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

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.

Concept

The capacitive reactance X_C for a series RC circuit is given by:

\(X_C={1\over 2\pi fC}\)

From the above expression, we found that the frequency is inversely proportional to the capacitive reactance.

So, if the frequency of a series R-C circuit is increased, than the capacitive reactance will decrease.

Concept:

Wire wound resistors are a type of precision resistor in which resistance is provided by a metal wire wound around an insulating core. The wire material must have high resistivity and stability over a wide range of temperatures.

Calculation

The most commonly used material for winding in wire wound resistors is a nickel-chromium alloy (such as Nichrome). This alloy offers high electrical resistance, good thermal stability, corrosion resistance, and mechanical strength, making it ideal for resistor applications.

Other materials like carbon are used in carbon composition resistors, while chromium cobalt and pure nickel are not commonly used for wire wound resistor windings.

Concept

The amount of charge stored in a capacitor is given by:

\(Q=CV\)

where, Q = Charge

C = Capacitance

V = Voltage

Calculation

Given, C = 5 μF

V = 12 volts

\(Q=5 \times 12=60\space \mu C\)

Explanation:

Inductive Coil and Rate of Change of Current

Definition: An inductive coil, often simply referred to as an inductor, is a passive electrical component that stores energy in its magnetic field when electrical current passes through it. It resists changes in the current flowing through it due to its property known as inductance.

Working Principle: When a voltage is applied across an inductor, it creates a time-varying magnetic field, resulting in an induced electromotive force (emf) that opposes the change in current. This phenomenon is described by Faraday's Law of Electromagnetic Induction and Lenz's Law.

Correct Option Analysis:

The correct option is:

Option 2: At the start of current flow.

This option accurately describes the behavior of an inductive coil when a voltage is first applied. At the initial moment of current flow, the rate of change of current is at its maximum. This can be understood by examining the fundamental principles of inductance and the behavior of an RL (resistor-inductor) circuit.

When a voltage \( V \) is applied to an inductive coil with inductance \( L \) and resistance \( R \), the current \( I(t) \) through the coil at any time \( t \) is given by the differential equation:

\( V = L \frac{dI(t)}{dt} + IR \)

At the moment the voltage is applied (\( t = 0 \)), the current \( I(0) \) is zero, and the rate of change of current \( \frac{dI(0)}{dt} \) is at its peak because the entire voltage is initially dropped across the inductor:

\( V = L \frac{dI(0)}{dt} \)

Thus, the maximum rate of change of current occurs at \( t = 0 \), immediately after the voltage is applied.

Additional Information

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

Option 1: After one time constant.

This option is incorrect. The time constant (\( \tau = \frac{L}{R} \)) of an RL circuit represents the time required for the current to reach approximately 63.2% of its maximum value. After one time constant, the rate of change of current is not at its maximum; it has significantly decreased as the current approaches its steady-state value.

Option 3: Near the final maximum value of current.

This option is also incorrect. As the current approaches its final steady-state value, the rate of change of current diminishes and approaches zero. At the maximum current value, the inductor behaves almost like a short circuit, and there is no further change in current.

Option 4: At 36.8% of its maximum steady-state value.

This option is incorrect. At 36.8% of its maximum steady-state value (which corresponds to \( 1 - \frac{1}{e} \)), the rate of change of current is not at its maximum. The maximum rate of change occurs at the very start, not at this particular value of current.

Conclusion:

Understanding the behavior of inductive coils in electrical circuits is essential for analyzing how current changes over time. The rate of change of current in an inductor is highest at the start of current flow, immediately after a voltage is applied. This behavior is crucial for the design and analysis of circuits involving inductive components, including those used in power supplies, transformers, and various types of electronic equipment.

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.

Concept:

The resistance of most of the metals increases in a linear way with temperature as shown and can be represented by the equation:

RT = R₀(1 + αT)

RT = Resistance at a temperature T

R₀ = Resistance at 0°C

α = Temperature coefficient of the resistance

Calculation:

Given R_40° = 4Ω and R_80° = 6Ω. We can write:

4 = R₀(1 + 40α)   ---(1)

6 = R₀(1 + 80α)   ---(2)

Multiplying Equation (1) with 4 and Equation (2) with 2, we get:

16 = R₀(4 + 160α)   ---(3)

12 = R₀(2 + 160α)   ---(4)

Subtracting (4) with (3), we get:

16 - 12 = 4R₀ - 2R₀

2R₀ = 4

R₀ = 2 Ω 

CONCEPT:

The resistance (R) of a conducting wire depends upon the length of wire, cross-sectional of the wire, and its resistivity.

Resistance is directly proportional to the length of the conductor and Inversely proportional to the cross-sectional area.

\(R = ρ \frac{l}{A}\)

Where l is the length of the conductor, a is the cross-sectional area of the conductor, ρ is resistivity. 

Resistivity: The resistivity is the resistor depends upon the nature of the conductor and temperature. ∴ Material 

Resistivity is the same for given conducting material at a given temperature and does not depend upon dimensions of wire.

If a wire is stretched, its volume remains the same. 

