Series and Parallel Connection of Resistance MCQ Quiz - Objective Question with Answer for Series and Parallel Connection of Resistance - Download Free PDF

Explanation:

Equivalent Resistance of Resistors in Parallel

Definition: When resistors are connected in parallel, the total or equivalent resistance of the network is less than the smallest resistor in the parallel arrangement. The formula for calculating the equivalent resistance (R_eq) of resistors connected in parallel is:

Formula:

Here, R₁, R₂, R₃, ..., R_n are the individual resistances of the resistors in the parallel network.

Problem Analysis:

In this problem, we are given:

• 5 resistors, each with a resistance of 10 Ω.
• These resistors are connected in parallel.

Step-by-Step Calculation:

Step 1: Write the formula for equivalent resistance in parallel:

Step 2: Substitute the values of resistors:

Since all resistors have the same resistance (10 Ω), we can simplify the formula:

Step 3: Add the reciprocals:

Step 4: Simplify the fraction:

Step 5: Take the reciprocal to find R_eq:

Final Answer:

The equivalent resistance of the network is 2 Ω.

Advantages of Parallel Connection:

• In parallel circuits, the total resistance is reduced, which allows more current to flow through the circuit.
• If one resistor fails, the other resistors continue to function, ensuring that the circuit remains operational.

Correct Option: The correct answer is Option 4, which states that the equivalent resistance is 2 Ω.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 10 Ω

This option is incorrect because the equivalent resistance of resistors in parallel is always less than the smallest individual resistance in the network. Since all resistors have a resistance of 10 Ω, the equivalent resistance must be less than 10 Ω.

Option 2: 50 Ω

This option is incorrect because 50 Ω is the sum of the resistances if they were connected in series. In a parallel connection, the resistances combine differently, resulting in a much smaller equivalent resistance.

Option 3: 500 Ω

This option is incorrect as it is significantly larger than the resistance values of individual resistors. The equivalent resistance in parallel is always less than the smallest resistor in the network, and 500 Ω is not plausible in this case.

Option 4: 2 Ω

This is the correct answer, as calculated above.

Conclusion:

Understanding how resistors combine in parallel circuits is essential for analyzing electrical networks. In this problem, the resistors combine to produce an equivalent resistance of 2 Ω, which is less than the resistance of any individual resistor in the network. This reduction in resistance is a key characteristic of parallel circuits, making them beneficial in applications requiring lower resistance and higher current flow.

Explanation:

Problem Statement:

When 'n' resistances, each of value 'r', are connected in parallel, the resultant resistance is 'x'. When these 'n' resistances are connected in series, the total resistance is to be determined.

Solution:

To solve this problem, we need to calculate the equivalent resistance when the resistances are connected in series. First, let’s analyze the situation for parallel and series connections.

Step 1: Equivalent Resistance in Parallel Connection

When 'n' resistances, each of value 'r', are connected in parallel, the formula for the resultant resistance 'x' is given by:

1/x = 1/r + 1/r + 1/r + ... (n terms)

Since all resistances have the same value 'r', the equation becomes:

1/x = n/r

Rearranging the terms, we get:

x = r/n

This means the equivalent resistance 'x' for the parallel connection is r/n.

Step 2: Equivalent Resistance in Series Connection

When the same 'n' resistances are connected in series, the total resistance is simply the sum of the individual resistances:

Total Resistance = r + r + r + ... (n terms)

This can be written as:

Total Resistance = n × r

Substituting the value of 'r' from the parallel connection equation, where r = x × n (from Step 1), we get:

Total Resistance = n × (x × n) = n² × x

Thus, the total resistance when 'n' resistances are connected in series is n² × x.

Final Answer:

The total resistance in series is n² × x. Therefore, the correct option is Option 1.

Additional Information

To further analyze the problem, let us evaluate the other options:

Option 2: x/n

This option suggests that the total resistance in series is x divided by n. However, this is incorrect because, in a series connection, the resistances add up linearly, and the correct formula is n² × x, not x/n.

Option 3: n²/x

This option suggests that the total resistance in series is n² divided by x. This is incorrect because the formula for series resistance involves multiplying n² with x, not dividing by x.

Option 4: n × x²

This option suggests that the total resistance in series is n multiplied by x². However, this is incorrect as there is no squaring of 'x' in the formula for total resistance in series.

Option 5: Not provided

This option does not contribute to the solution and is irrelevant in this context.

Conclusion:

The correct answer is Option 1: n² × x. This result aligns with the formula for calculating total resistance in a series connection, where the resistances add up linearly, and the relationship between the parallel and series configurations of the resistors is accurately established.

Concept

The current in an RL circuit is given by:

where, Z = Impedance

V = Voltage

I = Current

R = Resistance

ω = Angular frequency

Calculation

Given, V = 120∠30° 

X_L = ωL = 2πf = 120 Ω  

 = 134.60∠ 63.43° 

I = 0.895 ∠-33.43°A 

CONCEPT:

• Resistances in series combination: When two or more resistances are connected one after another such that the same current flows through them are called resistances in series.

