Thevenin's Theorem, Superposition & Norton's Theorem

  • #27
The Electrician
Gold Member
1,308
172
All, here is another way of going about this. The Nodal method basically says to model the currents leaving or entering a particular node, in this case the node on top of the load, which I call here as V3. If a current is entering a node, it is positive. If leaving, it is negative. Summing all the currents should equal zero. It is fair to assume for the moment that V1 provides a current into the node, V2 provides a current into the node, and then both of these current sum and leave the node, going to the load. If you convert the currents in its equivalent Ohm's Law of voltage over impedance, you will be left with an equation that has only one variable to solve. It's a lot of algebra to isolate it, but you could use Symbolab.com to crunch complex numbers and just let Symbolab solve for V3 or x in this equation solver. Of course, you won't likely be able to use Symbolab during a test.
View attachment 237649
x is found as 167.9+236.9i.
View attachment 237650
If you take this voltage and divide it by the impedance of the load, you get a load current of 5.0 + .8i

Where did you get that value (39.9 + 40.7 i) in the denominator of the middle term of your equation? The correct value is (35 + 35.70714 i) as given in post #1.
 
  • #28
48
11
The Electrician, I just realized I wrote down 57 ohms for the load impedance and not 50 ohms. Thanks for catching that. Dan.
 
  • #29
1
0

Homework Statement



FIGURE 1 shows a 50 Ω load being fed from two voltage sources via
their associated reactances. Determine the current i flowing in the load by:

(a) applying Thévenin’s theorem
(b) applying the superposition theorem
(c) by transforming the two voltage sources and their associated
reactances into current sources (and thus form a pair of Norton
generators)
View attachment 237468

The Attempt at a Solution


(a)
Converting V1 & V2 to RMS values and V1 to a sin value:
V1 = 415sin(100πt+90)
V2 = 415sin(100πt)
and from Asin(ωt+Φ)
V1 = 415∠90V or 0+j415 V
V2 = 415∠0V or 415+j0 V

Then current around the circuit:
I =V/Z = (V2-V1) / (Z2+Z1) = (415-j415) / (j10)
This as a polar form division:
586.8986∠-45 / 10∠90 = 58.68986∠-135 A or -41.5-j41.5

Vt = V2 - Z2*I = 415+j0 - (0+j6)(-41.5-j41.5) = 415+j0-(-j249-249j^2) as j^2 = -1 then:
Vt = 415-(249-j249) = 166+j249V

ZL = 50@0.7lag = 50∠-45.572998 = 35+j35.70714Ω
Zt = j4in parallel with j6= j2.4
so load current:
I = V / ZL + Zt = 166+j249 / ( 0+j2.4 + 35+j35.70714 ) = 166+j249 / 35+j38.10714
this as a polar division I = 299.26∠56.310 / 51.741∠47.434 = 5.784∠8.876A

I have only completed part (a), once I know I'm on the right track I'll complete b & c. All help is greatly appreciated as always.
Hi,
Can you confirm or anyone is this answer correct after you received feedback from the tutor.
Because i have asked my tutor about mystery around "50 ohm 0.7 lag " and he replied - " IT IS INDUCTANCE". I am little bit confused.
 

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