# Question : Thevenins Equivalent

• gatsbycollege
In summary, the circuit shown is trying to find the Thevenins Equivalent at the terminals a and b. The attempt at a solution involved using nodal analysis for Vth, but it was unclear if the 5 ohms and 8 ohms resistors had current flowing through them. For Rth, the current source was changed into an open circuit and solved by finding the equivalent resistance using 40 and 20 in series and parallel with 10 ohms. The final values for Vth and Rth were 143.52 and 78 ohms, respectively. Mesh analysis was then used to find the short circuit current (Isc) and plug it into the equation Vth = Isc * Rth, resulting in Is
gatsbycollege

## Homework Statement

http://img716.imageshack.us/img716/8021/69925791.jpg

Find The Thevenins Equivalent of the circuit shown .
at the terminal a and b i think

Vth = Rth(Ith)
V= IR

## The Attempt at a Solution

For Vth
i tried using Nodal Analysis Having 4 Node Voltages

but i don't know if the 5 ohms resistor and 8 ohms resistor is having flow of current in there?

For Rth

the Current source is change into open ckt.
then solve by having 40 and 20 in series then parallel in 10 ohms
therefor [(40+20)x10]/70 + 45 + 5 + 8 = 78 ohms = RthSo I am wondering how should i get the Vth now,,
And is Vth = Va + 35V and Vth = Vb +15 V ? or I = (Va + 35 - Vth) / 5 ohms

Last edited by a moderator:
For Vth
i tried using Nodal Analysis Having 4 Node Voltages
It might be easier to find the short circuit current, ISC, using mesh analysis: 3 currents (1 of them you already know-->2 equations). Then VTH = ISCRTH.

For Rth

the Current source is change into open ckt.
then solve by having 40 and 20 in series then parallel in 10 ohms
therefor [(40+20)x10]/70 + 45 + 5 + 8 = 78 ohms = Rth
Not sure what you are doing here. Like, where's the 10 ohms coming from? The 70? Somehow, your answer is correct.

lewando said:
It might be easier to find the short circuit current, ISC, using mesh analysis: 3 currents (1 of them you already know-->2 equations). Then VTH = ISCRTH.

Not sure what you are doing here. Like, where's the 10 ohms coming from? The 70? Somehow, your answer is correct.

oh sorry that 10 should be 30 ,,

thx for the answer,, i got my Isc = 1.84 and Vth = 143.52

tried to use that mesh analysis

i got equations

60V - 90 Ia - 40V +30 Isc = 0
35V - 15V - Isc (5 +30 +45 +8) + 30 Ia +90 =0
then by elimination
igot Isc , correct me if I am wrong thanks

## 1. What is Thevenin's equivalent circuit?

Thevenin's equivalent circuit is a simplified representation of a complex network that contains an independent voltage source, a series resistor, and an output load. It is used to analyze the behavior of a circuit without having to consider the internal details of the network.

## 2. How do I calculate Thevenin's equivalent resistance?

The Thevenin equivalent resistance is calculated by removing all sources from the original circuit and measuring the resistance between the output terminals. This can be done using a multimeter or by using Ohm's law to calculate the resistance.

## 3. What is the purpose of finding Thevenin's equivalent circuit?

Thevenin's equivalent circuit is used to simplify complex networks and make it easier to analyze their behavior. It can also be used to determine the maximum power that can be delivered to a load and to find the most efficient matching load for a given network.

## 4. How is Thevenin's equivalent circuit different from Norton's equivalent circuit?

Thevenin's equivalent circuit uses a voltage source and a series resistor, while Norton's equivalent circuit uses a current source and a parallel resistor. Both circuits provide a simplified representation of a complex network, but they are mathematically equivalent and can be converted into one another.

## 5. Can Thevenin's equivalent circuit be applied to all types of circuits?

Thevenin's equivalent circuit can be applied to linear circuits, meaning that the response of the circuit is directly proportional to the input. It is not applicable to non-linear circuits, where the response is not proportional to the input, or to circuits with active components such as transistors or operational amplifiers.

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