The net resistance/equivalent resistance (R) of resistances in series is given by:

Equivalent resistance, R = R1 + R2

• Resistances in parallel combination: When two or more resistances are connected across different branches such that the potential drop across them are the same, they are said to be in parallel connection.

The net resistance/equivalent resistance (R) of resistances in parallel is given by:

Equivalent resistance (R): 

EXPLANATION:

Initially, N resistors were connected in parallel. The equivalent resistance is x.

⇒  

⇒     

The simplified equivalent circuit is:

10Ω and 15Ω are in parallel.

6Ω and 8Ω are in series.

R = 6+8 = 14 Ω 

When two equal resistances are connected in parallel, their equivalent resistance is always the half of resistance.

The equivalent resistance is given by

R_eq = 15/8 Ω

Current in the circuit 

Resistances in parallel combination: When two or more resistances are connected across different branches such that the potential drop across them are the same, they are said to be in parallel connection.

The net resistance/equivalent resistance (R) of resistances in parallel is given by:

Equivalent resistance (R): 

EXPLANATION:

If n resistance connected in parallel the,

Hence,

R_eq = 

Concept:

When resistors are connected in delta, the circuit as shown below.

Power consumed, 

When resistors are connected in star, the circuit as shown below.

By using star-delta conversion, the circuit becomes

Power consumed, 

Therefore, WΔ = 3WY

So that power consumed by three identical resistors in star configuration is 1/3 times of power consumption in delta configuration.

For  RA  

(RA)² + RA = 2 + 3RA

(RA)² - 2RA - 2  = 0 

On solving we'll get:

R_A = 1 ± √3 ----(1)

(RB)2 + 2RB - 2  = 0 

On solving we'll get:

R_B = -1 ± √3   ---(2)  

From the given equation

RA ≠ RB & RA , RB ≠ 0

also RA > RB 

Hence options (D) is correct

Concept:

To convert Star from Delta, we follow the following procedure:

Branch resistance = product of connected resistance/sum of all resistances

For branch resistance, PS, Ra, and Rb are connected, so:

For branch resistance QS, Ra and Rc are connected, so:

For branch resistance, SR, Rb, and Rc are connected, so:

Calculation:

Covert Delta Network to star Network we get

Covert star Network to delta Network we get

R_eq = 32/60 = 8/15 Ω 

Minimum Number of Resistance required for Series Connection = 2

Minimum Number of Resistance required for Parallel Connection = 2

Minimum Number of Resistance required for series-parallel Connection = 3

Wheatstone Bridge:

• A Wheatstone bridge is a special arrangement of 4 resistors. It can be used to find an unknown resistance.
• If the Wheatstone bridge is balanced, there will be no current flowing through the galvanometer.
• Wheatstone bridge is used to measure the resistance with the help of a comparison method.
• The Wheatstone bridge work on the principle of null deflection.

​The bridge is balanced when:

∴ A Wheatstone bridge is balanced if the ratio of resistors on one side of the bridge equals the ratio of resistors on the other side.

Resistance:

• The measurement of the opposition of the flow of electric current through a conductor is called the resistance of that conductor. It is denoted by R.

There are mainly two ways of the combination of resistances:

1. Resistances in series:

• When two or more resistances are connected one after another such that the same current flows through them are called resistances in series.
• The net resistance/equivalent resistance (R) of resistances in series is given by:

Equivalent resistance, R = R1 + R2

2. Resistances in parallel:

• When the terminals of two or more resistances are connected at the same two points and the potential difference across them is equal is called resistances in parallel.
• The net resistance/equivalent resistance(R) of resistances in parallel is given by:

Calculation:

Here opposite arms are balanced having a resistance of 24 Ω,

so the circuit is balanced therefore no current flows through 6Ω resistance(middle one).

The resistances 6 Ω and 4Ω are in series from two sides,

so the resultant series resistance is 6 + 4 = 10 Ω each.

This 10Ω resistance are in parallel to each other, 

so, the resultant parallel resistance is 10 ∥ 10 = 

Voltage source (V) = Current (I) × R_eq

= 5× 5 = 25 V

Concept:

1^st mesh

2^nd mesh

3^rd mesh

The simplified equivalent circuit is:

10Ω and 15Ω are in parallel.

6Ω and 8Ω are in series.

R = 6+8 = 14 Ω 

When two equal resistances are connected in parallel, their equivalent resistance is always the half of resistance.

Different Types of Connections are Series Connection and Parallel Connection.

Calculation:

Given, four 100 Ω resistors are connected in parallel.

The circuit can be drawn as,

Where, R = 100 Ω

From the above concept,

∴