Recent content by johnsy1312

Finding the Norton Equivalent circuit for the circuit attached I attempted this solution but i am unsure if my Norton's current is correct: R_N = \frac{1}{\frac{1}{6}+\frac{1}{12}+\frac{1}{12}}=3ohms I'_N = \frac{12*2}{12+4}=1.5A, I''_N = 2A I_N = 2A - 1.5A = 0.5A

For this circuit, does V2=E?

so...

Is that new resistance positioned where R2 is or where R1 is?

Are R3 an R4 parallel?

Sorry, i did but it wouldn't allow me since i already posted it in another post. So: R1||2=\frac{20*5}{20+5}=4ohms

Ok, so R2 AND R4 are not in parallel. Are R2 and R1 in parallel?

I did :) They look like they are in parallel though :/

I have to determine the Thevenin's equivalent circuit for the circuit attached. I am stuck on the first and essential part, finding the thevenin's resistance. Answer should be 10ohms, my attempt: R_2||4 = \frac{5*16}{5+16} = 3.81ohms R_1,3,5 = 20 + 12 + 2 = 34ohms

Thanks! I also have this other problem which i cannot see to get right and i have attempted it 100 times :/ I have to determine the Thevenin's equivalent circuit for the circuit attached. I am stuck on the first and essential part, finding the thevenin's resistance. Answer should be...

Disregard previous post: Correct me if i am wrong. 6kohm<90^o=(6sin(90^o)+6cos(90^o))j =6j 4kohm<-90^o=(4sin(-90^o)+4cos(-90^o))j =-4j Therefore: Z= \frac{1}{\frac{1}{6j}-\frac{1}{4j}} =-12j

z=\frac{12}{(2+3)j} =12*(\frac{1}{(2+3)j})

So Z=\frac{1}{\frac{1}{6kohm<90^o}+\frac{1}{4kohm<-90^o}}?

So with what your saying would the total impedance for the first branch be Z=(6-4)j, how do we obtain an angle from that when they are both on the same axis? Isn't the angle between them 0?

When you add the two reactive components together, the angle becomes 90, since the inductor is 6kohm<90 and capacitor is 4kohm<-90?