Concept:
\(P = {I^2}R = \frac{{{V^2}}}{R}\).............( i )

P = power across resistance

I = current

R = resistance

V = voltage

Calculation:

R = 2.5 kΩ , P = 1 Watt

From equation (i)

\(1=I^2\times2500=\frac{V^2}{2500}\)

\(I=\frac{1}{\sqrt{2500}}=20\ mA\)

\(V=\sqrt {2500} =50\ V\)

Concept:

The below equation shows the method to find the resistance:

Resistor Colour Coding uses colored bands to easily identify a resistor resistive value and its percentage tolerance. The resistor color code markings are always read one band at a time starting from the left to the right, with the larger width tolerance band oriented to the right side indicating its tolerance. 

The value of the resistance is given in the form:

R = AB × C ± D%

Where,

‘A’ and ‘B’ indicates the first two significant figures of resistance (Ohms).

‘C’ indicates the decimal multiplies.

‘D’ indicates the tolerance in percentage.

The table for the resistor colour code is given below:

Calculation:

From the above resistance colour codes,

Y → 4

V → 7

‘Y’ and ‘V’ defines ‘A’ and ‘B’

Red, R → 10²

‘Y’ defines C

Now, the resistance is:

R = 47 × 10² Ω 

R = 4700 ohms

Shortcut Trick

BB ROY Great Britain Very Good Watch Gold and Silver

B - Black (0)

B - Brown (1)

R - Red (2)

O - Orange (3)

Y - Yellow (4)

G - Green (5)

B - Blue (6)

V - Violet (7)

G - Grey (8)

W - Whilte (9)

Tolerance - Gold(5%) and Silver(10%)

Concept:

• Conductance is the measure of how easily electricity flows along a certain path through an electrical element. Represented with letter G.
• Conductance is the reciprocal of the Resistance and is measured in Siemens or mhos.

Equivalent Conductance in Series connection is represented as,

\({G_{eq}} = \frac{{{G_1}{G_2}}}{{{G_!} + {G_2}}}\)

Equivalent Conductance in parallel connection is represented as,

G_eq = G₁ + G₂

Calculation:

In the given figure,

6 S, 12 S are in series apply series combination formula

4 S, 8 S are in parallel apply parallel combination formula

Circuit reduces to,

Then 2 S and 4 S are in parallel add them,

Then, \({G_{eq}} = \frac{{12 \times 6}}{{12 + 6}}\; = 4\;S\)

Concept:

Capacitor: 

• A capacitor is an electrical component with two terminals used to store charge in the form of an electrostatic field. 
• It consists of two parallel plates each possessing equal and opposite charges, separated by a dielectric constant.
• Capacitance is the ability of a capacitor to store charge in it. The capacitance C is related to the charge Q and voltage V across them as:

\(⇒ C =\frac{Q}{V} \)​

Equivalent capacitance of capacitors:

Connected in series: When n capacitors C1, C2, C3, ... Cn are connected in series, the net capacitance (Cs) is given by:

\(⇒ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}+ \frac{1}{C_3} + ... \frac{1}{C_n}\)

Connected in parallel: When n capacitors C1, C2, C3, ... Cn are connected in parallel, the net capacitance (Cp) is given by:

⇒ Cp = C1 + C2  + C3 +...  Cn

Calculation: 

Here 2F is in parallel with 2F,

So equivalent capacitor is 4F,

This 4F is again parallel with 4F, then equivalent is 8F

Again this 8F in series with 8F, then equivalent is 4F.

This 4F is parallel with 2F so equivalent is 6F,

This 6F is series to 2F, then equivalent capacitance across A and B is,

\(C_{AB} = \frac{(6\times2)}{8} = 1.5 F\) 

Concept:

Capacitance:

• Capacitance is defined as it is the capability of an element to store electric charge within it. It is also defined the amount of charge
  required to create a unit p.d. between its plates.
• Capacitance is the property that possesses the capacitor. Its unit is Farad.
• It opposes the sudden change in voltage.

It is given by,

\(C=\frac{Q}{V} Farad\)

Inductance: 

• Inductance is the property of a material by virtue of which it opposes any changes of magnitude or direction of current passing through the conductor.
• Inductance is said to be one henry when the current through a coil of the conductor changes at the rate of one ampere per second induction one volt across the coil.
• The unit of inductance is Henry. 

CONCEPT: 

Resistance:

• The measurement of the opposition of the flow of electric current through a conductor is called 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:

\(⇒\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\)

Calculation:

Reciprocal of resistance is conductance, we convert the conductances into resistances and the circuit will look like this

Given: R₁ = 1 Ω , R₂ = 1 / 3 Ω , R₃ = 1 / 2 Ω

Taking eqivalent resistance of R1 and R₂ 

\(R_{12} = \dfrac{1\times \frac13}{1+\frac13}=\frac14\; \Omega \)

Using current division:

I = \(I_{\frac12\Omega }=\dfrac{\frac14}{\frac14 + \frac12}\times 6 = 2\; A\)

Concept:

Conductance in Series:

\(\frac{1}{G_s}=\frac{1}{G_1}+\frac{1}{G_2}+\frac{1}{G_3}+......\frac{1}{G_n}\)

Conductance in Parallel:

G_P = G₁ + G₂ + G₃ + ..... G_n

Application:

Given circuit:

Here, 8 S is connected parallel with 4 S and, 6 S is connected series with 12 S

Hence, an equivalent circuit can be drawn as,

Now, 2 S is connected parallel with 4 S, equivalent circuit can be drawn as,

Hence,

\(\frac{1}{G_{eq}}=\frac{1}{12}+\frac{1}{6}=4\ S\)