Hi, I have the following circuit And i need to calculate the norton equivalent circuit. After i shorted the voltage source and changed the current source into an open circuit, the thevenin resistance Applying the superposition principle the short circuit current is 2A plus the current generated by the voltage source. This is where i had to stop and think. Are there any current traveling through the 6ohm resistor?
Thanks, i didn't know you could add them like that. Is there any other way besides source transformation?
Hi, Hm, the only way i get 1A from the voltage source is if i consider the 6ohm and 3ohm resistor in parallel. and then However arent they supposed to be in series when the voltage source appear in between?
Hmm, To find Norton equivalent First: Calculate the output current, IAB, with a short circuit as the load. Then find Rth Replace all voltage sources with short circuits and all current sources with open circuits. So first we need to find this short current by using superposition. I first remove the current source And then by inspection I write Isc1 = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A And now we replace voltage source with a short circuit Is2 = 2A * (6Ω||2Ω) / ( 6Ω||2Ω + 2Ω ) = 2A * 2/4 = 1A So our final solution is Isc = Ics1 + Isc2 = 1.5A and Rth= 4Ω And of course we can get the same answer with source transformation method, or nodal/mesh analysis .
Hi, Thank you very much. There is one part i don't really understand Isc1 = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A First, the total current is 6V/(3Ω||2Ω + 6Ω) = 6/7.2 = 0.83A Then we use current division. 3/(3+2) * 0.83 = 0.498A
You use current divider rule, but I use Ohms law. First I find total current Itot = [ 6V/(3Ω||2Ω + 6Ω) ] next I find voltage drop across 3Ω||2Ω V = Itot* 3Ω||2Ω =[ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω And finally, if I know the voltage across 2Ω resistor I can easily find current that flow through 2Ω resistor. I = Isc1 = V/2Ω = ( [ 6V/(3Ω||2Ω + 6Ω) ] * 3Ω||2Ω ) / 2Ω = ( (6/(1.2 + 6)) * 1.2 ) / 2 = 0.5A I hope you understand now. And maybe at the end I show how you can use nodal equation to find Isc for this circuit: 2A = Va/6Ω + Vb/3Ω + Vb/2Ω (1) Vb - Va = 6V ----> Vb = 6V + Va (2) 2 = Va/6 + 2Vb/6 + 3Vb/6 2 = (Va + 5Vb)/6 2= ( Va + 5*( Va+6 ) )/6= ( 6Va + 30) / 6 2 = (6Va +30) / 6 12 = 6Va + 30 -18 = 6Va Va = -18/6 =-3V Vb = 6V + (-3V) = 3V And finaly Isc = Vb/2Ω = 3/2 = 1.5A So as you can see you can use any method you want to solve this circuit.