# Superposition help

1. Apr 17, 2009

### James889

Hi,
I need some help with some circuit analysis.
http://img208.imageshack.us/img208/866/circuitj.png [Broken]

Sorry for the shoddy picture
Problem I need to calculate the voltage between the point P and GND.

So i know that i have to replace the current source with and open circuit
like such:
http://img411.imageshack.us/img411/355/circuit2.png [Broken]

Is it just a matter of using the voltage division principle?
Also im not really sure about the 12,16 and the 8 ohm resistances, are they all in series?

/James

Last edited by a moderator: May 4, 2017
2. Apr 17, 2009

### The Electrician

At this stage of your analysis, the 8 ohm resistor is doing nothing; ignore it.

So you have a couple of cascaded voltage dividers. Do you know how to solve that much?

3. Apr 18, 2009

### James889

I think so.
The total resistance is ((10+5)*(12+16))/(15+28) = 9,76$$\Omega$$
It's a little low, are you sure you don't have to account for the 8 $$\Omega$$ when calculating the total resistance?

So im not sure about the total resistance in the circuit but KVL would yield:

I*10 + I*5 = 12 and
I*10 + I*12 + I*16 = 12

4. Apr 18, 2009

### The Electrician

The right end of the 8 ohm resistor isn't connected to anything, so it can't possibly have any effect on the circuit.

Your calculation for the total resistance seen by the 12 volt source is incorrect.

Calculate it like this:

The 12 and 16 ohm resistors can be replaced by a single 28 ohm resistor, which is in parallel with the 5 ohm resistor. Then the 5 ohm resistor can be replaced by a (28*5)/(28+5) = 4.2424 ohm resistor. This is in series with the 10 ohm resistor, so the total resistance seen by the 12 volt source is 14.2424 ohms. The current out of the 12 volt source is therefore 12/14.2424 = .84255 amps. This current causes an 8.4255 volt drop across the 10 ohm resistor, so the voltage at the junction of the 10, 5 and 12 ohm resistors is 3.5745 volts. Then the 12 and 16 ohms resistors act as a voltage divider, so that the voltage at point P is 3.5745*(16)/(12+16) = 2.04255 volts.

Or, you could solve your problem using the loop method.

If you are are going to have 2 KVL equations, then you must have two separate currents. You can't just call all the currents by the single symbol I; you need to use I1 and I2.

You haven't shown which loops you are using, but I think I can guess which they are.

(I1+I2)*10 + (I1)*5 = 12
(I1+I2)*10 + (I2)*(12+16) = 12

which can be rearranged to give:

(10+5)*I1 + (10)*I2 = 12
(10)*I1 + (10+12+16)*I2 = 12

If you solve these, you get I2 (this current passes through the 16 ohm resistor). Its value is .12766 amps. Then the voltage at point P is .12766 * 16 = 2.04255 volts.

This is the first part of your superposition problem. Next you need to replace the 12 volt source with a short and calculate the voltage at point P due to the 5 amp current source, which you add to 2.04255 volts.

5. Apr 18, 2009

### James889

Thank you very much.
Just one last thing regarding the loop method.
So if there were three meshes you'd need to have three equations with three different currents?

6. Apr 18, 2009

### The Electrician

Generally speaking, yes. If there are dependent sources or some other complication, that may change.

Using the nodal method, you will generally need as many equations (and variables) as there are nodes.