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QUESTION: 1

[T] = ?

Solution:

Let I_{3} be the clockwise loop current in center loop

V_{2} = 4(I_{2} + I_{3}) ⇒ I_{3} = 0.25V_{2} - I_{2}

⇒ I_{1} = 0.35V_{2} - I_{2} .........(i)

V_{1} = 4I_{1} - 0.2V_{1} + V_{2}

1.2V_{1} = 4(0.35V_{2} - I_{2}) + V_{2} = 2.4V_{2} - 4I_{2}

⇒ V_{1} = 2V_{2} - 3.33I_{2} .........(ii)

QUESTION: 2

[H] = ?

Solution:

..........(i)

QUESTION: 3

[Y] = ?

Solution:

I_{1} = (V_{1} - V_{2}) Y_{ab} + V_{1}Y_{ab}

⇒ I_{1} = V_{1}(Y_{a} + Y_{ab}) - V_{2}Y_{ab} .....(i)

I_{2} = (V_{2} - V_{1})Y_{ab} + V_{2}Y_{b} = -V_{1}Y_{ab} + V_{2}(Y_{b} + Yab) .....(ii)

QUESTION: 4

The y-parameters of a 2-port network are

A resistor of 1 ohm is connected across as shown in fig. The new y –parameter would be

Solution:

y-parameter of 1Ω resistor network are

New y-parameter

QUESTION: 5

For the 2-port of fig.

The value of v_{o}/v_{s} is

Solution:

V_{1} = 600I_{1} + 100I_{2} , V_{2} = 100I_{1} + 200I_{2}

V_{s} = 60I_{1} + V_{1} = 660I_{1} + 100I_{2} , V_{2} = V_{o} = -300I_{2}

_{}

QUESTION: 6

The T-parameters of a 2-port network are

If such two 2-port network are cascaded, the z –parameter for the cascaded network is

Solution:

.......(i)

........(ii)

QUESTION: 7

[Y] = ?

Solution:

QUESTION: 8

Which among the following is regarded as short circuit forward transfer admittance?

Solution:

QUESTION: 9

h_{21} = ?

Solution:

QUESTION: 10

In the circuit shown in fig. P.1.10.34, when the voltage V_{1} is 10 V, the current I is 1 A. If the applied voltage at port-2 is 100 V, the short circuit current flowing through at port 1 will be

Solution:

Interchaning the port

QUESTION: 11

For a 2-port symmetrical bilateral network, iftransmission parameters A = 3 and B = 1Ω , the value ofparameter C is

Solution:

For symmetrical network A = D = 3

For bilateral AD - BC = 1, 9 - C = 1,C = 8 S

QUESTION: 12

A network has 7 nodes and 5 independent loops. The number of branches in the network is

Solution:

QUESTION: 13

The circuit shown in fig. is reciprocal if a is

Solution:

V_{1} = 0.5V_{1} + I_{1} + 2(I_{1} + I_{2}) + aI_{1}

⇒ V_{1} = (6 + 2a)I_{1} + 4I_{2} ......(i)

V2 = 2(I_{1} + I_{2}) + aI_{1} ⇒ V_{2} = (2 + a)I_{1} + 2I_{2} ......(ii)

For reciprocal network

z_{12} = z_{21}, 4 = 2 + a ⇒ a = 2

QUESTION: 14

Z_{in} = ?

Solution:

I_{1} = 4 x 10^{3}V_{1} -0.1 x 10^{-3}V_{2}

I_{2} = 50 X 10^{-3}V_{1} + 10^{-3}V_{2} ,V_{2} = -10^{3}I_{2}

- 10^{-3}V2 = 50 x 10^{-3}V_{1} + 10^{-3}V_{2}, V_{2} = -25V

QUESTION: 15

v_{1} ,v_{2} = ?

Solution:

I_{1} = 10 x 10^{-3}V_{1} -5 x 10^{-3}V_{2},

100 = 25I_{1} + V_{1}

100 - V_{1} = 0.25V_{1} - 0.25V_{2} ⇒ 800 = 10V_{1} - V_{2 } ....(i)

I_{2} = 50 x 10^{-3}V_{1} + 20 x 10^{-3}V_{2}, V_{2} = -100I_{2}

V_{2} = -5V_{1} -V_{2} ⇒ 3V_{2} + 5V_{1} = 0 ....(ii)

From (i) and (ii) V_{1} = 68.6 V, V2 = -114.3 V.

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