Thevenin equivalent circuit help

[full123]

http://img855.imageshack.us/img855/5369/eehmwk.jpg


I haven't done this before with 2 voltage sources so I'm not sure how I would go about doing it.

If someone could show me step by step how to understand this, I would appreciate it.

Equivalent resistance = 2.5ohms, right?
But what about Vs?

 

[gneill]

http://img855.imageshack.us/img855/5369/eehmwk.jpg


I haven't done this before with 2 voltage sources so I'm not sure how I would go about doing it.

If someone could show me step by step how to understand this, I would appreciate it.

Equivalent resistance = 2.5ohms, right?
But what about Vs?

There are several different ways to solve this circuit. What circuit theorems have you been introduced to?

 

[NascentOxygen]

But what about Vs?

Can you determine the voltage between a and b on the main circuit? That becomes Vs in the Thevenin equivalent.

 

[Su3liminal]

2 or 3 voltages, it doesn't matter. It is the same approach most of the time
Find the voltage across AB, that is: Vab (using nodal analysis). Then find the short-circuit current Isc, that is in the direction of the voltage DROP across AB (using mesh current method). Then it is a matter of using Ohm's law to find Rth and constructing Thevenin equivalent circuit.
Of course you don't have to have to find the current Isc (which is cumbersome used often), since you can replace the Voltage sources by short circuits and then find the equivalent resistance Rth. But I was just suggesting to use the formal approach.

 

[DeeNos]

Thevenin equivalent circuit is a useful concept in circuit analysis where we can simplify a complex circuit into a single equivalent circuit. This is particularly useful when we want to analyze the behavior of a circuit without having to deal with its intricate details. In this case, the Thevenin equivalent circuit helps us to understand the behavior of a circuit with two voltage sources.

To find the Thevenin equivalent circuit, we need to find the equivalent resistance (Rth) and the equivalent voltage source (Vth). The equivalent resistance is calculated by shorting all voltage sources and open-circuiting all current sources in the original circuit. In this case, the equivalent resistance is indeed 2.5 ohms, as you correctly mentioned.

To find the equivalent voltage source, we need to consider the voltage at the open-circuited terminals of the original circuit. In this case, the open-circuited terminals are the two points where the 2.5 ohm resistor meets the two voltage sources. Since there are two voltage sources, we need to use the voltage divider rule to find the voltage at these points. The voltage divider rule states that the voltage at a point is equal to the ratio of the resistance connected to that point to the total resistance in the circuit, multiplied by the total voltage in the circuit. In this case, the voltage at these points will be:

Veq = (2.5 ohms / (2.5 ohms + 2.5 ohms)) * 10V = 5V

Therefore, our Thevenin equivalent circuit will have an equivalent resistance of 2.5 ohms and an equivalent voltage source of 5V. This means that the original circuit can be simplified into a single voltage source of 5V in series with a 2.5 ohm resistor.

I hope this helps you understand the concept of Thevenin equivalent circuit better. If you have any further questions, please feel free to ask.