What's wrong with this calculation simplified transformation circuit?

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Discussion Overview

The discussion revolves around the application of source transformation in a circuit involving a voltage source and series-connected impedance. Participants are attempting to find the equivalent current source and parallel-connected impedance, while addressing specific calculations and simplifications in the process.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the equivalent current source as I=2<-90 A + 4<90 A = 6<0 A, but another participant challenges this addition, suggesting that the current sources should be expressed as complex numbers for proper addition.
  • There is a discussion about the treatment of the 15 ohm resistor, with one participant stating it should be combined with the other impedances using the formula for parallel impedances, while another questions why it is not considered in the parallel combination.
  • Another participant presents a calculation involving the 15 ohm and 30 ohm resistors in parallel, concluding with an equivalent impedance of 10 ohms.
  • Further calculations are provided, including the combination of current sources and the final impedance calculation, which leads to a result of 5 + j5.

Areas of Agreement / Disagreement

Participants express disagreement regarding the addition of current sources and the treatment of the 15 ohm resistor in the calculations. The discussion remains unresolved as different approaches and interpretations are presented without consensus.

Contextual Notes

Participants reference specific formulas and methods for combining impedances and current sources, but there are unresolved aspects regarding the correct application of these methods and the treatment of certain components in the circuit.

pokie_panda
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Homework Statement



Use source transformation on the voltage source and series-connected impedance for the circuit shown here to find the equivalent current source and parallel-connected impedance. Continue the simplification by combining the two parallel current sources into an equivalent current source, and by combining the three parallel impedances into a single equivalent impedance.

Homework Equations



1/((1/Zr)+(1/Zl))
I=V/R

The Attempt at a Solution



First we simplify the voltage source into a current source
I=V/R
30<-90/15
=2<-90 A
to find the new current source we have I1+I2
=2<-90 A+4<90
=6<0A
Now to combine the resistor with the inductor
1/((1/Zr)+(1/Zl))
Therefore the new Z is
3+j9
 

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pokie_panda said:

Homework Statement



... to find the new current source we have I1+I2
=2<-90 A+4<90
=6<0A

That is not the correct addition. Express the two current sources as complex numbers and add per the rules of complex addition.
Now to combine the resistor with the inductor
1/((1/Zr)+(1/Zl))
Therefore the new Z is
3+j9

What happened to the 15 ohm resistor?
 
Because we are simplifying we combine the resistor 15
Using 15 + (ZR*ZL)/(ZL+ZR)
 
pokie_panda said:
Because we are simplifying we combine the resistor 15
Using


15 + (ZR*ZL)/(ZL+ZR)

" ... and by combining the three parallel impedances into a single equivalent impedance." That's what the problem said. So it gave you a hint right there.

The 15 ohm resistor is now in parallel with the 30 ohm and the inductor, so why are you adding its impedance to the parallel impedance of the other two components?
 
so is this correct calculation

30V<-90 = I * 15 ohms
I = 2A<-90

The 15 ohm || 30 ohm = 10 ohms,

the 2A<-90 || 4A<90 = -j2 +j4 = j2 = 2<90

the 10 ohm || +j10 ohm = ((10) * (+j10)) / (10 +j10) = (+j100) / (10 +j10) = 5 +j5
 
Last edited:
pokie_panda said:
30V<-90 = I * 15 ohms
I = 2A<-90

The 15 ohm || 30 ohm = 10 ohms,

the 2A<-90 || 4A<90 = -j2 +j4 = j2 = 2<90

the 10 ohm || +j10 ohm = ((10) * (+j10)) / (10 +j10) = (+j100) / (10 +j10) = 5 +j5

Ah, much better.
 

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