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NETWORKS

1. In a series RLC high Q circuit, the current peaks at a frequency

(a) Equal to the resonant frequency

(b) Greater than the resonant frequency

(c) Less than the resonant frequency

(d) None of these.

2. Two two-port networks are connected in cascade. The combination is to be represented as a single two-port network.

The parameters of the network are obtained by multiplying the

individual

(a) z-parameter matrix

(b) h-parameter matrix

(c) y-parameter matrix

(d) ABCD parameter matrix

3. The necessary and sufficient condition for a rational function of s. T(s) to be driving point impedance of an RC network is

that all poles and zeros should be

(a) Simple and lie on the negative axis in the s-plane

(b) Complex and lie in the left half of the s-plane

(c) Complex and lie in the right half of the s-plane

(d) Simple and lie on the positive real axis of the s-plane.

4. (a) Find the laplace transform of the waveform x(t) shown in figure.

(b) The network shown in figure is initially under steady state

condition with the switch in position 1. The switch is moved

from position 1 to position 2 at t 0. Calculate the current i(t)

through R1 after switching.

5. Relative to a given fixed tree of a network,

(a) Link currents form an independent set

(b) Branch voltage from an independent set

(c) Link currents from an independent set

(d) Branch voltage from an independent set

6. For a 2-port network to be reciprocal (a) z11 = z12 (b) y21 = y22

(c) h21 = h12 (d) AD BC = 0

7. For the series R-L circuit of figure (a), the partial fissure diagram at a certain frequency is shown in figure (b). The

operating frequency of the circuit is:

(a) Equal to the resonance frequency

(b) Less than the resonance frequency

(c) Greater than resonance frequency

(d) Not zero.

8. For the compensated attenuator of figure, the impulse response under the condition R1C1 = R2C2 is :

(a) 1 1t

R C2

1 2

R1 e

R Ru(t)

(b)2

1 2

R(t)

R R

(c) 2

1 2

Ru(t)

R R

(d) 1 1t

R C2

1 2

R1 e .u(t)

R R

9. Of the four networks, N1, N2, N3 and N4 of figure, the network shaving identical driving point functions are

(a) N1 and N1 (b) N2 and N4

(c) N1 and N3 (d) N1 and N4.

10. A dc circuit shown in figure has a voltage source Vt a current source I and several resistors. A particular resistor R dissipates

a power of 4 watts when V alone is active. The same resistor R

dissipates a power of 9 watts when I alone is active. The

power dissipated by R when both sources are active will be

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(a) 1 W (b) 5 W (c) 13 W (d) 25 W.

11. In the series circuit shown in figure for series resonance, the value of the coupling K will be

(a) 0.25 (b) 0.5 (c) 0.999 (d) 1.0.

12. If the secondary winding of the ideal transformer shown in the circuit of the figure has 40 turns, the number of turns in the

primary winding for maximum power transfer to the

2 resistor will be

(a) 20 (b) 40 (c) 80 (d) 160.

13. A network contains linear resistors and ideal voltage sources. If values of all the resistors are doubled, then the voltage

across each resistor is

(a) halved (b) doubled

(c) increased by four times (d) not changed.

14. The two electrical sub network N1 and N2 are connected through three resistors as shown in figure. The voltage across

5 ohm resistor 5 ohm resistor and 1 ohm resistor are given to

be 10 V and 5V, respectively. Then voltage across 15 ohm

resistor is

(a) 105v (b) + 105v

(c) 15 v (d) +15 v

15. In figure, A1, A2 and A3 are ideal ammeters, if A1 reads 5A, A2 reads 12 A, then A3 should read.

(a) 7 A (b) 12 A (c) 13 A (d) 17 A.

16. An infinite grid is built up by connecting resistors in the manner indicated in figure, where each branch represents one

ohm resistor. Calculate the effective resistance between the

nodes A and B.

17. Find the Y-parameters (short circuit admittance parameters)for

the network shown in figure.

18. In the following circuit the capacitance varies as C = KQ,

where K is a constant equal to 0.5 Farads/Coulomb and Q, the

charge on the capacitor in coulombs. Determine the current

through the circuit and sketch the voltage waveform across the

capacitor (Vc) for a step input V1 as shown in figure. (initial

voltage across capacitor is + 2V)

19. The Laplace transform of a unit ramp function starting at t = a,

is

(a) 2

1

(s a) (b)

as

2

e

(s a)

(c)

as

2

e

s (d)

2

a

s.

20. A series LCR circuit consisting of R = 10 , LX = 20 and

cX = 20 , is connected across an a.c. supply of 200V rms.

The rms voltage across the capacitor is

(a) 200 90 V (b) 200 + 90 V

(c) 400 90 V (d) 400 + 90 V.

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ANSWERS 1. (a) At resonant frequency , series RLC circuit exhibit

minimum impedance condition, which lead to maximum

current in the circuit.

z = Ro + 1

jwLjwc

, At w = w0 (Resonant frequency),

z = Ro, w0 = 1

LC

2. (d) ABCD parameter matrix. These transmission parameter serve to relate the voltage

and current at one port to voltage and current at the other

port.

3. (b) The poles and zero must be conjugate and they should be in the half of S-plane.

4. (a) Since the given waveform exist from t = 0 to t = 5T .Thus

z(s) =

1T 4T ST

ST st st

0 2T 4T

E dt 2e dt 2e dt

Solving this we get, z(s) = sT 2sT 4sT 5sT1 3e 4e 4e 2e

S

(b) Initial circuit was

Under steady state condition, inductor will offer zero resistance

path. Thus initial current will be 10

5= 2amp.

After moving the switch the position 2

Circuit will become like

The time constant now will be 10

5

Thus

i(t) = ifinal +

Rt

Linitial final finali i e , i 0

Thus

i(t) = 0 + (2 0) e 5t = 2. e 5t amps.

5. (a) 6. Both (B) and (C). 7. When XC > XL, then the frequency is going to be less than

resonance frequency because XC is inversely proportional to

frequency because Xc is inversely proportional to frequency.

8. (b)

9. Z1(S) =

1 12 (1 2S)

S S

1 1 1 12 2 (1 2S) (1 2S)

S S S S

2 3

1 2S 1(1 2S) 2 1

S S 2 SZ (S) , Z (S)

1 2S 12 (1 2S) 1

S S 2 S

4

(1 2S), Z (S)

11 2S

S

Solve the Z1(S), Z2(S), Z3(S), Z4(S) and get the solution.

10.(d) 25 W.

11.(b) 0.25.

12.(c) 80.

13.(d) V

RI

. If R is doubled I becomes half. So, V= 2R 1

2 =

IR

V remains unchanged.

14.(a) 105 V

15.(c) 13 A.

16. Take a rectangular mesh out of the infinite grid. Applying a

test voltage V across A and B

Due to symmetry of infinite grid in all directions current

distribution is as shown.

Applying KVL to loop ACBA. We get

V = I

4R +

I

4R =

IR

2

eq

VR

Iacross A and B = R/2 = 1/2 [ R = 1 ]

17.

18.

19.(c)

as

2

e

s

20.(c) 2

a

s.

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