Network Synthesis MCQ Quiz in தமிழ் - Objective Question with Answer for Network Synthesis - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The coil is a circuit with Resistance and Inductance connected in series as shown:

The resistance offered by an inductor when an alternating current flows through it is called inductive reactance (XL) and is given by:

\(X_L=ω L\)

With ω = 2πf, the inductive reactance becomes:

\(X_L=2\pi f L\)

Analysis:

For a DC supply, ω = 0.

∴ XL = 0 Ω (Inductor will act as a short circuit)

∴ The steady-state current, with the inductor short-circuited, will be:

\(I=\frac{V}{R}=\frac{20}{2}\)

I = 10 A

The resistance offered by a capacitor when an alternating current flows through it is called capacitive reactance (XC) and is given by

\(X_C=\frac{1}{ω C}=\frac{1}{2 π f C}\)          

Here, f is the frequency

L is the inductance

C is the capacitance

For a DC supply, ω = 0.

∴ The capacitive reactance becomes infinity, i.e. a capacitor acts as an open circuit for DC supplies.

In a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

\(X_C=\frac{1}{\omega C}\)

The resonance curve of the series RLC circuit is given as:

Observation:

• Below the resonance frequency, XL is low and XC is high → circuit behaves like a capacitive circuit (leading power factor)
• At the resonance frequency, XL equals to XC → circuit behaves like a purely resistive circuit (Unity power factor)
• At the above resonance frequency, XL is high and XC is low → circuit behaves like an inductive circuit (lagging power factor)

Calculation:

Given that, 

R = 50 Ω

X_L = 100 Ω 

X_C = 50 Ω 

Z = 50 + j (100-50) Ω 

|Z| = √(50² + 50²) = 70.71 Ω 

Power Factor = cos ϕ = R/Z = 50/70.71 = .7071 = 1/√2

Concept:

The capacitor voltage for a discharging circuit is:

Vc(t) = V(0⁺)e−t/τ 

The capacitor current for a discharging circuit is:

I_c(t) = \(C {dV_c(t) \over dt}\)

Calculation:

Vc(t) = 56e−250t 

τ = RC = \( {1 \over 250}\) ...........(i)

\(I_c(t)=-C {d[56e^{-250t} ] \over dt}\) (minus sign as current is leaving from the positive side of capacitor)

\(I_c(t)=-C (-250)56\space e^{-250t}\)

 7e−250t = \(C (14\times 10^3)\space e^{-250t}\)

C = 0.5 mf

Putting the value of C in equation (i), we get:

\( {1 \over 250}=R(0.5 \times 10^{-3})\)

R = 8 Ω 

Concept:

Considered a sinusoidal Alternating wave of voltage and current

From the waveform:

\( v = {V_m}\sin \left( {\omega t} \right) \)

And,

\(i = {I_m}\sin \left( {\omega t + ϕ } \right)\)

Where Vm and Im are the maximum value of instantaneous voltage and current respectively.

v, i is the instantaneous value of voltage and current at any instant t.

ω is the angular frequency in radian/second.

And, ω = 2πf  

f is the frequency in Hz

ϕ is the phase difference between voltage and current

From the above three equations instantaneous value of voltage and current can be written as: is

\(v = {V_m}\sin \left( {2\pi ft} \right) \)

And,

\(i = {I_m}\sin \left( {2\pi ft + ϕ } \right) \)

Calculation:

Sinusoidal wave is given as,

i(t) = 50 cos (100πt + 10°)

Comparing to the given equation

\(i = {I_m}\sin \left( {2\pi ft + ϕ } \right) \)

∴ I_m = 50 A

The circuit after the switch is closed can be drawn as follows:

\(I\left( s \right) = \frac{{10\left( {s\; + \;10} \right)}}{{s\; + \;20}} + 10\)

\( = \frac{{20\;s\; + \;300}}{{s\; + \;20}}\)

\(I\left( s \right) = \frac{{V\left( {s\; + \;20} \right)}}{{s\left( {20\;s\; + \;300} \right)}}\)

\({I_L}\left( s \right) = \frac{{10}}{{10\; + \;s\; + \;10}}I\left( s \right)\)

