Use Thevenin’s and Norton’s theorems to find the current in RL for the

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The discussion focuses on using Thevenin’s and Norton’s theorems to determine the current in a load resistor (RL) in a given circuit. The user successfully calculated the Thevenin equivalent voltage (VTH) and impedance (ZTH), resulting in VTH=33.793-j5.517 and ZTH=19.068+j11.172. They then converted these values to polar form and calculated the current through the load resistor, yielding I=0.489<-18.401. A query regarding the Norton equivalent was raised, confirming that IN can be derived from VTH/ZTH, allowing for the subsequent calculation of load current (IL). The calculations and methodology presented are validated by other participants in the discussion.
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Homework Statement


Use Thevenin’s and Norton’s theorems to find the current in RL for the circuit

TASK1_A.jpg


Homework Equations


The Attempt at a Solution


Thevenin equivalent
THEV.jpg


I have calculated VTH and ZTH
VTH=33.793-j5.517
ZTH=19.068+j11.172
Then I converted them in polar forms
Thefore I calculated the current through the load resistor I=VTH/ZTH+RL
I=0.489<-18.401

Homework Statement


My problem is Norton equivalent.Can it be represented like this:
nortas.jpg


and then IN=VTH/ZTH,once I find IN can easily calculate IL=IN(ZTH/ZTH+RL)
 
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shaltera said:

Homework Statement


Use Thevenin’s and Norton’s theorems to find the current in RL for the circuit

TASK1_A.jpg


Homework Equations





The Attempt at a Solution


Thevenin equivalent
THEV.jpg


I have calculated VTH and ZTH
VTH=33.793-j5.517
ZTH=19.068+j11.172
Those values look fine.

Then I converted them in polar forms
Thefore I calculated the current through the load resistor I=VTH/ZTH+RL
I=0.489<-18.401
The current magnitude looks good. Maybe a typo in the second decimal place of the angle?

Homework Statement


My problem is Norton equivalent.Can it be represented like this:
nortas.jpg


and then IN=VTH/ZTH,once I find IN can easily calculate IL=IN(ZTH/ZTH+RL)
Yes, that's correct.
 
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