Parallel Circuits Alıştırma Soruları — Armed Services Vocational Aptitude Battery (ASVAB)

Soru 1Soru →

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.

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Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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.

Tahmini Süre:2m 0s

Soru 2Soru →

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.

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Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Parallel Circuit Governing Rules and Fault Characteristics

Tahmini Süre:1m 15s

Soru 3Soru →

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 ( $\Omega$ )?

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Cevap: 20

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Total equivalent resistance of parallel resistors

Alternatif Yöntem

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

.

Tahmini Süre:45s

Soru 4Soru →

An automotive technician tests a lighting circuit consisting of two auxiliary lamps wired in parallel. The resistance of the first lamp is $10\ \Omega$ , and the resistance of the second lamp is $15\ \Omega$ . What is the total equivalent resistance of this parallel circuit?

25\ \Omega

6\ \Omega

12.5\ \Omega

5\ \Omega

Cevabı ve açıklamayı göster

Cevap: $6\ \Omega$

Cevap

The total equivalent resistance of the parallel circuit is

6\ \Omega

.

The equivalent resistance of two resistors connected in parallel is calculated using

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

. Substituting

10\ \Omega

and

15\ \Omega

gives

\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega

.

Adım Adım Çözüm

1

Identify the formula for equivalent resistance of two parallel resistors

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

or

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

Resistors in parallel provide multiple paths for current, resulting in an equivalent resistance smaller than any individual branch.

2

Substitute the given resistances (

R_1 = 10\ \Omega

and

R_2 = 15\ \Omega

) into the product-over-sum formula

R_{\text{eq}} = \frac{10 \times 15}{10 + 15} = \frac{150}{25}

Using the product-over-sum shortcut provides a direct calculation for two parallel branches.

3

Calculate the final equivalent resistance

R_{\text{eq}} = 6\ \Omega

Dividing

150

by

25

yields the total resistance of

6\ \Omega

.

Anahtar Kavram

Equivalent resistance in a parallel circuit is always less than the smallest individual branch resistance and is calculated using the reciprocal or product-over-sum rule.

Tahmini Süre:45s

Soru 5Soru →

Which electrical principle or formula defines each parameter in a standard parallel circuit? Match each circuit parameter on the left with its corresponding characteristic on the right.

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Cevap

Voltage across each branch matches 'Remains equal across every branch and matches the source voltage'; Total circuit current matches 'Equals the sum of all individual branch currents'; Total equivalent resistance matches 'Is always less than the resistance of the smallest branch resistor'; Current through an individual branch matches 'Calculated by dividing source voltage by the branch resistance (

I_n = \frac{V}{R_n}

)'.

In parallel circuits, voltage is identical across every branch. Total current is the sum of all individual branch currents. Total equivalent resistance decreases as more branches are added and is always less than the smallest individual resistor. Each branch current is calculated independently using Ohm's Law with the common voltage.

Adım Adım Çözüm

1

Determine the voltage rule for parallel connections

Branch voltage is identical to source voltage (

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

)

All parallel branches share common input and return connection points.

2

Determine the total current rule for parallel connections

Total current is the sum of branch currents (

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

)

Current divides among all available parallel conducting branches according to Kirchhoff's Current Law.

3

Determine the equivalent resistance behavior

Equivalent resistance is less than the smallest branch resistance (

R_{\text{eq}} < R_{\text{smallest}}

)

Each additional parallel path increases overall circuit conductance.

4

Apply Ohm's law to calculate individual branch current

Branch current is

I_n = \frac{V}{R_n}

Ohm's law applies directly across each branch using the common source voltage.

Anahtar Kavram

Fundamental voltage, current, and equivalent resistance relationships in parallel circuits

Soru 6Soru →

Four identical $40\ \Omega$ heating elements are connected in parallel across a $120\text{ V}$ power line. What is the total equivalent resistance of this circuit?

160\ \Omega

10\ \Omega

40\ \Omega

20\ \Omega

Cevabı ve açıklamayı göster

Cevap: $10\ \Omega$

Cevap

The total equivalent resistance of the circuit is

10\ \Omega

.

