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QUESTION: 1

i_{1} = ?

Solution:

If we solve this circuit direct, we have to deal with three variable. But by simple manipulation variable can be reduced to one. By changing the LHS and RHS in Thevenin equivalent

QUESTION: 2

A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question..

Q. As viewed from terminal x and x' is

Solution:

We Thevenized the left side of xx' and source transformed right side of yy'

QUESTION: 3

A circuit is given in fig. P.1.4.12–13. Find the Thevenin equivalent as given in question..

As viewed from terminal y and y' is

Solution:

Thevenin equivalent seen from terminal yy' is

QUESTION: 4

A practical DC current source provide 20 kW to a 50 Ω load and 20 kW to a 200 Ω load. The maximum power, that can drawn from it, is

Solution:

QUESTION: 5

In the circuit of fig. P.1.4.15–16 when R = 0 Ω , the current i_{R} equals 10 A.

Q. The value of R, for which it absorbs maximum power, is

Solution:

Thevenized the circuit across R, R_{TH} = 2 Ω

QUESTION: 6

In the circuit of fig. P.1.4.15–16 when R = 0 Ω , the current i_{R} equals 10 A.

Q. The maximum power will be

Solution:

QUESTION: 7

If v_{s1} = 6 V and v_{s 2} = -6 V then the value of v_{α} is

Solution:

Since both source have opposite polarity, hence short circuit the all straight-through connection as shown in fig. S.1.4.33

QUESTION: 8

A network N feeds a resistance R as shown in fig.P1.4.34. Let the power consumed by R be P.If an identical network is added as shown in figure, the power consumed by R will be

Solution:

Let Thevenin equivalent of both network

QUESTION: 9

A certain network consists of a large number of ideal linear resistors, one of which is R and two constant ideal source. The power consumed by R is P_{1} when only the first source is active, and P_{2} when only the second source is active. If both sources are active simultaneously, then the power consumed by R is

Solution:

using superposition

QUESTION: 10

A battery has a short-circuit current of 30 A and an open circuit voltage of 24 V. If the battery is connected to an electric bulb of resistance 2 Ω, the power dissipated by the bulb is

Solution:

0.8

146.93

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