Test: Step Response of Second Order Circuits - Electrical Engineering (EE) MCQ

Concept:

The voltage across the resistance is given by:

v_r(t) = i(t) × R

The current i(t) is given by:

Calculation:

Given, f = 50 Hz and C = 20 × 10^-6 F

X_C = 159.23 Ω 

i(t) = 0.501 A

P = i²R

P = (0.501)² × 120

P = 30.20 W

Concept:

For Parallel RLC Circuit,

The damping factor is given by:

         ----(1)

Where ξ = damping factor

R = Resistance

C = Capacitance

L = Inductance

Explanation:

For Critically Damped SYstem, ξ = 1

Given:

C = 20 μf

L = 8 mH

We can find the value of R by putting the value of C, L & ξ in equation (1):

R = 10 Ω 

Concept:

Consider the parallel RLC circuit as shown:

This circuit produces a parallel resonance when the current through the circuit when the resultant current through the parallel combination is in phase with the supply voltage.

• At the resonance, there will be large circulating currents between the inductor and capacitor due to energy oscillations.
• A parallel resonant circuit stores the circuit energy in the magnetic field of the inductor and the electric field of the capacitor.
• This energy is constantly being transferred back and forth between the inductor and the capacitor which results in zero current and energy being drawn from the supply.

This is because the corresponding instantaneous values of I_L and I_C will always be equal and opposite and therefore the current drawn from the supply is the vector addition of these two currents and the current flowing in I_R.

The total impedance of a parallel resonance circuit at resonance becomes just the value of the resistance in the circuit and  Z = R.

Conclusion:

The power factor is unity at the resonance.(θ = 0°) Current in phase with the supply voltage.

Parallel RLC circuit:

The characteristic equation is given as,

The bandwidth of a parallel RLC network is given as:

Dynamic resistance of a tuned circuit:

The value of the resistance at which the circuit becomes independent of frequency is known as the dynamic resistance.

Under such conditions the value of the time constant of two parallel branches becomes equal.

If the value of 'R' increases, then the dynamic resistance (R_d) decreases.

Concept:

Under steady-state:

• When a capacitor, is present with a D.C supply, it behaves as an open circuit.
• When an inductor is present with D.C supply, it behaves as a short circuit

At t = 0⁺ :

Inductor replace with current source I_L (0⁺) = I_L (0^-)

The inductor does NOT allow the sudden change in current”

The capacitor is replaced with the voltage source

V_c (0⁺) = V_c (0^-)

The capacitor does NOT allow the sudden change in voltage.

Calculation:

The circuit at t = 0^-

V_c (0^-) = 5 V

Circuit at t = 0⁺

I = 5/4 = 1.25mA

Reference vector for series circuit:

• In any series circuit with combinations of RC, RL, or RLC connected in series, the current across all the elements is the same (since in series, the current is same and in parallel voltage is same) and the voltage across individual elements and its phase differs.
• Since the current is the common factor, it is considered as reference vector.

Consider RLC series circuit: 

An electrical circuit constituting an inductor (L), capacitor (C), and resistor (R) connected in series or parallel is called an LCR circuit.

• For an inductor (L), if we consider the current (I) to be the reference axis, then voltage leads by 90°.
• For the capacitor (C), the voltage lags by 90°.
• This is represented by the phasor diagram. ​

The angle between the phasor and current is called the phase angle and is denoted by θ.

omplex conjugate (non-multiple): oscillatory (sustained oscillations)
Complex conjugate (multiple): unstable (growing oscillations).

Maximum overshoot, rise time and delay time are the major factor of the transient behaviour of the system and determines the transient characteristics.

The output is given as a function of time. The final value of the output is lim_n->∞c(t) = 1; .

Hence Td (at 50% of the final value) is the solution of 0.5 = 1-e^-Td, and is equal to ln 2 or 0.693 sec.

C(s)/R(s) = 1/s²+s+1
C(s)/R(s) = w/ws² + 2Gws + w²
Compare both the equations,
w = 1 rad/sec
2Gw = 1
Mp = 16.3 %

Large overshoot refers to the maximum peak in the response of the closed loop system and this is mainly due to the high positive correcting torque.

C(s) = 1/s-1/s+5-5/(s+5)^2
C(s) = 25/s(s²+10s+25)
R(s) = 1/s
G(s) = 25/(s²+10s+25 )
w= √25
w = 5 rad/sec
G = 1.

C(s)/R(s) = 2/s+1
R(s) = 1/s (step input)
C(s) = 2/s(s+1)
c(t) = 2[1-e^-t]
1.96 = 2[1-e^-T]
T= 4sec.

 s + 6/K[s² + s/K + 6/K]
Comparing with s² + 2Gw + w²
w = √6/K
2Gw = 1/K
2 x 0.5 x √6/K = 1/K
K = 1/6.