Network Synthesis MCQ Quiz in தமிழ் - Objective Question with Answer for Network Synthesis - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 11, 2025
Latest Network Synthesis MCQ Objective Questions
Top Network Synthesis MCQ Objective Questions
Network Synthesis Question 1:
Calculate the total impedance for a circuit having the following values:
Resistance = 5 Ω
Inductance reactance = 20 Ω
Capacitance reactance = 8 Ω
Answer (Detailed Solution Below)
Network Synthesis Question 1 Detailed Solution
Concept:
In a series RLC circuit, the impedance is given by
\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}\)R is resistance
XL is inductive reactance = ωL
XC is capacitive reactance = -1/ωC
(XL – XC) is net reactance
At the resonant frequency, inductive reactance is equal to capacitive reactance i.e. XL = XC
So, at this condition the impedance is minimum, and it is equivalent to R.
Explanation:
Given that, resistance (R) = 5 Ω
Capacitive reactance (XC) = 20 Ω
Inductive reactance (XL) =8 Ω
Net Impedance Z is calculated as:
\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} \)
\(Z= \sqrt {{5^2} + {{\left( {20 - 8} \right)}^2}} = 13\;{\rm{\Omega }}\)
Network Synthesis Question 2:
Consider the circuit shown.
Given:
Vc(t) = 56e−250t for t > 0
i(t) = 7e−250t for t > 0
Find the values of R and C respectively.
Answer (Detailed Solution Below)
Network Synthesis Question 2 Detailed Solution
Concept:
The capacitor voltage for a discharging circuit is:
Vc(t) = V(0+)e−t/τ
The capacitor current for a discharging circuit is:
Ic(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
Network Synthesis Question 3:
An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω and a resistor of 50 Ω are connected in series. What is the power factor of the circuit?
Answer (Detailed Solution Below)
Network Synthesis Question 3 Detailed Solution
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
XL = 100 Ω
XC = 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
Network Synthesis Question 4:
If i(t) = 50 cos (100πt + 10°) is the expression of a sinusoidal current, find the maximum amplitude.
Answer (Detailed Solution Below)
Network Synthesis Question 4 Detailed Solution
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
\(i = {I_m}\sin \left( {2\pi ft + ϕ } \right) \)
∴ Im = 50 A
Network Synthesis Question 5:
The value of V that would result in a steady-state current of 1 A through the inductor in the above circuit is
Answer (Detailed Solution Below)
Network Synthesis Question 5 Detailed Solution
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 (Vt) = 10 × 1 = 10 V
Source current = 1 + 1 = 2 A
Source Voltage = 2 × 10 + Vt = 30 V
Network Synthesis Question 6:
The reactance offered by a capacitor to alternating current of frequency ‘f’ is 10 ohm. If the frequency is made to operate at ‘2f’, then the reactance becomes equal to ______ .
Answer (Detailed Solution Below)
Network Synthesis Question 6 Detailed Solution
The correct answer is option 4):(5 ohms)
Concept:
The reactance of the capacitor
Xc = \(1 \over 2 π × f × C\)
f is the frequency in Hz.
C is the capacitor in F
Xc is the reactance in ohms
Calculation:
Given
Xc1 = 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
Xc2 = \(1 \over 2 π × f_2 × C\)
By substituting C and f2
Xc2 = 10 × 2π × f1 × \(1 \over 2\times \pi \times f_1 \times 2\)
= 5 Ω
Network Synthesis Question 7:
The lowest and highest critical frequencies of RC driving point impedance respectively are ______
Answer (Detailed Solution Below)
Network Synthesis Question 7 Detailed Solution
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 ZRC(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 ZRC(s) at s = 0 is always greater than the value of ZRC(s) at s = ∞.
Network Synthesis Question 8:
An LC circuit cannot produce oscillations, when
Answer (Detailed Solution Below)
Network Synthesis Question 8 Detailed Solution
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.
Network Synthesis Question 9:
An electrical circuit that is operating at a certain frequency has an impedance of (10 – j10) Ω. If the frequency is made 200% of the original frequency then the new value of impedance is _________.
Answer (Detailed Solution Below)
Network Synthesis Question 9 Detailed Solution
Concept:
Series circuit element |
Impedance |
R – L |
Z = R + jXL Z = R + jωL |
R – C |
Z = R – jXC Z = R – j/ωC |
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
Network Synthesis Question 10:
What is the inverse of impedance also known as?
Answer (Detailed Solution Below)
Network Synthesis Question 10 Detailed Solution
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:
Impedance (Z) |
Admittance (Y) |
Series circuit |
Parallel circuit |
Z = R + jX |
Y = G + jB |
SI unit is ohm |
SI unit is siemen |
Current remains same |
Voltage remains same |