Networks Analysis and Synthesis 5. In a minimum function 7. The driving-point impedance Z(s) of a network
has a pole-zero locations as shown in the figure. If
1. In a parallel resonant circuit, the circuit current A. the degree of numerator and denominator are
Z(0) = 3, then Z (s) is
at resonance is maximum. equal
A. True B. False B. the degree of numerator and denominator are
unequal
In parallel resonance current is minimum.
C. the degree of numerator is one more than
2. The rms value of wave in figure is degree of denominator
D. the degree of numerator is one less than degree
of denominator
If degrees of numerator and denominator were
unequal, a pole or zero can be removed. Such a
function cannot be a minimum function.
Therefore degrees of numerator and denominator
A.
A. about 95 V B. about 80 V must be equal.
C. about 50 V D. about 25 V 6. When a current source I is suddenly connected
across a two terminal relaxed RC circuit at time t = B.
0, the voltage across the current source is shown in
figure. The RC circuit is
C.
3. An RLC series circuit is underdamped. To make it
overdamped, the value of R D.
A. has to be increased
From figure, The impedance function
B. has to be decreased
C. has to be increased to infinity
A. a series combination of R and C
D. has to be reduced to zero
B. a parallel combination of R and C
For overdamping R is more. and Z(0) = while in question Z(0) = 3
C. a series combination of R and parallel
4. Henry is equivalent to combination of R and C
A. Volts/Ampere B. Weber/Volt D. a pure capacitor =3 k = 2.
C. Weber/Ampere D. Weber/Ampere2 Initially capacitor behaves as short-circuit and
finally as open circuit.
Inductance is flux per ampere.
, 12. Wave A = 100 sin ωt and wave B = 100 cos ωt. 15. Two networks are cascaded through an ideal
8. Then buffer. If td1 and td2 are delay times of network,
the overall delay time is
A. rms values of the two waves are equal
and i(t) is a unit step, then V(t) in the steady state is A. sq. rt.(td1 + td2)
given by B. rms values of A is more than that of B
B. sq. rt. (t2d1 + t2d2)
A. 4/9 B. 4/3 C. 0 D. ∞ C. rms values of A is less than that of B
C. td1 + td2
D. rms values of the two waves may or may not be
equal
D.
Since peak values are equal, their rms values are
= 4/9. equal irrespective of phase difference. In cascaded networks delay times are added.
9. For a transmission line open circuit and short 13. Z(c) for the network shown in the figure is 16. A series RC circuit has R = 5 Ω and C = 10 μF.
circuit impedances are 20 Ω and 5 Ω. Then The current in the circuit is 5 sin 20000t. The
characteristic impedance is The value of C and R are, applied voltage is
respectively
A. 100 Ω B. 50 Ω C. 25 Ω D. 10 Ω A. 252 sin (20000t + 45°)
Z0 = sq. rt. of Zoc Zsc .
B. 252 sin (20000t - 45°)
10. The terms RMS and average values apply only
to sine waves. C. 252 sin 20000t
A. True B. False D. 252 sin (20000t - 90°)
These terms are applicable to all waveshapes.
A. 1/6 F and 4 Ω B. 2/9 F and 9/2 Ω
11. The drift velocity of electrons is
C. 2/3 F and 1/2 Ω D. 1/2 F and 1 Ω
A. very small as compared to speed of light
14. In an R-C series circuit excited by a voltage E,
B. equal to speed of light the charge across capacitor at t = 0+ is
Z = R2 + XC2 = 50
C. almost equal to speed of light A. 0 B. CE
V = 550 sin(20000 t - 45°)
D. half the speed of light C. CE (1 - e-t/RC) D. none of the above
V = 252 sin(20000 t - 45°)
Drift velocity of electrons is very small. Voltage across a capacitor cannot change
instantaneously.