For

N

identical resistors of equal resistance

R

connected in parallel, the total equivalent resistance is simply the resistance of one branch divided by the total number of branches:

R_{\text{eq}} = \frac{R}{N} = \frac{40\ \Omega}{4} = 10\ \Omega

.

Adım Adım Çözüm

1

Identify the number of parallel branches (

N

) and individual branch resistance (

R

).

N = 4

identical branches, each with

R = 40\ \Omega

.

Recognizing identical parallel branches allows the use of the equal-resistor shortcut formula.

2

Apply the parallel equivalent resistance formula for identical resistors:

R_{\text{eq}} = \frac{R}{N}

.

R_{\text{eq}} = \frac{40\ \Omega}{4} = 10\ \Omega

.

Connecting resistors in parallel creates multiple paths for electron flow, reducing the total circuit resistance below that of any individual branch.

Anahtar Kavram

Total Equivalent Resistance of Equal Parallel Resistors

Tahmini Süre:45s

Soru 7Soru →

A $36\text{ V}$ direct-current power supply is connected across a parallel circuit containing two branches. If the first branch contains an $18\ \Omega$ resistor and the second branch contains a $9\ \Omega$ resistor, what is the current flowing through the second branch in amperes?

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Cevap: 4

Cevap

4 A

In a parallel circuit, the voltage across all individual branches is identical to the supply voltage (

36\text{ V}

). Using Ohm's law for the second branch, the current is calculated as

I = \frac{V}{R} = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}

.

Adım Adım Çözüm

1

Determine the voltage drop across the second branch.

The voltage across the second branch is

36\text{ V}

.

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

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

).

2

Calculate the branch current using Ohm's law.

I_2 = \frac{36\text{ V}}{9\ \Omega} = 4\text{ A}

.

Current through an individual resistor branch is found by dividing the branch voltage by the branch resistance (

I = \frac{V}{R}

).

Anahtar Kavram

In a parallel circuit, the voltage across every parallel branch is equal to the source voltage, and the current through any branch depends solely on that branch's resistance via Ohm's law (

I = \frac{V}{R}

).

Tahmini Süre:45s

Soru 8Soru →

Two resistors with values of $12\ \Omega$ and $6\ \Omega$ are wired in parallel across a circuit board. What is the total equivalent resistance of this combination?

18\ \Omega

4\ \Omega

9\ \Omega

72\ \Omega

Cevabı ve açıklamayı göster

Cevap: $4\ \Omega$

Cevap

4\ \Omega

For two resistors in parallel, the equivalent resistance is calculated by dividing their product by their sum:

\frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega

. In parallel circuits, the equivalent resistance is always lower than any single branch resistance.

Adım Adım Çözüm

1

Identify the given branch resistances and select the parallel equivalent resistance formula.

R_1 = 12\ \Omega

,

R_2 = 6\ \Omega

, and

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

The product-over-sum formula determines the total equivalent resistance of two parallel branches.

2

Calculate the product and sum of the two resistors, then divide.

R_{eq} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\ \Omega

Completing the arithmetic yields the equivalent resistance, which must be less than the smallest individual branch.

Anahtar Kavram

Total equivalent resistance in a parallel circuit is always less than the lowest individual branch resistance and can be calculated using the reciprocal or product-over-sum formula.

Tahmini Süre:45s

Soru 9Soru →

A technician is evaluating the electrical characteristics of a direct-current parallel circuit network. Match each circuit parameter on the left with its corresponding behavior or rule on the right.

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Cevap

Branch Voltage matches 'Remains constant and equal across all connected branches'; Total Current matches 'Equals the sum of the individual currents flowing through all branches'; Equivalent Resistance matches 'Is always lower than the resistance of the smallest branch resistor'; Branch Current matches 'Varies inversely with the branch resistance according to

I = \frac{V}{R}

'.

In parallel circuits, all components share common connection points, maintaining equal voltage across each branch. Total current equals the sum of all individual branch currents. Total equivalent resistance decreases as more branches are added (remaining less than the smallest branch resistance). Individual branch current is calculated via Ohm's Law as voltage divided by that branch's resistance.

