Network Elements MCQ Quiz - Objective Question with Answer for Network Elements - Download Free PDF

Explanation:

Factors Determining the Rating of a Resistor

Definition: The rating of a resistor is a critical parameter that defines the maximum amount of electrical power it can dissipate without being damaged. This rating is essential for ensuring the reliability and longevity of the resistor in various electrical and electronic circuits.

Correct Option:

The correct option is:

Power dissipation capacity

This factor is primarily used to determine the rating of a resistor. The power dissipation capacity of a resistor indicates the maximum power it can handle before it overheats and potentially fails. This is calculated using the formula P = V²/R, where P is the power in watts, V is the voltage across the resistor, and R is the resistance in ohms. The power rating is usually specified in watts (W) and is a crucial parameter when selecting a resistor for a particular application.

Additional Information

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

Option 1: Material used for construction

While the material used for constructing a resistor is important for determining its properties, such as temperature stability and resistance value, it is not the primary factor for determining the resistor's rating. The material affects characteristics like tolerance and temperature coefficient but does not directly define the power dissipation capacity.

Option 3: Temperature coefficient

The temperature coefficient of a resistor indicates how its resistance changes with temperature. While this is an important parameter for precision applications, it is not the primary factor for determining the power rating. The power dissipation capacity is more directly related to the resistor's ability to handle electrical power without overheating.

Option 4: Colour code

The colour code on a resistor is a method of indicating its resistance value and tolerance. It does not provide information about the power rating. The colour code is a useful tool for identifying resistors quickly, but it does not determine the maximum power dissipation capacity.

Conclusion:

Understanding the various factors that influence the rating of a resistor is essential for selecting the appropriate component for a given application. The power dissipation capacity is the primary factor used to determine the rating of a resistor, as it defines the maximum power the resistor can handle without being damaged. Other factors, such as the material used for construction, temperature coefficient, and colour code, provide additional information about the resistor's characteristics but do not directly determine its rating. By focusing on the power dissipation capacity, engineers and designers can ensure the reliable and safe operation of resistors in their circuits.

Solution:

To solve this problem, we need to use Kirchhoff's Voltage Law (KVL) and the concept of series circuits. Let's go through the detailed steps to find the voltage drop across R2.

Step 1: Understanding Series Circuits

In a series circuit, the current flowing through each component is the same, but the voltage drop across each component can be different. The total resistance in a series circuit is the sum of the individual resistances.

Given:

• R1 = 5 Ω
• R2 = 10 Ω
• V (total voltage) = 15 V

Step 2: Calculate the Total Resistance

The total resistance (Rtotal) in a series circuit is the sum of the resistances of the individual resistors:

R_total = R₁ + R₂

Rtotal = 5 Ω + 10 Ω = 15 Ω

Step 3: Calculate the Total Current

Using Ohm's Law, we can calculate the total current (I) flowing through the circuit:

V = I × Rtotal 

15 V = I × 15 Ω

I = 15 V / 15 Ω

I = 1 A

Step 4: Calculate the Voltage Drop Across R2

Now that we have the current flowing through the circuit, we can calculate the voltage drop across R2 using Ohm's Law:

V_R2 = I × R₂

VR2 = 1 A × 10 Ω

VR2  = 10 V

Concept

The RMS value of an alternating current is the steady (D.C) current that produces the same amount of heat as the A.C. 

The RMS value of any signal is given by:

\(RMS=\sqrt{{1\over T}\int_{-\infty}^{\infty}x^2(t)\space dt}\)

where, T = Time period of the signal

Explanation

The RMS value of a purely resistive AC circuit is given by:

\(\rm V_{rrms}=\frac{V_m}{\sqrt2}\)

where, V_m = Peak value of the signal

Concept

The self-inductance of a coil is given by:

\(L={N\times ϕ\over I}\)

where, L = Inductance

N = No. of turns

ϕ = Flux

l  = Current

So, self-inductance is directly proportional to both magnetic flux and the number of turns in the coil.

The correct answer is option 4.

Function of the Armature Core in Magnetic Circuits

Magnetic Path Completion:

• The armature core is a key component in the magnetic circuit of machines like motors and generators.
• It provides a low-reluctance path for magnetic flux to travel between the field poles and yoke, completing the magnetic loop.
   

Efficient Magnetic Flux Conduction:

• Made of laminated soft iron or silicon steel to reduce eddy current losses.
• It efficiently conducts the alternating magnetic flux without significant energy losses.
   

Supports Induced EMF:

• As the armature rotates within the magnetic field, the armature windings (placed in slots on the core) cut through magnetic lines of force, inducing an electromotive force (EMF).
   

Mechanical Structure:

• It also serves as a mechanical support for the conductors/windings.

Ohm’s law: Ohm’s law states that at a constant temperature, the current through a conductor between two points is directly proportional to the voltage across the two points.

Voltage = Current × Resistance

V = I × R

V = voltage, I = current and R = resistance

The SI unit of resistance is ohms and is denoted by Ω.

