electrical junctions.

4

Determine current division rules for unequal resistances.

Current flows inversely to resistance (

I_1 = \frac{V}{R_1} > I_2 = \frac{V}{R_2}

when

R_1 < R_2

).

Lower resistance paths draw greater current from the constant branch voltage.

Key Concept

Parallel Circuit Voltage, Current, Resistance, and Fault Characteristics

Estimated Time:1m 15s

Question 24Question →

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

7.2\ \Omega

18\ \Omega

12\ \Omega

72\ \Omega

Show answer & explanation

Answer: $18\ \Omega$

Answer

The resistance of the second branch load is

18\ \Omega

.

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

36\text{ V}

. Using Ohm's Law, the first branch draws

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

. Subtracting this from the total current of

5\text{ A}

gives a second branch current of

I_2 = 2\text{ A}

. Dividing the

36\text{ V}

bus voltage by

2\text{ A}

yields a resistance of

18\ \Omega

for the second branch.

Step-by-Step Solution

1

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

I_1 = V / R_1

).

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

In a parallel circuit, the full supply voltage of

36\text{ V}

is present across every parallel branch.

2

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

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

).

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

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

3

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

R_2 = V / I_2

).

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

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

Key Concept

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

Alternative Method

Find the total circuit equivalent resistance first:

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

. Then apply the parallel resistance formula:

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

, which gives

R_2 = 18\ \Omega

.

Estimated Time:1m 0s

Question 25Question →

A marine auxiliary direct-current power circuit supplies a constant $48\text{ V}$ across three parallel branches. The first branch has a resistance of $16\ \Omega$ , the second branch has a resistance of $24\ \Omega$ , and the third branch has a resistance of $48\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), drawn from the power supply?

Show answer & explanation

Answer: 6

Answer

The total current drawn from the power supply is

6\text{ A}

.

Each parallel branch receives the full source voltage of

48\text{ V}

. Dividing the voltage by each branch resistance yields branch currents of

3\text{ A}

,

2\text{ A}

, and

1\text{ A}

. Summing these branch currents gives a total circuit current of

6\text{ A}

.

Step-by-Step Solution

1

Calculate the current passing through the first branch

I_1 = \frac{48\text{ V}}{16\ \Omega} = 3\text{ A}

The full source voltage of

48\text{ V}

appears across every parallel branch.

2

Calculate the current passing through the second branch

I_2 = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

Applying Ohm's Law (

I = \frac{V}{R}

) with the same supply voltage across the

24\ \Omega

branch.

3

Calculate the current passing through the third branch

I_3 = \frac{48\text{ V}}{48\ \Omega} = 1\text{ A}

Applying Ohm's Law (

I = \frac{V}{R}

) across the

48\ \Omega

branch.

4

Sum all individual branch currents to determine total current

I_{\text{total}} = 3\text{ A} + 2\text{ A} + 1\text{ A} = 6\text{ A}

In a parallel circuit, the total current equals the sum of all branch currents (

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

).

Key Concept

In a parallel DC circuit, the voltage across each branch is identical to the source voltage, and the total current supplied is the sum of the individual branch currents.

Alternative Method

Find the equivalent resistance (

R_{\text{eq}}

) first using the reciprocal formula:

\frac{1}{R_{\text{eq}}} = \frac{1}{16} + \frac{1}{24} + \frac{1}{48} = \frac{3}{48} + \frac{2}{48} + \frac{1}{48} = \frac{6}{48} = \frac{1}{8}\implies R_{\text{eq}} = 8\ \Omega

. Then, calculate total current via Ohm's Law:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{48\text{ V}}{8\ \Omega} = 6\text{ A}

.

Estimated Time:1m 0s

Question 26Question →

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

8\text{ A}

6\text{ A}

1.33\text{ A}

10\text{ A}

Show answer & explanation

Answer: $6\text{ A}$

Answer

The total current drawn from the power source is

6\text{ A}

.

The correct answer is

6\text{ A}

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

60\text{ V}

. Applying Ohm's Law to each branch yields

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

and

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

. Summing these branch currents gives a total current of

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

. Alternatively, finding the equivalent resistance (

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

) and applying Ohm's Law (

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

) confirms

6\text{ A}

.

Step-by-Step Solution

1

Determine the voltage across each parallel branch

V_1 = V_2 = 60\text{ V}

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

2

Calculate the individual branch currents using Ohm's Law (

I = \frac{V}{R}

)

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

,

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

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

3

Sum the branch currents to find the total circuit current (

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

)

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

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

Key Concept

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

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

).

Alternative Method

Find equivalent resistance first:

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

. Then calculate total current:

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

.

Estimated Time:1m 0s

Question 27Question →

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

12\ \Omega

18\ \Omega

36\ \Omega

45\ \Omega

Show answer & explanation

Answer: $45\ \Omega$

Answer

The resistance of the second resistor (

R_2

) is

45\ \Omega

.