Adım Adım Çözüm

1

Determine the voltage relationship across parallel branches.

Branch Voltage (

V_{\text{branch}}

) is equal across all branches (

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

).

All branches connect across the same two common nodes to the power supply.

2

Determine the total current relationship for a parallel network.

Total Current (

I_{\text{total}}

) equals the sum of branch currents (

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

).

Current splits at junction nodes according to Kirchhoff's Current Law.

3

Determine how total equivalent resistance behaves in a parallel circuit.

Equivalent Resistance (

R_{\text{eq}}

) is always smaller than the smallest branch resistance.

Adding parallel branches adds additional conductive paths for electron flow.

4

Determine how current through a specific branch is calculated.

Branch Current (

I_{\text{branch}}

) is calculated using Ohm's Law (

I_{\text{branch}} = \frac{V}{R_{\text{branch}}}

).

Branch current depends on the common branch voltage divided by the individual branch resistance.

Anahtar Kavram

Fundamental rules governing voltage, current, and resistance in DC parallel circuits.

Tahmini Süre:45s

Soru 10Soru →

A direct-current power distribution box supplies power to two parallel branches. The first branch has a resistance of $20\ \Omega$ , and the second branch has a resistance of $5\ \Omega$ . What is the total equivalent resistance of the network?

25\ \Omega

12.5\ \Omega

4\ \Omega

100\ \Omega

Cevabı ve açıklamayı göster

Cevap: $4\ \Omega$

Cevap

4\ \Omega

In a parallel circuit with two branches, the equivalent resistance is calculated using the product-over-sum formula

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

. Substituting

20\ \Omega

and

5\ \Omega

gives

\frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega

. Total resistance in parallel is always less than the lowest individual resistor value.

Adım Adım Çözüm

1

Identify the formula for the equivalent resistance of two parallel resistors.

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

or

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

Parallel branches provide multiple paths for current, resulting in a total resistance that is less than any individual branch resistance.

2

Substitute the given values (

R_1 = 20\ \Omega

and

R_2 = 5\ \Omega

) into the product-over-sum formula.

R_{\text{eq}} = \frac{20 \times 5}{20 + 5} = \frac{100}{25}

Simplifying the numerator and denominator allows direct calculation of equivalent resistance.

3

Divide the product by the sum.

R_{\text{eq}} = 4\ \Omega

Computing

\frac{100}{25}

yields the final equivalent resistance.

Anahtar Kavram

Total Equivalent Resistance in Parallel Circuits

Tahmini Süre:45s

Soru 11Soru →

A direct-current marine radio circuit contains two resistors connected in parallel: a $10\ \Omega$ resistor and a $40\ \Omega$ resistor. What is the total equivalent resistance of this parallel circuit, in ohms?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8 Ω

For two resistors in parallel with values of

10\ \Omega

and

40\ \Omega

, the total equivalent resistance is calculated using the product-over-sum formula:

R_{eq} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{10 \times 40}{10 + 40} = \frac{400}{50} = 8\ \Omega

. In any parallel circuit, the total equivalent resistance is always smaller than the smallest branch resistor.

Adım Adım Çözüm

1

Select the appropriate formula for calculating the equivalent resistance of two parallel resistors.

R_eq = (R_1 * R_2) / (R_1 + R_2) or 1/R_eq = 1/R_1 + 1/R_2

In parallel circuits, total resistance is found by taking the reciprocal of the sum of the reciprocals, which simplifies to the product-over-sum rule for two branches.

2

Substitute the given resistor values (10 Ω and 40 Ω) into the product-over-sum formula.

R_eq = (10 * 40) / (10 + 40) = 400 / 50

Multiplying the branch values gives the numerator, and adding them gives the denominator.

3

Divide the numerator by the denominator to find the equivalent resistance.

R_eq = 8 Ω

400 divided by 50 equals 8, which is less than the smallest branch resistance (10 Ω), confirming the validity of the parallel calculation.

