Finding Current in a Thevenin Circuit

In summary: With the load removed you are trying to find the voltage drop across R2 and you already know the current through it is zero.
  • #1
miguelcard
2
0

Homework Statement



Given a circuit with a total of 3 sources , 2 Voltage sources and 2 Current sources and 4 resostors, Find the Cirrent I2 passing through the branch where R2 and V2 are. And I am asked to give the formula for the value of the voltage in the current source. Attached the picture of the circuit .
Now, the problem is simple I don't know if I should I take the load just as R2 or as R2 plus the voltage source V2. and I am having trouble on how to get the voltage of the current source

Homework Equations


R thevenin and V thevenin

The Attempt at a Solution


for the thevenin resitance I turned Isources= open loop, Vs= short circuit, ad got an Rth =R1 but I am not sure if that is right.
for the thevenin voltage... I took just R2 as my load and when I take out the load and fond the voltage drop for U2 which is in the same branch it gives me a non sense
 

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  • #2
Hi miguelcard,
Welcome to Physics Forums!

It's not clear to me how useful Thevenin's Theorem will be for this problem since it appears that you need to find circuit values (a voltage and a current) in two different locations in the circuit. In using Thevenin you have to decide on where the load is and everything else gets reduced to a single voltage and resistance; you lose access to all the individual components and circuit paths inside the "black box".

If it were my choice I'd probably go for superposition theorem or nodal analysis.

Were you given any component values for the circuit or are you meant to give symbolic results only?
 
  • #3
gneill said:
Hi miguelcard,
Welcome to Physics Forums!

It's not clear to me how useful Thevenin's Theorem will be for this problem since it appears that you need to find circuit values (a voltage and a current) in two different locations in the circuit. In using Thevenin you have to decide on where the load is and everything else gets reduced to a single voltage and resistance; you lose access to all the individual components and circuit paths inside the "black box".

If it were my choice I'd probably go for superposition theorem or nodal analysis.

Were you given any component values for the circuit or are you meant to give symbolic results only?
Hi thanks for your reply! ... The exercise is meant to be solved by showing the symbolic equations only, and yea I know how to find the values through nodal analysis, but this exercise says "find current I2 strictly using a thevenin equivalent" and then as a minor point it asks for "Value of UI ? (formula)" so I thought this voltage value of a current source could be found easily after, but I don't know how to do it this way
 
  • #4
Okay, well then I suppose you can treat it as essentially two separate problems.

For the first problem (finding I2), your approach looks okay. But check your Thevenin resistance: you said it was equal to R3 in your post above, but I don't see that in your suppressed source circuit:

upload_2017-2-21_15-3-20.png


miguelcard said:
for the thevenin voltage... I took just R2 as my load and when I take out the load and fond the voltage drop for U2 which is in the same branch it gives me a non sense
Can you find the potential drop across R1 when the load is removed?
 
  • #5
gneill said:
Okay, well then I suppose you can treat it as essentially two separate problems.

For the first problem (finding I2), your approach looks okay. But check your Thevenin resistance: you said it was equal to R3 in your post above, but I don't see that in your suppressed source circuit:

View attachment 113537Can you find the potential drop across R1 when the load is removed?
I mean Rth= R1 sorry about that, because the current will go through the short circuit instead of going through the other 2 resistors making it the total resistance, correct me if I am wrong, and now that you say taking them like different loads, for finding the current I2 would it be better to take the whole branch R2 + Vsource U2 as a load or just R2?
 
  • #6
miguelcard said:
I mean Rth= R1 sorry about that, because the current will go through the short circuit instead of going through the other 2 resistors making it the total resistance, correct me if I am wrong, and now that you say taking them like different loads, for finding the current I2 would it be better to take the whole branch R2 + Vsource U2 as a load or just R2?
I'd just take R2 as the load.
 

Related to Finding Current in a Thevenin Circuit

1. What is a Thevenin circuit?

A Thevenin circuit is a type of electrical circuit that is commonly used to simplify complex circuits. It consists of a voltage source, a series resistor, and a load resistor. The Thevenin circuit is used to find the equivalent voltage and resistance of a more complex circuit.

2. How do I find the current in a Thevenin circuit?

To find the current in a Thevenin circuit, you can use Ohm's Law (I = V/R) where I is the current, V is the voltage, and R is the equivalent resistance of the circuit. You can also use the current divider rule, which states that the current in a branch of a parallel circuit is equal to the total current multiplied by the resistance of that branch divided by the total resistance of the circuit.

3. Can I use a Thevenin circuit to analyze AC circuits?

Yes, a Thevenin circuit can be used to analyze both DC and AC circuits. However, the calculations may be slightly different for AC circuits as they involve complex numbers and impedance.

4. What is the purpose of finding the current in a Thevenin circuit?

The primary purpose of finding the current in a Thevenin circuit is to determine the behavior of the circuit and to simplify complex circuits for analysis. This information can also be used to calculate power dissipation and to ensure that the circuit operates within safe limits.

5. Are there any limitations to using a Thevenin circuit to analyze circuits?

While Thevenin circuits are useful for simplifying complex circuits, they do have limitations. They are only accurate for linear circuits that contain only resistors, voltage sources, and current sources. They also assume that the circuit is in a steady-state, meaning that all voltages and currents have reached their maximum values and are not changing over time.

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