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

90\text{ V}

). The current in the first branch is

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

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

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

). Therefore, the current in the second branch is

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

. Applying Ohm's Law to the second branch gives

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

.

Step-by-Step Solution

1

Determine the current flowing through the first parallel branch (

I_1

)

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

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

90\text{ V}

).

2

Calculate the current flowing through the second branch (

I_2

) using Kirchhoff's Current Law

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

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

3

Calculate the resistance of the second resistor (

R_2

) using Ohm's Law

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

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

Key Concept

Parallel branch voltage equality and Kirchhoff's Current Law

Estimated Time:1m 15s

Question 28Question →

An emergency egress lighting system operates from a $50\text{ V}$ direct-current (DC) power source connected across two parallel branch circuits. Branch 1 contains an LED driver with an effective resistance of $25\ \Omega$ . If the total current drawn from the power source is $3\text{ A}$ , what is the resistance, in ohms ( $\Omega$ ), of Branch 2?

Show answer & explanation

Answer: 50

Answer

The resistance of Branch 2 is

50\ \Omega

.

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

50\text{ V}

. Branch 1 draws

I_1 = \frac{50\text{ V}}{25\ \Omega} = 2\text{ A}

. Since the total current is

3\text{ A}

, Branch 2 must draw

I_2 = 3\text{ A} - 2\text{ A} = 1\text{ A}

. Applying Ohm's Law to Branch 2 gives

R_2 = \frac{50\text{ V}}{1\text{ A}} = 50\ \Omega

.

Step-by-Step Solution

1

Determine the voltage across each parallel branch.

V_1 = V_2 = 50\text{ V}

In any parallel circuit configuration, the full source voltage appears equally across every individual parallel branch.

2

Calculate the current flowing through Branch 1 using Ohm's Law (

I = \frac{V}{R}

).

I_1 = \frac{50\text{ V}}{25\ \Omega} = 2\text{ A}

Ohm's Law allows solving for branch current when branch voltage and resistance are known.

3

Calculate the remaining current flowing through Branch 2 by subtracting Branch 1 current from total line current.

I_2 = 3\text{ A} - 2\text{ A} = 1\text{ A}

According to Kirchhoff's Current Law, total current entering a parallel network must equal the sum of all individual branch currents (

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

).

4

Calculate the resistance of Branch 2 using Ohm's Law (

R = \frac{V}{I}

).

R_2 = \frac{50\text{ V}}{1\text{ A}} = 50\ \Omega

Dividing the branch voltage by the branch current yields the branch's effective resistance.

Key Concept

Parallel circuit branch voltage equality, Kirchhoff's Current Law (

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

), and Ohm's Law (

R = \frac{V}{I}

).

Estimated Time:1m 0s

Question 29Question →

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

100\ \Omega

24\ \Omega

50\ \Omega

20\ \Omega

Show answer & explanation

Answer: $24\ \Omega$

Answer

24\ \Omega

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

R_{\text{eq}}

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

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

. Substituting the circuit values yields

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

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

24\ \Omega < 40\ \Omega

).

Step-by-Step Solution

1

Identify the parallel resistance formula for two branch loads

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

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

2

Calculate the product of the two branch resistances

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

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

3

Calculate the sum of the two branch resistances

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

Adding the branch values forms the denominator of the formula.

4

Divide the product by the sum to determine equivalent resistance

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

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

Key Concept

Parallel Equivalent Resistance

Estimated Time:1m 0s

Question 30Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

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

Step-by-Step Solution

1

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

Equivalent resistance decreases and total current increases

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

2

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

Total current drops while other branches continue normal operation

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

3

Determine how current distributes across branches with different resistances

Current divides inversely with resistance

With uniform branch voltages, Ohm's law (

I = \frac{V}{R}

) ensures that smaller resistances allow larger branch currents.

4

Verify voltage characteristics across parallel circuit elements

Voltage drop across every branch equals the source voltage

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

Key Concept

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

Estimated Time:1m 0s

Question 31Question →

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

27\ \Omega

6\ \Omega

13.5\ \Omega

2\ \Omega

Show answer & explanation

Answer: $6\ \Omega$

Answer

The total equivalent resistance of the parallel circuit is

6\ \Omega

.

The equivalent resistance of parallel resistors is found using the formula

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

or the shortcut

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

. Substituting

18\ \Omega

and

9\ \Omega

yields

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

, which is properly less than the smallest branch resistance (

9\ \Omega

).

