# Superposition Principle (EE)

1. Feb 16, 2014

### Tekneek

1. The problem statement, all variables and given/known data

I have to find voltage at point 2 using superposition method. So, I drew one circuit with no voltage source and another with no current source, and then tried to analyze each one separately.

(On the picture starting from left, its 10 volts, then 10ohms, then at top 5 ohms, then 1 amp current source pointing down, then on far right 20 ohms)

3. The attempt at a solution

For No Voltage Source:

Using Node Voltage analysis at V2:
(V2-0)/5 + 1 + (V2-0)/20 = 0
V2= -4 Volts

For No current Source:

Using Voltage divider formula:
V2 = (5/(5+20))*10 = 2 Volts

Adding the two V2's i get V2 = -2, which i think is wrong.

#### Attached Files:

• ###### IMG_20140216_155716.jpg
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2. Feb 16, 2014

### Staff: Mentor

Hi Tekneek. This line doesn't look right ...

3. Feb 16, 2014

### Tekneek

For the no current source circuit isn't V1 10 volts? Then since 5ohms and 20ohms were in series i used the voltage divider formula. Also the 10 ohms resistor and 5+20 ohms are in parallel so they would have the same voltage (10v)?

4. Feb 16, 2014

### FOIWATER

Voltage divider works 'oppositely' to current divider.

Last edited: Feb 16, 2014
5. Feb 16, 2014

### Tekneek

What do you mean by that? Isn't the total voltage across the resistors in series divided?

6. Feb 16, 2014

### FOIWATER

The rules require I can only offer moderate assistance. Based on where you managed to get in this problem, I think you understand this but are simply not seeing it. Look at what I have attached

**edit that 30 should be a 20, my eyes are not good

#### Attached Files:

• ###### voltage divider.png
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7. Feb 16, 2014

### Tekneek

Oh ok so V2 is the voltage across 20ohms resistor (for circuit with no current source). Thank You!

8. Feb 16, 2014

### FOIWATER

Yeah that should make sense looking at where you have placed the ground, right?

You're welcome, I am glad you came to that conclusion on your own as well!