Anahtar Kavram

Parallel Circuit Equivalent Resistance

Tahmini Süre:45s

Soru 12Soru →

An aircraft instrument panel contains three identical $12\ \Omega$ indicator lights connected in parallel across a direct-current power bus. What is the total equivalent resistance of this parallel circuit?

36\ \Omega

12\ \Omega

4\ \Omega

3\ \Omega

Cevabı ve açıklamayı göster

Cevap: $4\ \Omega$

Cevap

4\ \Omega

For parallel circuits containing equal resistors, the total equivalent resistance is found by dividing the individual resistance value by the number of parallel branches (

R_{\text{total}} = R / N

). Here,

12\ \Omega / 3 = 4\ \Omega

.

Adım Adım Çözüm

1

Identify the number of parallel branches and branch resistance

Branch resistance

R = 12\ \Omega

, number of branches

N = 3

When all resistors in parallel have identical resistance values, the equal-resistor shortcut formula can be directly applied.

2

Calculate total equivalent resistance using

R_{\text{total}} = \frac{R}{N}

R_{\text{total}} = \frac{12\ \Omega}{3} = 4\ \Omega

Connecting resistors in parallel provides multiple parallel paths for current, reducing the total equivalent resistance below any individual branch.

Anahtar Kavram

Equal-Resistor Parallel Shortcut Rule (

R_{\text{total}} = R / N

)

Tahmini Süre:45s

Soru 13Soru →

A technician is troubleshooting a direct-current parallel lighting system. Match each circuit modification on the left with its direct electrical effect on the system on the right.

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Cevap

The circuit modifications match their electrical effects as follows: adding an additional branch decreases total resistance and increases total supply current; disconnecting one branch increases total resistance while allowing remaining branches to operate normally; increasing resistance of a single branch reduces current only through that branch; creating a short circuit across one branch bypasses all branches and causes a dangerous surge in total current.

Each pair reflects core parallel circuit behavior: adding branches lowers equivalent resistance and raises total current; disconnecting a branch raises equivalent resistance without interrupting other branches; changing one branch resistance alters only that branch's current; and shorting across one branch shorts the entire common bus.

Adım Adım Çözüm

1

Determine the impact of adding a parallel branch

Total equivalent resistance drops and total supply current rises (

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

).

Adding a parallel branch creates an extra pathway for current flow.

2

Determine the impact of opening or disconnecting an active branch

Total equivalent resistance increases, but unaffected branches stay powered.

Removing a branch eliminates a current path while full source voltage remains across remaining branches.

3

Determine the impact of increasing a single branch's resistance

Current through that specific branch decreases (

I = V/R

) without changing the other branches.

Branch voltages in a parallel network are independent and equal to source voltage.

4

Determine the impact of short-circuiting one branch

The entire power source is shorted, bypassing all parallel loads.

All parallel branches connect directly to the same common positive and negative supply nodes.

Anahtar Kavram

Parallel Circuit Branch Behavior and Independence

Tahmini Süre:45s

Soru 14Soru →

A $60\text{ V}$ direct-current power supply feeds a circuit containing three parallel branches. The total current drawn from the supply is $10\text{ A}$ . If the first two branches have resistances of $20\ \Omega$ and $30\ \Omega$ , what is the resistance of the third branch?

6\ \Omega

12\ \Omega

18\ \Omega

50\ \Omega

Cevabı ve açıklamayı göster

Cevap: $12\ \Omega$

Cevap

The resistance of the third branch is

12\ \Omega

.

Because the circuit is wired in parallel, the full supply voltage of

60\text{ V}

is applied across each branch. Using Ohm's Law (

I = V / R

), the first branch draws

60\text{ V} / 20\ \Omega = 3\text{ A}

and the second branch draws

60\text{ V} / 30\ \Omega = 2\text{ A}

. The sum of these two branch currents is

5\text{ A}

. Since the total source current is

10\text{ A}

, the third branch must carry the remaining

10\text{ A} - 5\text{ A} = 5\text{ A}

. Applying Ohm's Law to the third branch yields

R_3 = 60\text{ V} / 5\text{ A} = 12\ \Omega

.