, 17. A capacitor stores 0.15C at 5 V. Its capacitance D. none of the above
is
i(t) is the incoming current. The currents leaving
A. 0.75 F B. 0.75 μF i(t) = 3e-4t - 2e-3t
node 1 are
C. 0.03 F D. 0.03 μF 23. In the figure shown, A1, A2, A3 are identical
20. In the circuit of figure, the switch is closed at t Ammeters. If A1 and A3 read 5 and 13 A
Q = CV or 0.15 = C(5) or C = 0.03 F. = 0. At t = 0+ the current through C is respectively, reading of A2 will be
18. A 0.5 μF capacitor is connected across a 10 V
battery. After a long time, the circuit current and
voltage across capacitor will be
A. 0.5 A and 0 V B. 0 A and 10 V
C. 20 A and 5 V D. 0.05 A and 5 V
When the capacitor is fully charged, i = 0 and A. 4 A B. 2.5 A
voltage across capacitor is equal to battery A. 8 B. 13 A
voltage. C. 3.1 A D. 0
Capacitor behaves as a short-circuit at t = 0. C. 18 D. 12 A
19. For node 1 in figure, KCL equation is
21. A cable has an insulation resistance of 10^10 A2 = sq. rt.(132 – 25) 12 A.
ohms. If the length is doubled, the insulation
resistance will be 24. Which of the following pairs are correctly
matched?
A. 2 x 10^10 Ω B. 0.5 x 10^10 Ω
1.Brune's realisation : realisation with ideal
C. 10^10 Ω D. 4 x 10^10 Ω transformer.
2.Cauer realisation : ladder realisation.
Insulation resistance of cable is inversely 3.Bott Duffin realisation : with non ideal
A. proportional to length. transformer.
Select the answer using the following codes:
22. , i(t) =
A. 1, 2 and 3 B. 2 and 3
B.
A. 3e^-4t – e^-5t B. 3e^-4t - 2e^-3t
C. 1 and 3 D. 1 and 2
C. 2e^-4t - 5e^-2t D. 3e^-4t – e^-t
Bott Duffins realisation does not use transformer.
C.
has a pole-zero locations as shown in the figure. If
1. In a parallel resonant circuit, the circuit current A. the degree of numerator and denominator are
Z(0) = 3, then Z (s) is
at resonance is maximum. equal
A. True B. False B. the degree of numerator and denominator are
unequal
In parallel resonance current is minimum.
C. the degree of numerator is one more than
2. The rms value of wave in figure is degree of denominator
D. the degree of numerator is one less than degree
of denominator
If degrees of numerator and denominator were
unequal, a pole or zero can be removed. Such a
function cannot be a minimum function.
Therefore degrees of numerator and denominator
A.
A. about 95 V B. about 80 V must be equal.
C. about 50 V D. about 25 V 6. When a current source I is suddenly connected
across a two terminal relaxed RC circuit at time t = B.
0, the voltage across the current source is shown in
figure. The RC circuit is
C.
3. An RLC series circuit is underdamped. To make it
overdamped, the value of R D.
A. has to be increased
From figure, The impedance function
B. has to be decreased
C. has to be increased to infinity
A. a series combination of R and C
D. has to be reduced to zero
B. a parallel combination of R and C
For overdamping R is more. and Z(0) = while in question Z(0) = 3
C. a series combination of R and parallel
4. Henry is equivalent to combination of R and C
A. Volts/Ampere B. Weber/Volt D. a pure capacitor =3 k = 2.
C. Weber/Ampere D. Weber/Ampere2 Initially capacitor behaves as short-circuit and
finally as open circuit.
Inductance is flux per ampere.
, 12. Wave A = 100 sin ωt and wave B = 100 cos ωt. 15. Two networks are cascaded through an ideal
8. Then buffer. If td1 and td2 are delay times of network,
the overall delay time is
A. rms values of the two waves are equal
and i(t) is a unit step, then V(t) in the steady state is A. sq. rt.(td1 + td2)
given by B. rms values of A is more than that of B
B. sq. rt. (t2d1 + t2d2)
A. 4/9 B. 4/3 C. 0 D. ∞ C. rms values of A is less than that of B
C. td1 + td2
D. rms values of the two waves may or may not be
equal
D.
