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Which theorem assists in replacement of an impedance branch over the network by the other network comprising different circuit components, without affecting the VI relations throughout the entire network?
A simple equivalent circuit of the 2 terminal network shown in fig. is
After killing all source equivalent resistance is R
Open circuit voltage = v_{1}
The short circuit current across the terminal is
For the calculation of R_{TH} if we kill the sources
then 20Ω resistance is inactive because 5 A source will
be open circuit
RTH = 30+25 = 55 Ω,
vTH = 5+5X30 = 155V
After killing the source, R_{TH} = 6 Ω
The Thevenin impedance across the terminals ab of the network shown in fig. is
After killing all source,
For In the the circuit shown in fig. a network and its Thevenin and Norton equivalent are given
The value of the parameter are
V_{TH }R_{TH} I_{R } R_{R}_{N}
v∝ = 2 x 2 + 4 = 8 V = v_{TH}
R_{TH} = 2 + 3 = 5Ω = R_{N},
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
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. Find the Thevenin equivalent as given in question..
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. Find theThevenin 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 50Ω load and 20 kW to a 200 Ω load. The maximum power, that can drawn from it, is
(r + 200)^{2} = 4(r + 50)^{2}
⇒ r = 100 Ω
i = 30A,
In the circuit of fig. P.1.4.15–16 when R = 0 Ω , the current i_{R} equals 10 A.
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 when R = 0 Ω , the current iR equals 10 A.
The maximum power will be
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 2Ω, the power dissipated by the bulb is
r_{battery }= V_{oc} / I_{sc} = 24/30 = 0.8 ohms
P = V^{2}_{oc} / ( r + 2)^{2 } * 2
= 24^{2 }/ ( .8 + 2)^{2 }* 2
= 146.9 W
The following results were obtained from measurements taken between the two terminal of a resistive network
The Thevenin resistance of the network is
A DC voltmeter with a sensitivity of 20 kΩ/V is used to find the Thevenin equivalent of a linear network. Reading on two scales are as follows
(a) 0  10 V scale : 4 V
(b) 0  50 V scale : 5 V
The Thevenin voltage and the Thevenin resistance of the network is
= 50 μ A
For 0  10 V scale R_{m} = 10 x 20k = 200 kΩ
For 0  50 V scale R_{m} = 50 x 20k = 1 MΩ
For 4 V reading
v_{TH} = 20μR_{TH} + 20µ x 200k = 4 + 20μR_{TH } ...(i)
For 5V reading
U_{TH }= 5μ x R_{TH} + 5μ x 1M = 5 + 5μR_{TH} ...(ii)
Solving (i) and (ii)
Consider the network shown in fig.
The power absorbed by load resistance R_{L} is shown in table :
Find the Thevenin equivalent.
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