Adım Adım Çözüm

1

Determine the voltage across each parallel branch.

Each branch receives the full source voltage of

60\text{ V}

.

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

2

Calculate the individual currents flowing through the first two known branches using Ohm's Law (

I = V / R

).

I_1 = 60\text{ V} / 20\ \Omega = 3\text{ A}

and

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

.

Branch currents must be determined to find how much of the total current is accounted for by the known branches.

3

Find the current flowing through the third branch by subtracting the known branch currents from the total supply current (

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

).

I_3 = 10\text{ A} - (3\text{ A} + 2\text{ A}) = 5\text{ A}

.

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

4

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

R_3 = V / I_3

).

R_3 = 60\text{ V} / 5\text{ A} = 12\ \Omega

.

Applying Ohm's Law to the third branch gives its required resistance value.

Anahtar Kavram

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

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

), and each branch resistance satisfies Ohm's Law (

R_n = V / I_n

).

Tahmini Süre:1m 30s

Soru 15Soru →

A $48\text{ V}$ direct-current telecommunications power bus supplies electricity to three parallel branches. The first branch draws a current of $2\text{ A}$ , the second branch contains a load with a resistance of $24\ \Omega$ , and the total current supplied by the power bus is $7\text{ A}$ . What is the electrical resistance of the third branch, in ohms?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The resistance of the third branch is

16\ \Omega

.

Because the circuit is parallel, the full supply voltage of

48\text{ V}

is present across every branch. The second branch with

24\ \Omega

draws

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

. Subtracting the known currents of Branch 1 (

2\text{ A}

) and Branch 2 (

2\text{ A}

) from the total current of

7\text{ A}

leaves

3\text{ A}

flowing through Branch 3. Using Ohm's Law,

R_3 = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

.

Adım Adım Çözüm

1

Calculate the current flowing through Branch 2

I_2 = \frac{V}{R_2} = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

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

V_1 = V_2 = V_3 = 48\text{ V}

).

2

Calculate the current flowing through Branch 3

I_3 = I_{\text{total}} - I_1 - I_2 = 7\text{ A} - 2\text{ A} - 2\text{ A} = 3\text{ A}

The total current entering a parallel network equals the sum of the individual branch currents (

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

).

3

Calculate the resistance of Branch 3

R_3 = \frac{V}{I_3} = \frac{48\text{ V}}{3\text{ A}} = 16\ \Omega

Applying Ohm's Law (

R = \frac{V}{I}

) using the common

48\text{ V}

supply voltage and the calculated

3\text{ A}

branch current yields the branch resistance.

Anahtar Kavram

In a parallel DC circuit, voltage remains constant across all branches, total current is the algebraic sum of individual branch currents, and branch resistance is determined by Ohm's Law (

R = \frac{V}{I}

).

Soru 16Soru →

An industrial control panel utilizes a direct-current bus to power three parallel branches. Branch 1 contains a $15\ \Omega$ resistor, Branch 2 contains a $30\ \Omega$ resistor, and Branch 3 contains a $10\ \Omega$ resistor. If an ammeter connected in Branch 2 measures a current of $3\text{ A}$ , what is the total current delivered by the power supply to the entire circuit?

4.5\text{ A}

6.0\text{ A}

9.0\text{ A}

18.0\text{ A}

Cevabı ve açıklamayı göster

Cevap: $18.0\text{ A}$

Cevap

18.0\text{ A}

The supply voltage across the parallel circuit is constant. Using Branch 2,

V = 3\text{ A} \times 30\ \Omega = 90\text{ V}

. The remaining branch currents are

I_1 = 90\text{ V} / 15\ \Omega = 6\text{ A}

and

I_3 = 90\text{ V} / 10\ \Omega = 9\text{ A}

. Summing all three branch currents yields

I_{\text{total}} = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

.

Adım Adım Çözüm

1

Calculate the common supply voltage across the parallel circuit using Ohm's Law on Branch 2.

V = I_2 \times R_2 = 3\text{ A} \times 30\ \Omega = 90\text{ V}

In a parallel circuit, the voltage drop across every individual branch is equal to the total supply voltage.

