Help with Superpostion Theorem

[Qwme5]

Can someone help me to find I1 and I2 for the following circuit by using superpostion thorem ?

 

[KerimF]

What do you know about the superposition theorem? Please don't tell me... nothing

Let us assume, you missed the class on this subject.

When superposition theorem is used, the response (current or voltage) due to only one independent source (current or voltage source) is obtained at a time.
The other sources are replaced, by either open-circuits (current sources) or short-circuits (voltage sources).
The final result of a response is the algebric sum of its calculated sub-results.

On the circuit here, there are two independent sources... and two responses (currents) to find out.

 

[errakeshpillai]

Superposition theorem states to replace all sources, except one, with equivalent resistance. Then finally add up them to get the total value. So where are you facing problem in this.?

 

[LvW]

Superposition theorem states to replace all sources, except one, with equivalent resistance. Then finally add up them to get the total value. So where are you facing problem in this.?

Are you sure? What is the "equivalent resistance" of a voltage source?

 

[errakeshpillai]

Are you sure? What is the "equivalent resistance" of a voltage source?

Yes, I am sure. But I would like to make a slight modification in my statement. It should be equivalent impedance, not resistance (to be more perfect).
And the equivalent impedance of a Voltage source is 0 ohm, i.e. it should be replaced with a short circuit.

 

[LvW]

Yes, I am sure. But I would like to make a slight modification in my statement. It should be equivalent impedance, not resistance (to be more perfect).
And the equivalent impedance of a Voltage source is 0 ohm, i.e. it should be replaced with a short circuit.

OK - so you mean not really an "equivalent" resistor but simply the internal source impedance of the source, correct?
I think, this clarification is necessary - otherwise the OP (who seems to be not very experienced) will get confused.

 

[errakeshpillai]

OK - so you mean not really an "equivalent" resistor but simply the internal source impedance of the source, correct?
I think, this clarification is necessary - otherwise the OP (who seems to be not very experienced) will get confused.

Yes, I meant exactly the same. Sorry for not being too clear the first time around.
The sources should be replaced with its own equivalent impedance.

 

[Qwme5]

Stop fighting im my post , i need help if u could help me i will be thankful otherwise plz dont reply TQ.

 

[The Electrician]

As a first step, remove the 1.2 amp source and the 1.2Ω resistor and calculate I1 and I2. What do you get?

 

[LvW]

Stop fighting im my post , i need help if u could help me i will be thankful otherwise plz dont reply TQ.

Dear Junior member Qwme5, you should not give such instruction to us.
We all are trying to help you.
However, if there is one answer which could cause misunderstanding on YOUR side, I think it is only YOUR benefit if this item is clarified, don`t you think so?

 

[KerimF]

As a first step, remove the 1.2 amp source and the 1.2Ω resistor and calculate I1 and I2

Could you draw then upload the simplified circuit, now having one voltage source, and give to each node a label as a, b, c etc?

 

[albbg]

As said in the post before, to apply superposition, you have to drawn as many circuit as many generators are present in you original circuit. In each "new" circuit you have to consider just one generator while all the others are replaced by short circuit if they are volatge generator, or open circuit if they are current generator. The for each circuit you will calculate the parameters you are interested in, then the solution of the original circuit will be given by the sum of all the partial solutions. In your case there are 2 generators, then we will have two circuits with just one generator:



I've added the currents at each node as well as the voltages at each branch. We can apply KCL and KVL.

The equation of circuit A are:

Iv=I2+I4
IL=I1+I4
I1=I5+I2
V1=V4-V2
Vg=V2-V5
Vg=V4+VL

Equation for circuit B are:

IL=I1+Ig+I4
I1=I2+I5-Ig
V1=V4-V2
V2=V5
V4=-VL

For both circuits:

V1=R1*I1
V2=R2*I2
V4=R4*I4
V5=R5*I5
VL=RL*IL

Now solving for I1 and I2 the system of circuit A and circuit B you will have I1(A), I2(A), I1(B), I2(B) then

I1 = I1(A)+I1(B)
I2 = I2(A)+I2(B)

please note that R3 plays no role, since it is in series with a current generator (however it dissipates power).

I hope not to have introduced errors in writing some of the equation or forgot some equation (please check with the circuit), however I tried to show you the general method.