v_{TH ,} R_{TH} = ?
I_{N , }R_{N} = ?
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
_{VTH,} RTH = ?
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
R_{TH }= ?
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},
v_{1} = ?
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
i_{1 = } ?
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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