# Superposition theorem problem

#### A.Zanev

Joined Feb 10, 2015
6
Hello guys i am studying electronics for half an year and i need help with the superposition theorem. I have been headbanging for about 2 days now and i cant get up with an adequate answer corresponding to what MultiSim (software) had shown me as a value. Can any of you please give me an explanation about that particular one. I have been reading about it but never saw anything close to my example.. :/ any help will be appreciated! What i get as a results: Total resistance 1955Ohm corresponding to both sources. Its1 = 0.001023 A Its2 =0.001534 A. please dont let those numbers fool you, those are my calculations and at this point i am a bit unsure in myself. Best regards from one new forum member!

#### IAHREIGN

Joined Feb 9, 2015
3
try eliminate dat 3v,replace it with a short, which means will be left with one power source (2v) and then find total resistance of the cct,find current flowing thru each resistor.
repeat the procedure using 3v battery alone.
FINALY SUM UP THE CURRENTS TOWARDS R4
hop this gona work

#### Dodgydave

Joined Jun 22, 2012
8,582
approx 460micro Amps

#### A.Zanev

Joined Feb 10, 2015
6
approx 460micro Amps
Can you please tell me how did you get there ?

#### tindel

Joined Sep 16, 2012
682
Superposition is one of the most important theorems in electrical engineering. I use it somewhat regularly in my engineering (maybe 2 to 3 times a month). It is very important that you understand this concept. I put superposition up there with ohms law, Norton and Thevenin equivalents, etc.

Follow these easy steps:
1. Set all but one power source to 0. Meaning that if you set a voltage source to 0V, that it's shorted, and if you set a current source to 0A, that it is open circuit.
2. Calculate the current or voltage at the point of interest, in your case R4.
3. Continue do do steps 1 and 2 for each power source in your circuit.
4. Sum the results that you get together.

Using these steps... how would you solve the problem? Please post your results in detail. You'll find that this forum is willing to help, but generally not willing to give you the answer - at least not with any details as to how we arrived at the answer.