Thevenin, Norton and superposition theorem

[rds1975]

Thanks to you all friends for the response. Sorry that I did not give you one more information. In the above circuit, I wanted to find current in 8 ohms resistor and not 10 ohms resistor. Sorry once again.
The answer should be some where around 0.24 A , but iam not getting the steps to do that. Could you please help me once again
Could you please explain me solving by thevenin, norton and superposition theorem.

Thanks

 

[crust]

I did it quick and dirty and got .23A through the 8ohm. I could have easily made a mistake however. The ideas of norton & thevenin is to make a transform that usually allows you to combine two components and thus lower the number of elements overall. You do this a couple of times until you get to something easy to work. Superposition is a property of linear systems that allows you to independently analyze the output variable by examining each forcing function's effect on it independently. So if you have a light bulb (1 ohm) in series with two voltage sources, say 10 and 20 volts. The total voltage across the bulb is 30V, so the current is 30A. But you could say the current in the bulb due to the 10V source is 10A and the current due to the 20V is 20A, so the total current must be 30A. In practice, you short the voltage sources and open current sources you are *not* analyzing and repeat.

In any case, this circuit is not too bad. From left to right, I reduce the 4V/5O to its norton, combined it with the 15O. Then convert that to a thev and combine with the 10O. Then convert that whole thing to a norton equivalent. From the right convert the 4V and 12O to a norton equivalent. Now you have a circuit with two sources and 3 parallel Rs in the middle. You can combine the Rs to find the total resistance. Now the voltage at the top node is the current from the left source * Rp + current from the right source * Rp. With the voltage at the top node, you can just divide by 8 to find the current through that resistor.