Please help me to solve this circuit

[satiz]

Hi
Please explain me to get the vout for the attached circuits

 

[BradtheRad]

Voltage at the central node is influenced by three voltages. Each sends a certain voltage through a given resistance.

Treat the resistors as proportions of an effective resistance. Calculate it similarly to the parallel resistance formula:

1 / (1/R1+1/R2+1/R3)

Divide this by each resistance. This gives you a weighting proportion for each leg. It is different for each leg. (Unless of course you start with equal resistors. Suppose they are all 100k, then each leg's weighting proportion is 1/3.)

Multiply each weighting proportion by the voltage coming through it.
Then add up all three values, to get resulting V at central node.

 

[E-design]

Here is the solution to the first circuit, you can calculate the other one from this.

{DC result:}
R1:=82E3, R2:=910, R3:=300E3, R4:=100E3, VS1:=3.3, VS2:=12
Vout :=(R4*R1)/(R2*R1+R3*R1+R4*R1+R4*R2+R4*R3)*VS2+(R2*R4+R3*R4)/(R2*R1+R3*R1+R4*R1+R4*R2+R4*R3)*VS1;

Vout=[3.1398]

 

[c_mitra]

I hope you are not getting your home work done from us. This is a very simple exercise given to school exams.

Follow the basic steps and try to understand the basic principles:

1. 910 and 300K are two resistors in series. They can be replaced by their equivalent resistances without any loss of generality. We replace the series combination as 300.910K

2. Ignore the 100K resistor for the time being; you have 3.3V and 12V sources connected: the voltage at the junction can be calculated from the current: (pot diff)/(tot res)=(12-3.3)/(82K+300.910K);
then the voltage drop across the 82K will be 82K*(12-3.3)/(82K+300.910K) and the actual potential will be 3.3+82K*(12-3.3)/(82K+300.910K) (say some value V1)

3. Looking into V1 you see two resistors in parallel (82K and 300.910K): this becomes the source resistance for V1.

4. I guess you can now work out the rest.