Transient Analysis MCQ Quiz - Objective Question with Answer for Transient Analysis - Download Free PDF

Explanation:

Current Impulse Signal in a Capacitor

Problem Statement: A current impulse signal of \(4 \delta(t)\) is forced through a capacitor \(C\). We are required to determine the voltage \(V_c(t)\) across the capacitor. The given options are:

• 1) \(\frac{4}{C}t\)
• 2) \(\frac{4.u(t)}{C}\)
• 3) \(4.u(t)-C\)
• 4) \(4.t\)

The correct answer is option 2: \(\frac{4.u(t)}{C}\).

Detailed Solution:

The voltage across a capacitor is related to the current through it by the following fundamental relationship:

Voltage-Current Relationship for a Capacitor:

\[ V_c(t) = \frac{1}{C} \int i(t) \, dt \]

Where:

• \(V_c(t)\) is the voltage across the capacitor at time \(t\).
• \(i(t)\) is the current through the capacitor.
• \(C\) is the capacitance of the capacitor.

Given that the current impulse signal is \(i(t) = 4 \delta(t)\), let us substitute this into the relationship above.

Step 1: Substituting Current \(i(t)\)

\[ V_c(t) = \frac{1}{C} \int 4 \delta(t) \, dt \]

The property of the impulse signal \(\delta(t)\) is that its integral is the unit step function \(u(t)\):

\[ \int \delta(t) \, dt = u(t) \]

Therefore, the integral of \(4 \delta(t)\) becomes:

\[ \int 4 \delta(t) \, dt = 4 \cdot u(t) \]

Step 2: Substituting the Integral

Substituting the result of the integral back into the voltage equation:

\[ V_c(t) = \frac{1}{C} \cdot (4 \cdot u(t)) \]

Thus:

\[ V_c(t) = \frac{4 \cdot u(t)}{C} \]

Step 3: Final Expression

The voltage across the capacitor is:

\[ V_c(t) = \frac{4.u(t)}{C} \]

Hence, the correct option is Option 2.

Additional Information

To further understand the analysis, let us evaluate the other options:

Analysis of Other Options:

Option 1: \(\frac{4}{C}t\)

This option suggests that the voltage across the capacitor increases linearly with time. However, the given current is an impulse signal \(4 \delta(t)\), which is a very short-duration signal. The voltage across a capacitor cannot increase linearly with time for an impulse input. This option is incorrect.

Option 3: \(4.u(t)-C\)

This option combines a scaled unit step function \(4.u(t)\) with a term \(-C\). However, the capacitance \(C\) is a constant and cannot be subtracted as a term in the voltage expression. The voltage should depend solely on the current and the capacitance, as derived earlier. This option is incorrect.

Option 4: \(4.t\)

This option suggests that the voltage increases linearly with time \(t\), independent of the capacitance \(C\). However, this does not align with the voltage-current relationship of a capacitor. For an impulse input, the voltage depends on the unit step function \(u(t)\), not a linear time function. This option is incorrect.

Conclusion:

From the above analysis, the correct voltage across the capacitor for the given current impulse signal is:

Option 2: \(\frac{4.u(t)}{C}\).

Explanation:

Parallel Resistance (R) and Capacitor (C) Circuit

Definition: A parallel resistance and capacitor circuit is an electrical circuit configuration where a resistor (R) and a capacitor (C) are connected in parallel to a common voltage source. Such circuits are widely used in AC (alternating current) applications to analyze impedance, phase angles, and time constants.

Working Principle: In an AC circuit, a capacitor introduces a phase shift because the current leads the voltage by 90 degrees, while the resistor does not introduce any phase shift (current and voltage are in phase). The combination of these two components results in a net phase angle (ϕ) between the total current and the source voltage, which is dependent on the relative values of resistance (R) and capacitive reactance (X_c).

Phase Angle (ϕ): The phase angle (ϕ) in a parallel R-C circuit is determined by the relationship between the current through the capacitor and the current through the resistor. Mathematically, the phase angle is given by:

tan(ϕ) = I_C / I_R

Where:

• I_C = Current through the capacitor
• I_R = Current through the resistor

Since the current through the capacitor is given by I_C = V / X_C and the current through the resistor is I_R = V / R, the phase angle can also be expressed as:

tan(ϕ) = X_C / R

Where:

• X_C = 1 / (2πfC), the capacitive reactance
• R = Resistance

From the above equation, it is evident that the phase angle ϕ depends on the ratio of X_C (capacitive reactance) to R (resistance).

Effect of Increasing Resistance (R):

When the resistance (R) in the parallel R-C circuit is increased:

• The denominator of the term X_C / R increases, resulting in a decrease in the value of tan(ϕ).
• As tan(ϕ) decreases, the phase angle ϕ also decreases since the tangent function is directly related to the angle.

Hence, when resistance (R) increases, the phase angle (ϕ) decreases.

