Recent content by gatsbycollege

oh sorry that 10 should be 30 ,, thx for the answer,, i got my Isc = 1.84 and Vth = 143.52 tried to use that mesh analysis i got equations 60V - 90 Ia - 40V +30 Isc = 0 35V - 15V - Isc (5 +30 +45 +8) + 30 Ia +90 =0 then by elimination igot Isc , correct me if I am wrong thanks

Homework Statement http://img716.imageshack.us/img716/8021/69925791.jpg Find The Thevenins Equivalent of the circuit shown . at the terminal a and b i think Homework Equations Vth = Rth(Ith) V= IR The Attempt at a Solution For Vth i tried using Nodal Analysis Having 4 Node Voltages but i...

Ahm ok :approve: Thx for Helping me :smile:

i double check the given here in my copy.. the circuit is right.. but not so sure about my computation :-p so what do you think.. maybe the answer given by my prof is wrong? Vo = -50V :confused:

by referring to what you said sir, i= (6v-0.004Vo)/ 2k BY KCL then 25(i)-Vo/40k - Vo/10k = 0 then substitute 25((6v-0.004Vo)/2l)) -Vo/40k -Vo/10k = 0 Therefore Vo = 428.57V Hmm :confused: still not the right answer ?

Aw.. my mistake,, 50v at the left should be 6v Sorry :-p Edit: So there it is i changed the circuit.

Homework Statement http://img202.imageshack.us/img202/939/95788682.jpg Find Vo Vo = -50V (given answer by my prof) Homework Equations KVL (Series) V=V1+V2 (Parallel) V=V1=V2 KCL Current toward the node = Current Away from the node V=IR The Attempt at a Solution i tried to solve for the i in...

am i right that in ur analysis u also neglect the presence of the 24V at current i4

ahm can you please show ur computation in doing it separately.. i might have another idea in solving future circuit problems... thx

the diagram is right. the problem statement.. "Find the voltage across a and b" and find the current "i" the 3A current source is placed like that.. there's nothing wrong with that the blue line is the original diagram and the current i1 , i2 , i3 and i4 is just my assumed direction..i...

Homework Statement Find Voltage across a and b and the missing current (i) so this is the circuit http://img3.imageshack.us/img3/184/circuitsf.jpg Homework Equations KCL and KVL current towards the node is + away is - Et = E1 + E2 + E3 in series Et = E1 = E2 = E3 in parallel Ohms law E= IR...

so i got this answers. i1= 2.44A i3= 7.304A then my P= 533.48 is this right?

At node B i2+i3+i1=0 at node C i3+Io=i4 now tryring what you said, ill inform you of my answer and correct me if I am wrong.

aw.. i forgot to put the value in E1 its 220V.. so itried using that Y-Delta Transformation,, then came to answer Rt= 15Ohms and the It= 14.67A.. so I am stuck again.. i don't know what should i use next to find i1.. should i use KCL by node or should i use the KCL in series and parallel ?

i don't know... but the figure is right.. the three current is going out of point B