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

Concept

The current through a capacitor is given by:

\(I_C=C{dV_C\over dt}\)

where, \({dV_C\over dt}=\) Rate of change of capacitor voltage

Calculation

Given, C = 0.5F

\(\rm V_c(s)=\frac{1}{s^2+1}=sin( t)\)

\({dV_C\over dt}= cos( t)\)

\(I_C=0.5cos(t)\)

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

\(I_C=0.5cos(0)\)

\(I_C=0.5\space A\)

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:

The time constant (τ) of an RC circuit is given by the formula:

τ = R_eqC

where R_eq is the equivalent resistance seen by the capacitor, and C is the capacitance.

While finding out the time constant the voltage source will be shorted.

Calculation:

Given:

Resistors: 3Ω and 2Ω will be in parallel
Capacitance: C = 10 F

Solution:

If the resistors are actually in parallel (common in RC circuits):

R_eq = (3Ω × 2Ω)/(3Ω + 2Ω) = 6/5Ω = 1.2Ω

τ = R_eqC = 1.2Ω × 10 F = 12 sec

Final Answer:

The correct time constant is 1) 12 sec when considering parallel resistors.

Explanation:

R-L-C Series Circuit

Definition: An R-L-C series circuit is an electrical circuit consisting of a resistor (R), an inductor (L), and a capacitor (C) connected in series. This type of circuit is characterized by its ability to resonate at a particular frequency known as the resonant frequency. At this frequency, the inductive reactance and capacitive reactance cancel each other out, resulting in a purely resistive impedance.

Working Principle: In an R-L-C series circuit, the total impedance (Z) is the sum of the resistive (R), inductive (XL), and capacitive (XC) reactances. The impedance can be expressed as:

Z = R + j(XL - XC)

Where:

• R is the resistance in ohms (Ω)
• XL is the inductive reactance in ohms (Ω), given by XL = 2πfL
• XC is the capacitive reactance in ohms (Ω), given by XC = 1/(2πfC)
• f is the frequency in hertz (Hz)
• j is the imaginary unit

At the resonant frequency (fr), the inductive reactance (XL) equals the capacitive reactance (XC), and the impedance is purely resistive:

fr = 1/(2π√(LC))

Behavior Above Resonant Frequency: When the supply frequency is more than the resonant frequency (f > fr), the inductive reactance (XL) becomes greater than the capacitive reactance (XC), resulting in a net inductive impedance. In this condition, the total impedance (Z) of the circuit is dominated by the inductive reactance, causing the supply current to lag the applied voltage.

Correct Option Analysis:

The correct option is:

Option 2: The supply current lags the applied voltage.

This option correctly describes the behavior of an R-L-C series circuit when the supply frequency is more than the resonant frequency. Due to the dominance of the inductive reactance in this scenario, the circuit exhibits inductive characteristics, causing the supply current to lag behind the applied voltage.

Additional Information

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

Option 1: Supply current leads the applied voltage.

This option is incorrect because it describes a scenario where the capacitive reactance dominates the impedance. When the supply frequency is less than the resonant frequency (f < fr), the capacitive reactance (XC) is greater than the inductive reactance (XL), causing the supply current to lead the applied voltage. However, this is not the case when the supply frequency is more than the resonant frequency.

Option 3: Supply current is in phase with the applied voltage.

This option is also incorrect because it describes the condition at the resonant frequency (f = fr). At resonance, the inductive reactance (XL) equals the capacitive reactance (XC), resulting in a purely resistive impedance. In this case, the supply current is in phase with the applied voltage. When the supply frequency is more than the resonant frequency, the current lags the voltage due to the net inductive reactance.

Option 4: Supply current becomes zero.

This option is incorrect because, in an R-L-C series circuit, the supply current will not become zero unless there is an open circuit or a fault condition. The supply current depends on the total impedance of the circuit. Even when the supply frequency is more than the resonant frequency, the current will not be zero; it will simply lag behind the applied voltage due to the inductive reactance.

Conclusion:

Understanding the behavior of an R-L-C series circuit at different frequencies is crucial for correctly identifying the phase relationship between the supply current and the applied voltage. When the supply frequency is more than the resonant frequency, the inductive reactance dominates the impedance, causing the supply current to lag behind the applied voltage. This characteristic is essential for various applications, including tuning circuits, filters, and oscillators, where the frequency response of the circuit is a critical factor.

Ans. (2) Sol.

The phase difference (ϕ) between the applied voltage and the current in an A.C. circuit containing a resistor (R) and inductor (with inductive reactance X_L) connected in series is given by: ϕ = tan⁻¹(X_L / R)

Given: Resistance, R = 5 Ω Inductive reactance, X_L = 5 Ω Substitute the values: ϕ = tan⁻¹(5 / 5) ϕ = tan⁻¹(1) ϕ = π / 4 radians

Therefore, the phase difference is π / 4 radians.

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) = \(R_{eq}C\)
• The time constant for an RL circuit (τ) = \( {L \over R_{eq}}\)

Calculation:

\(R_{eq}= {2R\times R \over 2R+R}\)

\(R_{eq}= {2R^2 \over 3R}\)

\(R_{eq}= {2R \over 3}\)

\(τ = R_{eq}C\)

\(τ= {2R\over 3}\times C\)

τ = \(\frac{{2RC}}{3}\)

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,

\(V=L\frac{di}{dt}\)

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.

\(I\left( {{0^ - }} \right) = 0\;A\)

After switch closed, \(I\left( {{0^ + }} \right) = I\left( {{0^ - }} \right) = 0\;A\)

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

\({V_t} = {V_0}\left( {1 - {e^{ - \frac{t}{{RC}}}}} \right)\)

V₀ = final voltage across the capacitor

RC = time constant

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

⇒ V_t = 0.632 V₀

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:

\(V\left( t \right) = V\left( \infty \right) - \left( {V\left( \infty \right) - V\left( {{0^ + }} \right)} \right){e^{ - \frac{t}{\tau }}}\)

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

\(\therefore V\left( t \right) = V - \left( {V - 0} \right){e^{ - \frac{t}{{RC}}}}\)

\(V\left( t \right) = V\left( {1 - {e^{ - \frac{t}{{RC}}}}} \right)\)

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 \(E = \frac{1}{2}L{I^2}\)

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 \(= \frac{1}{2}L{I^2} = 1000J\)

Copper losses = I²R = 2000 W

To get L/R 

\( \frac{\frac{1}{2}L{I^2}}{{}I^2R}= \frac{1000}{2000}\)

\( \frac{L}{2R}= \frac{1000}{2000}\)

⇒ 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, \(\rm I^2_{rms}=10\)

⇒ \(\rm I^2_{rms}=\frac{10}{5}=2\ A\)

⇒ I_rms = √2 A

V_s = 50 sin ωt

V_srms = 50 / √2 

∴ \(\rm z=\frac{V_{srms}}{I_{rms}}=\frac{50/√{2}}{√{2}} Ω=25\ Ω\)

r_eq = 5 + 10 = 15 Ω

∴ \(\rm \cos\phi=\frac{R_{eq}}{z_{eq}}=\frac{15}{25}=0.6\)

Therefore, correct option will be 3.

The correct answer is option

Concept:

The Bandwidth of Series RLC CIrcuit is given as

B.W = \(R \over L\)  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 = \(1000 \over 100 × 10^{-3}\)

= 10 × 10³  rad/s

Mistake Points

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