Correct Option:

The correct answer is:

Option 4: Decreases

As explained above, increasing the resistance reduces the phase angle in a parallel R-C circuit due to the inverse relationship between resistance and tan(ϕ).

Additional Information

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

Option 1: Increases

This option is incorrect because increasing resistance (R) in the parallel R-C circuit reduces the phase angle (ϕ). The relationship between tan(ϕ) and resistance is inversely proportional, so an increase in R leads to a decrease in tan(ϕ) and consequently a decrease in ϕ.

Option 2: Becomes zero

This option is incorrect because the phase angle (ϕ) does not become zero simply by increasing the resistance. The phase angle becomes zero only in a purely resistive circuit (when the capacitor is removed or its effect is negligible). In a parallel R-C circuit, the phase angle will decrease but will not reach zero as long as the capacitor is present.

Option 3: Remains the same

This option is incorrect because the phase angle (ϕ) is dependent on the resistance (R) and capacitive reactance (X_C). When R changes, the phase angle also changes. Therefore, the phase angle cannot remain the same when resistance is increased.

Conclusion:

In a parallel R-C circuit, the phase angle (ϕ) decreases when the resistance (R) increases. This is due to the inverse relationship between resistance and tan(ϕ), as shown in the mathematical derivation. Understanding the behavior of phase angles in R-C circuits is essential for analyzing AC circuits and their impedance characteristics.

Explanation:

Admittance in Phasor Form

Definition: Admittance, represented as Y, is the measure of how easily a circuit allows the flow of electric current when a voltage is applied. It is the reciprocal of impedance (Z) and is expressed in Siemens (S). In phasor form, admittance is a complex quantity, consisting of a real part and an imaginary part:

Y = G + jB,

where:

• G = Conductance (Real part of admittance)
• B = Susceptance (Imaginary part of admittance)

Components of Admittance:

• Conductance (G): It is the real part of admittance and indicates the ability of a circuit to conduct electric current. It is measured in Siemens (S).
• Susceptance (B): It is the imaginary part of admittance and represents the reactive component of the circuit. Susceptance is also measured in Siemens (S).

Relationship with Impedance:

Admittance is the reciprocal of impedance:

Y = 1/Z

In phasor form:

Y = 1/(R + jX),

where R is the resistance and X is the reactance of the circuit. By rationalizing the denominator, the admittance can be expressed as:

Y = G + jB

with:

• G = R / (R² + X²)
• B = -X / (R² + X²)

Correct Option Analysis:

The correct option is:

Option 4: Conductance; Susceptance

In phasor form, the real part of admittance (Y) is called Conductance (G), and the imaginary part is called Susceptance (B). Conductance measures the ability of the circuit to allow current flow due to its resistive elements, while susceptance represents the reactive behavior of the circuit caused by its capacitive or inductive components.

Important Information:

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

Option 1: Impedance; Reactance

This option is incorrect because impedance and reactance are not components of admittance. Impedance (Z) is the total opposition to current flow in a circuit and is the reciprocal of admittance. Reactance (X) is the imaginary part of impedance, not admittance.

Option 2: Reactance; Conductance

This option is also incorrect because it mixes the terms incorrectly. Reactance is associated with impedance, not admittance. Conductance is indeed the real part of admittance, but the imaginary part is not reactance; it is susceptance.

Option 3: Resistance; Reactance

This option is incorrect as it pertains to the components of impedance, not admittance. Resistance (R) is the real part of impedance, and reactance (X) is its imaginary part. These terms do not describe admittance.

Conclusion:

The real part of admittance in phasor form is called Conductance (G), and the imaginary part is called Susceptance (B). These components provide critical insight into how a circuit behaves under the influence of applied voltage. Understanding the distinction between admittance, impedance, and their respective components is crucial for analyzing and designing electrical circuits.

Concept

The current through a capacitor is given by:

where,  Rate of change of capacitor voltage

Calculation

Given, C = 0.5F

The value of the current through the capacitor at t = 0+ will be: 

Concept Used:

In an AC circuit with an inductor and capacitor , the impedance of the elements affects the voltage and current distribution.

Impedance (Z): The total opposition offered by circuit components to AC current, given by:

Z = R + jX

The current in an AC circuit follows Ohm's Law for AC circuits :

I = V / Z

Calculation:

Given,

Angular frequency, ω = 80 rad/s

Using the provided values:

Impedance in first branch:

Z₁ = √(125² +50²)= 25√29

Impedance in second branch:

Z₂ = √(80²+ 50²) = 10√89  

⇒ Current in the first branch:

I₁ = (80 / Z₁)

⇒ I₁ = (16 / 5√29) A at 45° leading

⇒ Current in the second branch:

I₂ = 80 / Z₂

⇒ I₂ = (8 / √89) A at 45° lagging

The voltage drop 1H is 

V_H= I₂ × X_L= (8 / √89) × 80 =640/√89 = 67.8 V 

Concept:

• In a series R-L circuit, the voltage drop across the resistor depends upon the resistance value, and the voltage drop across the inductor depends on the rate of change of the current through it. 
• Initially (at t = 0), the voltage drop across the inductor is maximum (inductor acts as an open circuit) and the current flowing through the circuit is 0. 
• The time constant is defined as the time in which the current reaches 63% of its steady-state value (maximum value).
• For the LR series circuit, the time constant τ = L/R. 
• The current reaches steady-state in 5 time-constants (5τ). 
• At steady-state inductance of the coil is reduced to zero acting more like a short circuit. 

