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I'm trying to analyze this circuit and spend almost 2 days to make a formula for it but I don't get it , anyone know what it's called or how can I make formula for it ? I want to calculate I1,I2,I3 currents but their rotation change at certain points .I used the mesh currents method and branch currents method but when the current rotation change equation change too . My goal is make a program in my calculator (R1,R2,R3,VIN,VSUPPLY adjustable) so I can get results in future enter image description here

enter image description here

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Your circuit can be solved using the superposition principle. First set Vbatt equal to 0 and calculate the effect of Vm which is simply a voltage divider consisting of R5 and the parallel combination of R6 and Ro. Then set Vm equal to 0 and calculate the effect of Vbatt which is simply a voltage divider consisting of R6 and the parallel combination of R5 and Ro. What you have is basically an adding circuit where the output is a weighted sum of Vbatt and Vo.

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  • \$\begingroup\$ Do you know name of the circuit ? I will be really upset if solution is that simple , I was trying to calculate it for a long time .Thanks \$\endgroup\$
    – Mordecai
    Commented Aug 26, 2019 at 2:17
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    \$\begingroup\$ As I already mentioned, this configuration is usually called an adding circuit since the output is a weighted sum of the two inputs. \$\endgroup\$
    – Barry
    Commented Aug 26, 2019 at 2:33
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I solved the righthand side (below) picture.

Well, in order to calculate \$\text{R}_\text{th}\$ we get:

$$\text{R}_\text{th}=\frac{1}{\frac{1}{1000}+\frac{1}{1200}+\frac{1}{4700}}=\frac{282000}{577}\approx488.735\space\Omega\tag1$$

And, the short current is:

$$\text{I}_\text{th}=\frac{5}{1000}+\frac{6}{1200}=\frac{1}{100}=0.01\space\text{A}\tag2$$

So:

$$\text{V}_\text{th}=\text{I}_\text{th}\cdot\text{R}_\text{th}=\frac{1}{100}\cdot\frac{282000}{577}=\frac{2820}{577}\approx4.88735\space\text{V}\tag3$$

I solved the lefthand side (above) picture.

Well, in order to calculate \$\text{R}_\text{th}\$ we get:

$$\text{R}_\text{th}=\frac{1}{\frac{1}{1000}+\frac{1}{1200}+\frac{1}{4700}}=\frac{282000}{577}\approx488.735\space\Omega\tag4$$

And, the short current is:

$$\text{I}_\text{th}=\frac{5}{1000}-\frac{6}{1200}=0\space\text{A}\tag5$$

So:

$$\text{V}_\text{th}=\text{I}_\text{th}\cdot\text{R}_\text{th}=0\cdot\frac{282000}{577}=0\space\text{V}\tag6$$

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  • \$\begingroup\$ This looks like you applied Norton’s theorem to the 3 resistor voltage divider. \$\endgroup\$
    – Daniel K.
    Commented Feb 12 at 11:28
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Applying the superposition principle (as described in Barry's answer), you get the 3 Resistor Voltage Divider Formula:

$$\text{U}_\text{out}=(\frac{\text{U}_1}{\text{R}_1}+\frac{\text{U}_2}{\text{R}_2})(\text{R}_1 \parallel \text{R}_2 \parallel \text{R}_3) $$

with

$$ \text{R}_1 \parallel \text{R}_2 \parallel \text{R}_3 = \frac{1}{\frac{1}{\text{R}_1} + \frac{1}{\text{R}_2} + \frac{1}{\text{R}_3} } = \frac{\text{R}_1 \text{R}_2 \text{R}_3}{\text{R}_1 \text{R}_2 + \text{R}_1 \text{R}_3 + \text{R}_2 \text{R}_3} $$

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