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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
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
We Thevenized the left side of xx' and source transformed right side of yy'
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
Thevenin equivalent seen from terminal yy' is
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
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
Thevenized the circuit across R, R_{TH} = 2 Ω
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
If v_{s1} = 6 V and v_{s 2} = 6 V then the value of v_{α} is
Since both source have opposite polarity, hence short circuit the all straightthrough connection as shown in fig. S.1.4.33
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
Let Thevenin equivalent of both network
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
using superposition
A battery has a shortcircuit 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
0.8
146.93
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