​Transient curves for an LR series circuit is shown in the figure below:

Calculation:

Given that,

Resistance (R) = 2 Ω

Inductance (L) = 200 mH

Time constant 

t (τ) = L/R = 200/2 = 100

Time taken by the inductor reach its maximum steady state value = 5τ = 5 × 100

= 500 ms

= 0.5 sec

Time constant:

• The time constant is always calculated for the circuit at t>0.
• The time constant for an RC circuit (τ) = 
• The time constant for an RL circuit (τ) = 

Calculation:

τ = 

At time t = 2 sec

First switch will be in the open condition because it closes at 5^th second

Second switch will be in a closed state because it gets opened at 3^rd second.

So circuit at t = 2 s reduces to

∴ i = 6 / 3000 A

i = 2 mA

Concept:

Transients are present in the network when the network is having any energy storage elements.

1. Inductor does not allow sudden change of current and it stores energy in the form of the magnetic field.

(Inductor allows sudden change of voltage).

2. Capacitor does not allow sudden change of voltage and it stores energy in the form of the electric field.

(Capacitor allows sudden change of current).

Now as,

For a sudden change of current, we require dt = 0,

then V = ∞, but practically this much voltage not possible.

Hence inductor does not allow sudden change of current.

Analysis:

Initially the switch is open, hence the current flowing through the circuit is zero.

After switch closed, 

Important Points

Before Switching             After Switching

• When a battery is connected to a series resistor and capacitor, the initial current is high as the battery transports charge from one plate of the capacitor to the other
• The charging current asymptotically approaches zero as the capacitor becomes charged up to the battery voltage
• Charging the capacitor stores energy in the electric field between the capacitor plates
• The rate of charging is typically described in terms of a time constant RC

 

The voltage across the capacitor at time t is given by

V₀ = final voltage across the capacitor

RC = time constant

After one time constant i.e. at t = RC

⇒ V_t = 0.632 V₀

During capacitor charging, the voltage actually rises to 63.2 per cent of its final value.

At t = 0⁺, the inductor acts as an open circuit with current reflected back.

At t = ∞, the inductor acts as a short circuit

Important Points:

At t = 0⁺, a capacitor with zero initial condition acts as a short circuit with voltage reflected back.

Concept:

Transients are caused because of the following:

• The load is suddenly connected to or disconnected from the supply.
• The sudden change in applied voltage from one finite value to the other.
• The inductor and the capacitor store energy in the form of the magnetic field and electric field respectively, and hence these elements have transients.
• Circuits containing only resistive element has no transients because resistors do not store energy in any form. It dissipates energy in form of heat coming from I2R loss.

Example:

The above circuit will show transient because of the presence of the capacitor as the capacitor does not allow a sudden change in voltage.

The expression for the voltage is given by:

With V(0+) = V(0-) = 0 V, and V(∞) = V, we get:

Similarly, RL and RLC circuits show transient in the same way.

Note:

• The linear circuit is an electric circuit and the parameters of this circuit are resistance, capacitance, inductance, etc.
• As a linear circuit consists of energy storing elements so, transient present in the linear circuits.

Concept:

Energy stored in an iron cored coil 

Copper losses = I²R

The time constant of RL circuit = L/R

Calculation:

Let the current in the iron core is I.

Energy stored in iron cored coil 

Copper losses = I²R = 2000 W

To get L/R 

⇒ L/R = 1

Concept:

The given circuit is R-L series circuit. Therefore current will be equal in all the element.

Given, Power consumed by 5 Ω resister is 10 W.

i.e, 

⇒ 

⇒ I_rms = √2 A

V_s = 50 sin ωt

V_srms = 50 / √2 

∴ 

r_eq = 5 + 10 = 15 Ω

∴ 

Therefore, correct option will be 3.

The correct answer is option

Concept:

The Bandwidth of Series RLC CIrcuit is given as

B.W =   rad/s

Where R is the resistance in ohm

L is the Inductance in henry

Calculation:

Given

R = 1000 Ω, L = 100 mH

B.W = 

= 10 × 10³  rad/s

Mistake Points

• Bandwidth= \(\frac{R}{L} \ rad/s\)
• Bandwidth=  \(\frac{R}{2\pi L}\ Hz\)