It helps to calculate the power, efficiency, current, voltage, and resistance of an element of an electrical circuit.

Limitations of ohms law:

• Ohm’s law is not applicable to unilateral networks. Unilateral networks allow the current to flow in one direction. Such types of networks consist of elements like a diode, transistor, etc.
• Ohm’s law is also not applicable to non – linear elements. Non-linear elements are those which do not have current exactly proportional to the applied voltage that means the resistance value of those elements’ changes for different values of voltage and current. An example of a non-linear element is thyristor.
• Ohm’s law is also not applicable to vacuum tubes.

Concept:

Ideal voltage source: An ideal voltage source have zero internal resistance.

Practical voltage source: A practical voltage source consists of an ideal voltage source (VS) in series with internal resistance (RS) as follows.

An ideal voltage source and a practical voltage source can be represented as shown in the figure.

Ideal current source: An ideal current source has infinite resistance. Infinite resistance is equivalent to zero conductance. So, an ideal current source has zero conductance.

Practical current source: A practical current source is equivalent to an ideal current source in parallel with high resistance or low conductance.

Ideal and practical current sources are represented as shown in the below figure.

• When an ideal voltage source and an ideal current source in series, the combination has an ideal current sources property.
• Current in the circuit is independent of any element connected in series to it.

Explanation:

In a series circuit, the current flows through all the elements is the same. Thus, any element connected in series with an ideal current source is redundant and it is equivalent to an ideal current source only.

In a parallel circuit, the voltage across all the elements is the same. Thus, any element connected in parallel with an ideal voltage source is redundant and it is equivalent to an ideal voltage source only.

Concept:

When resistances are connected in parallel, the equivalent resistance is given by

\(\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \ldots + \frac{1}{{{R_n}}}\)

When resistances are connected in series, the equivalent resistance is given by

\({R_{eq}} = {R_1} + {R_2} + \ldots + {R_n}\)

Calculation:

Given that R₁ = R₂ = R₃ = 6 Ω and all are connected in parallel.

\(\frac{1}{{{R_{eq}}}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6}\)

⇒ R_eq = 2 Ω

When capacitors are connected in series across DC voltage:

• The charge of each capacitor is the same and the same current flows through each capacitor in the given time.
• The voltage across each capacitor is dependent on the capacitor value.

When capacitors are connected in parallel across DC voltage:

• The charge of each capacitor is different and the current flows through each capacitor in the given time are also different and depend on the value of the capacitor.
• The voltage across each capacitor is the same.

The circuit after removing the voltage source

The total resistance of the new circuit will be the equivalent resistance of the network.

R_eq = R_t = 3 + 2 + 2 = 7 Ω 

The equivalent resistance of the network is 7 Ω.

 Mistake PointsWhile finding the equivalent resistance of the network, don't consider the internal resistance of the voltage source. Please read the question carefully it is mentioned in the question as well.

There are two kinds of voltage or current sources:

Independent Source: It is an active element that provides a specified voltage or current that is completely independent of other circuit variables.

Dependent Source: It is an active element in which the source quantity is controlled by another voltage or current in the circuit.

Concept:

The resistance of conductor changes when the temperature of that conductor changes.

New resistance is given by:

\({{R}_{t}}={{R}_{0}}\left( 1+\alpha \text{ }\!\!\Delta\!\!\text{ }T \right)\)

Where R_t = the resistance of the conductor after temperature changes

R₀ = the resistance of the conductor before temperature changes

α = temperature coefficient

ΔT = final temperature – initial temperature

Calculation:

R₀ = ?

α = 0.00125/°C

T₁ = 300 k = 300 - 273 = 27°C

T₂ = 1100 k = 1100 – 273 = 827°C

Resistance at T₁ =27°C

R_27°C =  R0  {1+ (0.00125 × 27)}

R0  = 1 / {1+ (0.00125 × 27)}  

R₀ = 0.967

Now at T₂  =  827 *C

R = 0.967 * {(1+ 0.00125 × 827)

R = 1.967 ohms 

Here the nearest option is 2ohm.

Concept-

The dimensional formula is defined as the expression of the physical quantity in terms of mass, length, time and ampere.

Explanation-

Power – It is defined as rate of doing work.

 \(\therefore P = \frac{W}{t}\)

Where, P = power, W = work done and t = time.

Now,

Dimensional formula of work (W) = [ML²T^-2]

Dimensional formula of time (t) = [T¹]

\(P = \frac{{M{L^2}{T^{ - 2}}}}{{{T^1}}} = \frac{{M{L^2}}}{{{T^3}}}\)

∴ The dimensional formula of power P is [ML²T^-3].

Concept:

Electric current: If the electric charge flows through a conductor, we say that there is an electric current in the conductor.

If Q charge flow through the conductor for ‘t’ seconds, then the current given by that conductor is \(I=\frac{Q}{t}\)

Q = I × t

I = current

t = times

Calculation:

Given I = 5 amp

t = 3 min = 180 sec

Q = I × t