Step-by-Step Solution

1

Identify the standard equivalent resistance relationship for two unequal parallel branches

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

or

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

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

2

Substitute the given resistances (

R_1 = 18\ \Omega

and

R_2 = 9\ \Omega

) into the product-over-sum formula

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

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

3

Divide the product by the sum to determine the equivalent resistance

R_{eq} = 6\ \Omega

Simplifying

\frac{162}{27}

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

Key Concept

Total Equivalent Resistance of Unequal Parallel Branches

Estimated Time:1m 0s

Question 32Question →

A $24\text{ V}$ direct-current (DC) power bus supplies two cooling fans connected in parallel inside a communications cabinet. The primary fan draws a steady branch current of $3\text{ A}$ , while the secondary fan has an internal resistance of $12\text{ }\Omega$ . What is the total current, in amperes ( $\text{A}$ ), delivered by the power bus to the parallel circuit?

Show answer & explanation

Answer: 5

Answer

The total current delivered by the power bus is

5\text{ A}

.

In a parallel DC circuit, the source voltage (

24\text{ V}

) is applied directly across each branch. The secondary branch current is found using Ohm's law:

I_2 = \frac{24\text{ V}}{12\text{ }\Omega} = 2\text{ A}

. Adding the known primary branch current (

3\text{ A}

) to the secondary branch current (

2\text{ A}

) yields the total source current of

5\text{ A}

.

Step-by-Step Solution

1

Determine the voltage across the secondary branch

Voltage across the secondary fan is

24\text{ V}

In any parallel circuit, every branch experiences the identical full source voltage (

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

).

2

Calculate the secondary branch current (

I_2

) using Ohm's law

I_2 = \frac{24\text{ V}}{12\text{ }\Omega} = 2\text{ A}

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

I = \frac{V}{R}

).

3

Sum all individual branch currents to determine total circuit current

I_{\text{total}} = 3\text{ A} + 2\text{ A} = 5\text{ A}

Total current in a parallel circuit is the sum of the individual branch currents (

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

).

Key Concept

Parallel Circuit Current and Ohm's Law

Estimated Time:1m 0s

Question 33Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

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

Step-by-Step Solution

1

Analyze the impact of adding a parallel branch

R_{\text{eq}}

decreases and

I_{\text{total}}

increases

Each additional parallel path increases overall circuit conductance (

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

), lowering equivalent resistance.

2

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

I_{\text{branch}}

drops,

I_{\text{total}}

drops, and

R_{\text{eq}}

rises

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

3

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

Defective branch current goes to zero while intact branches continue functioning

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

4

Evaluate a short-circuit condition across any parallel branch

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

and total current surges drastically

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

Key Concept

Independent branch operation and equivalent resistance behavior in parallel DC circuits

Estimated Time:1m 0s

Question 34Question →

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

1\text{ A}

6\text{ A}

10\text{ A}

12\text{ A}

Show answer & explanation

Answer: $10\text{ A}$

Answer

10\text{ A}

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

120\text{ V}

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

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

,

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

, and

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

. Summing these branch currents yields a total current of

10\text{ A}

(

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

). Alternatively, calculating equivalent resistance gives

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

, so

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

, yielding

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

.

Step-by-Step Solution

1

Calculate the individual branch currents using Ohm's Law (

I = \frac{V}{R}

).

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

,

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

, and

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

.

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

120\text{ V}

.

2

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

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

).

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

.

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

Key Concept

Parallel circuit branch current calculation and Kirchhoff's Current Law

Estimated Time:1m 0s

Question 35Question →

A $72\text{ V}$ direct-current (DC) power supply energizes an industrial sensor interface consisting of two resistors connected in parallel. The first resistor ( $R_1$ ) has a resistance of $24\ \Omega$ . If the power supply delivers a total current of $5\text{ A}$ to the circuit, what is the resistance of the second resistor ( $R_2$ )?

9.6\ \Omega

36\ \Omega

14.4\ \Omega

28.8\ \Omega

Show answer & explanation

Answer: $36\ \Omega$

Answer

36\ \Omega

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

72\text{ V}

). The current through the known resistor is

I_1 = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

. Since the total current entering the parallel junction is

5\text{ A}

, the remaining current flowing through the second branch must be

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

. Using Ohm's law for the second branch yields

R_2 = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega

.

Step-by-Step Solution

1

Calculate the current flowing through the first branch resistor (

R_1

)

I_1 = \frac{V}{R_1} = \frac{72\text{ V}}{24\ \Omega} = 3\text{ A}

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

72\text{ V}

.

2

Determine the current flowing through the second branch (

I_2

)

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

By Kirchhoff's Current Law, total current in a parallel circuit equals the sum of the individual branch currents.

3

Calculate the resistance of the second resistor (

R_2

) using Ohm's law

R_2 = \frac{V}{I_2} = \frac{72\text{ V}}{2\text{ A}} = 36\ \Omega

Applying Ohm's law (

R = \frac{V}{I}

) with the branch voltage and branch current gives the branch resistance.

Key Concept

Parallel branch voltage equality, Kirchhoff's Current Law, and Ohm's law branch calculations

Estimated Time:1m 0s

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