\( = \left( {\frac{{10}}{{s\; + \;20}}} \right)\frac{{v\left( {s\; + \;20} \right)}}{{s\left( {20\;s\; + \;300} \right)}}\)

\( = \frac{{10\;V}}{{s\left( {20\;s\; + \;300} \right)}}\)

The steady state value of current through inductor:

\(i\left( s \right) = \mathop {\lim }\limits_{s \to 0} s\;{I_L}\left( s \right)\)

\(1 = \mathop {\lim }\limits_{s \to \infty } \frac{{s\;10\;V}}{{s\left( {20\;s\; + \;300} \right)}}\)

10 V = 300

V = 30 volt

Current through right-most branch = 1 A

Both branches have the same resistance of 10 Ω, so the current through it will be the same.

Voltage across 10 Ω branch (V_t) = 10 × 1 = 10 V

Source current = 1 + 1 = 2 A

Source Voltage = 2 × 10 + V_t = 30 V

The correct answer is option 4):(5 ohms)

Concept:

The reactance of the capacitor

X_c = \(1 \over 2 π × f × C\)

f is the frequency in Hz.

C is the capacitor in F

Xc is the reactance in ohms

Calculation:

Given

X_c1 = 10 ohm at f₁

10 = \(1 \over 2 π × f_1 × C\)

10 × 2π × f₁ = \(1 \over C\)

The reactance at the frequency f₂ 

 f2 = 2f₁

X_c2 = \(1 \over 2 π × f_2 × C\)

By substituting C and  f2

Xc2 = 10 × 2π × f1 × \(1 \over 2\times \pi \times f_1 \times 2\)

= 5 Ω 

A simple RC driving point impedance circuit is as shown below.

Now driving point impedance is calculated as

\(z\left( s \right) = R + \frac{1}{{sC}}\)

After writing in pole-zero form, we get

\(z(s)= \frac{{R\left( {s + \frac{1}{{RC}}} \right)}}{s}\)

Pole at s = 0, zero at \(s = - \frac{1}{{RC}}\)

Properties of RC Driving point impedance functions:

• The poles and zeros are simple. There are no multiple poles and zeros
• The poles and zeros are located on the negative real axis
• The poles and zeros interlace (alternate) each other on the negative real axis
• The poles and zeros are called critical frequencies of the network. The critical frequency nearest to the origin is always a pole. This may be located at the origin.
• The critical frequency at the greatest distance away from the origin is always a zero, which may be located at ∞ also.
• The partial fraction expansion of Z_RC(s) gives the residues which are always real and positive.
• There is no pole located at infinity.
• The slope of the graph of Z(σ) against σ is always negative.
• There is no zero at the origin.
• The value of Z_RC(s) at s = 0 is always greater than the value of Z_RC(s) at s = ∞.

LC Oscillator

• Whenever we connect a charged capacitor to an inductor the electric current and charge on the capacitor in the circuit undergo LC Oscillations.
• The process continues at a definite frequency and if no resistance is present in the LC circuit, then the LC Oscillations will continue indefinitely.
• This circuit is known as an LC oscillator.
• The oscillations are produced due to the transfer of energy between the inductor and capacitor.
• An LC circuit cannot produce oscillations when the resistance is large.
• The presence of a large value of resistance in the circuit will damp out the oscillation. 

Concept:

R = resistance

XL = ωL = Inductive reactance

XC = 1/ωC = Capacitive reactance

L = Inductance

C = Capacitance

Calculation:

Given that 

Impedance Z = (10 - j10) Ω ⇒ R - jXC 

The given impedance show the series circuit has resistance and capacitance.

If the frequency is made 200% of the original frequency then

the frequency(f) is doubled(f' = 2f), then Impedance will be

\(Z' = R - \frac{j}{{2\pi f'C}} = R - \frac{j}{{2\left( {2\pi f} \right)C}}\)

\(Z' = R - \frac{{j{X_C}}}{2} = 10 - j\frac{{10}}{2}\)

Z’ = (10 – j5) Ω 

The relationship between the impedance and admittance is given by:

\(Z= {1\over Y}\)

where Z = Impedance

Y = Admittance

The impedance is analogous to admittance in the following ways: