Step Response of Second Order Circuits MCQ Quiz in తెలుగు - Objective Question with Answer for Step Response of Second Order Circuits - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

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

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.

The correct answer is 10.5 sec. 

Concept:

The peak time of the system is given by 

n = 1,3,5 for overshoot.

n = 2,4,6 for undershoot.

Solution: 

t_p = 3.50 for n = 1

∴ 

 time at the second overshoot is 

= 3× 3.50

=10.5 sec.

In the given figure the voltage across the circuit can be calculated by using below formula:

Where V is the voltage across the circuit

  is the voltage across the Resistor

  is the voltage across the Inductor

  is the voltage across the capacitor

In the figure, it is clear that =3V, 

V_RL

=4V, =8V

 On substituting n the above equation

  = 5 V

At , switch is opened. The Network is in steady state and the inductor is short circuited and capacitor acts like an open circuit.

Now at , switch is closed and network becomes

Apply Nodal analysis, we get

At  when switch is closed

Source free parallel RLC circuit bandwidth

And

For critical damping

∴