# Find Thevenin Equiv at (a,b): Troubleshoot Currents

• gfd43tg
In summary, when you add the 1 volt test voltage at terminals a-b, you now have 5 meshes. You need the current in that 5th mesh to calculate Rth.
gfd43tg
Gold Member

## Homework Statement

Find the Thevenin equivalent circuit at terminals (a,b) of the circuit below:

## The Attempt at a Solution

my currents aren't adding up right. When I try to solve for Ix using my two equations that should find it, I get different answers.

#### Attachments

• 3.10.png
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• 3.10 attempt 1.pdf
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When you add the 1 volt test voltage at terminals a-b, you no longer have 4 meshes; you now have 5.

Your top schematic shows 4 meshes, and this allows you to calculate Vth, because the 2Ω resistor has no effect when the output is open circuited. Vth is just the voltage across the 8Ω resistor.

But as soon as you add the 1 volt test voltage as in the bottom schematic, you have 5 meshes. You need the current in that 5th mesh to calculate Rth.

You could solve it with only 4 meshes by assuming a short across a-b; this then replaces the 8Ω resistor with the parallel combination of 2Ω and 8Ω. Solving your 4 equations would give you the current through that parallel combination. Then use the current divider rule to calculate the current through the 2Ω resistor; that would be the short circuit current through a-b.

If you solve this using the nodal method, you only have 3 equations to deal with. That's what I meant when I said "You usually can use any network analysis method. One method may be easier for a given circuit than another."

I'm doing the proper amount of meshes, and now my voltage is high and the resistance seems way too high. I just put those equations into a 5x5 and 4x4 solver. Here is my 2nd attempt at it

#### Attachments

• 3.10 attempt 2.pdf
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Your 4th equation is incorrect. Your units are wrong. I3 is a current and 8*I4 is a voltage.

What you need is to say I4-I3 = 2*Ix, then set Ix = I1-I2

Give that a try, and make the same correction in your 5 mesh equations.

Your 5th equation for the 5 mesh system should be -8*I4 + 10*I5 = -1

With the dependent current source there I don't know how to express the voltage drop across it. I redid it and got 75.4 V. my first 3 equations are unchanged but the last two are now

2 i1 - 2 i2 + i3 - i4 = 0

then for the 5 mesh system similarily, I changed to the same and what you said about -8 i4 + 10i5 = -1

Last edited:
The voltage across the dependent current source is the same as the voltage across the 8Ω resistor in the 4 mesh case; that would be 8*I4 in the 4 mesh case. Your last two equations look good now, but I don't get 75.4 volts for that voltage.

Show the 4 equations for the 4 mesh case and the values you get for the 4 currents, and I should be able to see what the difficulty is.

Also, what do you get for the currents in the 5 mesh case?

## 1. What is the Thevenin equivalent circuit?

The Thevenin equivalent circuit is a simplified representation of a complex circuit that contains only a voltage source and a resistor in series. It can be used to analyze the behavior of a circuit at a particular point, without having to consider the rest of the circuit.

## 2. How do I find the Thevenin equivalent at a specific point in the circuit?

To find the Thevenin equivalent at a specific point (a,b) in the circuit, you need to follow these steps:
1. Disconnect all the sources in the circuit.
2. Calculate the equivalent resistance between point (a,b) and the rest of the circuit.
3. Reconnect the sources and calculate the open-circuit voltage at point (a,b).
4. The Thevenin voltage is the open-circuit voltage, and the Thevenin resistance is the equivalent resistance calculated in step 2.

## 3. Why is it important to find the Thevenin equivalent?

The Thevenin equivalent allows us to simplify a complex circuit and analyze its behavior at a specific point. This can be useful in troubleshooting and designing circuits, as well as understanding the overall behavior of the circuit.

## 4. Can the Thevenin equivalent be used in all types of circuits?

No, the Thevenin equivalent can only be used in linear circuits, which means that the current-voltage relationship follows Ohm's law. Non-linear elements, such as diodes and transistors, cannot be represented using the Thevenin equivalent.

## 5. How do I troubleshoot currents in a circuit using the Thevenin equivalent?

To troubleshoot currents in a circuit using the Thevenin equivalent, you can follow these steps:
1. Find the Thevenin equivalent at the point where you need to troubleshoot the current.
2. Use the Thevenin equivalent to calculate the current in the circuit.
3. Compare the calculated current with the measured current to determine if there is a problem in the circuit.
4. If the calculated current is significantly different from the measured current, you can use the Thevenin equivalent to determine which component or connection in the circuit may be causing the issue.

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