2

Determine the current flowing through Branch 1 and Branch 3 using the common voltage of

90\text{ V}

.

I_1 = \frac{90\text{ V}}{15\ \Omega} = 6\text{ A}

and

I_3 = \frac{90\text{ V}}{10\ \Omega} = 9\text{ A}

Current in each individual branch is determined by dividing the constant branch voltage by that branch's resistance.

3

Apply Kirchhoff's Current Law to sum the currents of all parallel branches.

I_{\text{total}} = I_1 + I_2 + I_3 = 6\text{ A} + 3\text{ A} + 9\text{ A} = 18\text{ A}

The total current leaving the power supply equals the sum of the individual branch currents in a parallel circuit.

Anahtar Kavram

In a parallel circuit, voltage remains constant across all branches (

V_{\text{total}} = V_1 = V_2 = V_3

), and total circuit current is the sum of all individual branch currents (

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

).

Alternatif Yöntem

Calculate total equivalent resistance first:

\frac{1}{R_{\text{eq}}} = \frac{1}{15\ \Omega} + \frac{1}{30\ \Omega} + \frac{1}{10\ \Omega} = \frac{2 + 1 + 3}{30} = \frac{6}{30} = \frac{1}{5\ \Omega}

, giving

R_{\text{eq}} = 5\ \Omega

. With

V = 90\text{ V}

, total current is

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{90\text{ V}}{5\ \Omega} = 18\text{ A}

.

Tahmini Süre:1m 30s

Soru 17Soru →

An industrial control panel is powered by a $36\text{ V}$ DC source. The main circuit splits into three parallel branches. Branch one powers a cooling unit that dissipates $108\text{ W}$ of power. Branch two contains an $18\ \Omega$ limiting resistor. Branch three contains a sensor module with an unknown resistance. If a technician measures the total equivalent resistance of this entire parallel network to be exactly $4\ \Omega$ , what is the resistance of the sensor module?

4\ \Omega

9\ \Omega

12\ \Omega

18\ \Omega

Cevabı ve açıklamayı göster

Cevap: $9\ \Omega$

Cevap

The resistance of the sensor module in the third branch is

9\ \Omega

.

To find the resistance of a single parallel branch, you need the voltage across it and the current through it. The voltage is the source voltage (

36\text{ V}

). The total current is

9\text{ A}

(

36\text{ V} / 4\ \Omega

). Subtracting the current from the first branch (

108\text{ W} / 36\text{ V} = 3\text{ A}

) and the second branch (

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

) leaves

4\text{ A}

flowing through the third branch. Using Ohm's Law (

R = V/I

),

36\text{ V} / 4\text{ A}

equals

9\ \Omega

.

Adım Adım Çözüm

1

Calculate the total current (

I_{total}

) drawn by the entire circuit.

I_{total} = \frac{V}{R_{total}} = \frac{36\text{ V}}{4\ \Omega} = 9\text{ A}

Ohm's law applies to the circuit as a whole just as it does to individual components.

2

Calculate the current flowing through Branch 1 (

I_1

).

I_1 = \frac{P_1}{V} = \frac{108\text{ W}}{36\text{ V}} = 3\text{ A}

In a parallel circuit, the voltage across each branch is equal to the source voltage. We use the power formula (

P = V \times I

) rearranged to solve for current.

3

Calculate the current flowing through Branch 2 (

I_2

).

I_2 = \frac{V}{R_2} = \frac{36\text{ V}}{18\ \Omega} = 2\text{ A}

Using Ohm's law (

I = V/R

) with the known voltage and resistance for this specific branch.

4

Determine the current flowing through the unknown Branch 3 (

I_3

).

I_3 = I_{total} - (I_1 + I_2) = 9\text{ A} - (3\text{ A} + 2\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current entering a parallel network must equal the sum of the currents in individual branches.

5

Calculate the resistance of Branch 3 (

R_3

).

R_3 = \frac{V}{I_3} = \frac{36\text{ V}}{4\text{ A}} = 9\ \Omega

Applying Ohm's law once more to the specific values of the third branch.