Since peak values are equal, their rms values are
= 4/9. equal irrespective of phase difference. In cascaded networks delay times are added.
9. For a transmission line open circuit and short 13. Z(c) for the network shown in the figure is 16. A series RC circuit has R = 5 Ω and C = 10 μF.
circuit impedances are 20 Ω and 5 Ω. Then The current in the circuit is 5 sin 20000t. The
characteristic impedance is The value of C and R are, applied voltage is
respectively
A. 100 Ω B. 50 Ω C. 25 Ω D. 10 Ω A. 252 sin (20000t + 45°)
Z0 = sq. rt. of Zoc Zsc .
B. 252 sin (20000t - 45°)
10. The terms RMS and average values apply only
to sine waves. C. 252 sin 20000t
A. True B. False D. 252 sin (20000t - 90°)
These terms are applicable to all waveshapes.
A. 1/6 F and 4 Ω B. 2/9 F and 9/2 Ω
11. The drift velocity of electrons is
C. 2/3 F and 1/2 Ω D. 1/2 F and 1 Ω
A. very small as compared to speed of light
14. In an R-C series circuit excited by a voltage E,
B. equal to speed of light the charge across capacitor at t = 0+ is
Z = R2 + XC2 = 50
C. almost equal to speed of light A. 0 B. CE
V = 550 sin(20000 t - 45°)
D. half the speed of light C. CE (1 - e-t/RC) D. none of the above
V = 252 sin(20000 t - 45°)
Drift velocity of electrons is very small. Voltage across a capacitor cannot change
instantaneously.
, 17. A capacitor stores 0.15C at 5 V. Its capacitance D. none of the above
is
i(t) is the incoming current. The currents leaving
A. 0.75 F B. 0.75 μF i(t) = 3e-4t - 2e-3t
node 1 are
C. 0.03 F D. 0.03 μF 23. In the figure shown, A1, A2, A3 are identical
20. In the circuit of figure, the switch is closed at t Ammeters. If A1 and A3 read 5 and 13 A
Q = CV or 0.15 = C(5) or C = 0.03 F. = 0. At t = 0+ the current through C is respectively, reading of A2 will be
18. A 0.5 μF capacitor is connected across a 10 V
battery. After a long time, the circuit current and
voltage across capacitor will be
A. 0.5 A and 0 V B. 0 A and 10 V
C. 20 A and 5 V D. 0.05 A and 5 V
When the capacitor is fully charged, i = 0 and A. 4 A B. 2.5 A
voltage across capacitor is equal to battery A. 8 B. 13 A
voltage. C. 3.1 A D. 0
Capacitor behaves as a short-circuit at t = 0. C. 18 D. 12 A
19. For node 1 in figure, KCL equation is
21. A cable has an insulation resistance of 10^10 A2 = sq. rt.(132 – 25) 12 A.
ohms. If the length is doubled, the insulation
resistance will be 24. Which of the following pairs are correctly
matched?
A. 2 x 10^10 Ω B. 0.5 x 10^10 Ω
1.Brune's realisation : realisation with ideal
C. 10^10 Ω D. 4 x 10^10 Ω transformer.
2.Cauer realisation : ladder realisation.
Insulation resistance of cable is inversely 3.Bott Duffin realisation : with non ideal
A. proportional to length. transformer.
Select the answer using the following codes:
22. , i(t) =
A. 1, 2 and 3 B. 2 and 3
B.
A. 3e^-4t – e^-5t B. 3e^-4t - 2e^-3t
C. 1 and 3 D. 1 and 2
C. 2e^-4t - 5e^-2t D. 3e^-4t – e^-t
Bott Duffins realisation does not use transformer.
C.