Anahtar Kavram

Parallel Circuit Current and Power Laws

Soru 18Soru →

An aircraft auxiliary power bus supplies a constant $24\text{ V}$ DC across three parallel branch circuits. Branch 1 draws a current of $3\text{ A}$ , Branch 2 contains an unknown resistor $R_2$ , and Branch 3 contains a $12\ \Omega$ resistor. If the total electrical power consumed by all three branches combined is $192\text{ W}$ , what is the resistance of resistor $R_2$ ?

4\ \Omega

8\ \Omega

6\ \Omega

3\ \Omega

Cevabı ve açıklamayı göster

Cevap: $8\ \Omega$

Cevap

The resistance of resistor

R_2

is

8\ \Omega

.

The correct resistance is

8\ \Omega

. First, finding total current from the total power gives

I_{\text{total}} = \frac{192\text{ W}}{24\text{ V}} = 8\text{ A}

. Since the voltage across each parallel branch is

24\text{ V}

, Branch 3 draws

I_3 = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}

. Subtracting the known branch currents (

3\text{ A}

and

2\text{ A}

) from the total current (

8\text{ A}

) leaves

3\text{ A}

flowing through Branch 2. Applying Ohm's Law yields

R_2 = \frac{24\text{ V}}{3\text{ A}} = 8\ \Omega

.

Adım Adım Çözüm

1

Calculate the total circuit current using total power and supply voltage.

I_{\text{total}} = \frac{P_{\text{total}}}{V} = \frac{192\text{ W}}{24\text{ V}} = 8\text{ A}

In any DC circuit, total power is the product of total voltage and total current (

P = V \times I

).

2

Calculate the current drawn by Branch 3 using Ohm's Law.

I_3 = \frac{V}{R_3} = \frac{24\text{ V}}{12\ \Omega} = 2\text{ A}

In a parallel circuit, full supply voltage (

24\text{ V}

) appears across every parallel branch.

3

Determine the current flowing through Branch 2 using Kirchhoff's Current Law for parallel circuits.

I_2 = I_{\text{total}} - I_1 - I_3 = 8\text{ A} - 3\text{ A} - 2\text{ A} = 3\text{ A}

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

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

).

4

Calculate the resistance of

R_2

using Ohm's Law.

R_2 = \frac{V}{I_2} = \frac{24\text{ V}}{3\text{ A}} = 8\ \Omega

Branch resistance is found by dividing the branch voltage by the branch current (

R = \frac{V}{I}

).

Anahtar Kavram

Parallel circuit rules: constant branch voltage (

V_{\text{total}} = V_1 = V_2 = V_3

), additive branch currents (

I_{\text{total}} = \sum I_n

), and power relationships (

P = V \times I

).

Alternatif Yöntem

Calculate individual branch powers directly: Branch 1 consumes

P_1 = V \times I_1 = 24\text{ V} \times 3\text{ A} = 72\text{ W}

. Branch 3 consumes

P_3 = \frac{V^2}{R_3} = \frac{576}{12} = 48\text{ W}

. The power dissipated in Branch 2 is

P_2 = P_{\text{total}} - P_1 - P_3 = 192\text{ W} - 72\text{ W} - 48\text{ W} = 72\text{ W}

. Using the power-resistance formula,

R_2 = \frac{V^2}{P_2} = \frac{576}{72} = 8\ \Omega

.

Tahmini Süre:1m 30s

Soru 19Soru →

A mobile radar subsystem is powered by a $24\text{ V}$ direct-current source. Its internal circuitry contains a parallel configuration with three distinct branches.

The total equivalent resistance of this parallel section is exactly $3\ \Omega$ .
A technician measures the current through the first branch to be $3\text{ A}$ .
The second branch has a known resistance of $24\ \Omega$ .

What is the resistance of the third branch in ohms?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The resistance of the third branch is

6\ \Omega

.

The correct value is derived by finding the total circuit current (

8\text{ A}

) and subtracting the known branch currents (

3\text{ A}

and

1\text{ A}

) to find the current in the third branch (

4\text{ A}

). Dividing the common

24\text{ V}

source by this

4\text{ A}

current yields

6\ \Omega

.

Adım Adım Çözüm

1

Calculate the total circuit current using the total voltage and equivalent resistance.

I_{total} = 24\text{ V} / 3\ \Omega = 8\text{ A}

Ohm's law applies to the entire circuit, linking total voltage, total current, and equivalent resistance.

2

Determine the current flowing through the second branch.

I_2 = 24\text{ V} / 24\ \Omega = 1\text{ A}

In a parallel circuit, the full source voltage drops across every branch. Ohm's law (

I = V/R

) gives the branch current.

3

Calculate the current flowing through the third branch.

I_3 = 8\text{ A} - (3\text{ A} + 1\text{ A}) = 4\text{ A}

Kirchhoff's Current Law states that the total current is the sum of the individual branch currents (

I_{total} = I_1 + I_2 + I_3

).

4

Calculate the resistance of the third branch.

R_3 = 24\text{ V} / 4\text{ A} = 6\ \Omega

Apply Ohm's law (

R = V/I

) specifically to the third branch using the source voltage and the calculated branch current.

Anahtar Kavram

Applying Ohm's law to both individual branches and the entire circuit, combined with parallel circuit rules for voltage and current.

Alternatif Yöntem

Use the parallel resistance reciprocal formula:

1/R_{eq} = 1/R_1 + 1/R_2 + 1/R_3

. First, find

R_1 = V/I_1 = 24\text{ V} / 3\text{ A} = 8\ \Omega

. Then, substitute the known values:

1/3 = 1/8 + 1/24 + 1/R_3

. Finding a common denominator of 24 gives

8/24 = 3/24 + 1/24 + 1/R_3

, which simplifies to

8/24 = 4/24 + 1/R_3

. Therefore,

1/R_3 = 4/24 = 1/6

, meaning

R_3 = 6\ \Omega

.

Tahmini Süre:1m 30s

Soru 20Soru →

A direct-current power supply in an automotive workshop delivers $24\text{ V}$ across two parallel branches. The first branch contains an $8\ \Omega$ inspection lamp, and the second branch contains a $24\ \Omega$ ventilation motor. What is the total current drawn by the circuit from the power supply?

0.75\text{ A}

4\text{ A}

3\text{ A}

6\text{ A}

Cevabı ve açıklamayı göster

Cevap: $4\text{ A}$

Cevap

4\text{ A}

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

24\text{ V}

). The current in each branch is calculated using Ohm's Law: the

8\ \Omega

load draws

3\text{ A}

(

24\text{ V} / 8\ \Omega

), and the

24\ \Omega

load draws

1\text{ A}

(

24\text{ V} / 24\ \Omega

). Summing these branch currents gives a total supply current of

4\text{ A}

.

Adım Adım Çözüm

1

Calculate the current flowing through the first branch (

8\ \Omega

resistor).

I_1 = \frac{V}{R_1} = \frac{24\text{ V}}{8\ \Omega} = 3\text{ A}

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

24\text{ V}

.

2

Calculate the current flowing through the second branch (

24\ \Omega

resistor).

I_2 = \frac{V}{R_2} = \frac{24\text{ V}}{24\ \Omega} = 1\text{ A}

Ohm's Law applies independently to the second branch using the same

24\text{ V}

potential.

3

Sum the individual branch currents to find the total circuit current.

I_{\text{total}} = I_1 + I_2 = 3\text{ A} + 1\text{ A} = 4\text{ A}

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

Anahtar Kavram

In a parallel circuit, total current is the sum of the individual branch currents, each determined by dividing the constant source voltage by the branch resistance.

Alternatif Yöntem

Find the equivalent parallel resistance first:

R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{8 \times 24}{8 + 24} = \frac{192}{32} = 6\ \Omega

. Then apply Ohm's Law to the whole circuit:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{24\text{ V}}{6\ \Omega} = 4\text{ A}

.

Tahmini Süre:1m 0s

Parallel Circuits

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