Finding current/impedance in circuit

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Discussion Overview

The discussion revolves around finding current and impedance in a circuit using the law of superposition and Ohm's law. Participants explore various methods for circuit reduction and analyze the implications of short-circuiting voltage sources.

Discussion Character

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

Main Points Raised

  • One participant attempts to reduce the circuit using the law of superposition, calculating equivalent resistances and currents for different configurations.
  • Another participant suggests using Kirchhoff's Current Law (KCL) to find currents in specific branches of the circuit.
  • Some participants challenge the calculations, particularly regarding the equivalent impedance seen by the voltage sources, indicating errors in the assumptions made during circuit reduction.
  • There is a discussion about the necessity of short-circuiting voltage sources and the implications of doing so on the circuit's impedance.
  • Participants debate the voltage across specific resistors when certain voltage sources are set to zero, leading to confusion about the correct interpretation of circuit conditions.
  • Some participants assert that the current through the circuit must account for contributions from both voltage sources, emphasizing the need for superposition in calculations.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to calculating currents and impedances in the circuit. There is no consensus on the validity of the initial calculations, and several participants highlight potential errors without agreeing on a definitive solution.

Contextual Notes

Some calculations depend on the interpretation of circuit configurations and the application of superposition. Participants note that certain assumptions, such as the treatment of short-circuited voltage sources, may affect the outcomes of their analyses.

Who May Find This Useful

This discussion may be useful for students or individuals studying circuit analysis, particularly those interested in the application of the law of superposition and the implications of circuit simplifications.

johndough999
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Homework Statement


upload_2015-5-5_16-9-49.png


Homework Equations


Law of Superposition
I=V/R

The Attempt at a Solution


I am reducing the circuit.
i) By law of Superposition, short circuit V2
100 ohms on the leftmost branch and 50 ohms on the rightmost branch, both are in parallel so
100*80/(100+80)=44.4 ohms seen by V1(t)

ii)
I=V/R
I=2/44.4
I=0.045 Aiii) short circuit V1
reduce the circuit,
60+40+80=180 ohms as all the resistors are in series.

iv)
I=V/R
I=10/180
I=1/18 A

v)Using law of super position
By shorting V2 I obtain I=0.02A
By shorting V1 I obtain I= 1/18A
I = 0.02+1/18 = 0.076A

Please correct me where I am going wrong. Your help is appreciated. Thanks!
 

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Say there is 0V at the bottom. Then there must be V1 at the top: I60+40Ω = V1 / R60+40Ω.

The voltage over 80Ω = V1-V2. Calculate I80Ω.

Now you can calculate the current in the leftmost branch by KCL.
 
Last edited:
(iii) is wrong: Zseen = 80Ω + ((60Ω+40Ω) || ZV1). Thus (iv) is also wrong.
 
For part iii) do I not have to short circuit V1? hence would i not be left with a series circuit?
 
johndough999 said:
For part iii) do I not have to short circuit V1?

Yes, you may short circuit V1, but this is not necessary because it is already "short circuited" as to impedance. It is 0Ω as well as for V2.

Contrary the impedance in a current source is infinite.
 
Zseen = 80Ω + ((60Ω+40Ω) || ZV1)

From the formula you mentioned does that mean 80Ω is in parallel with 60 and 40 Ω?
 
johndough999 said:
Zseen = 80Ω + ((60Ω+40Ω) || ZV1)

By " || " , parallel is meant.

ZV1 = 0, so Zseen = 80Ω + (100Ω || 0Ω) = 80Ω
 
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johndough999 said:

Homework Statement


View attachment 83092

Homework Equations


Law of Superposition
I=V/R

The Attempt at a Solution


I am reducing the circuit.
i) By law of Superposition, short circuit V2
100 ohms on the leftmost branch and 50 ohms on the rightmost branch, both are in parallel so
100*80/(100+80)=44.4 ohms seen by V1(t)

ii)
I=V/R
I=2/44.4
I=0.045 Aiii) short circuit V1
reduce the circuit,
60+40+80=180 ohms as all the resistors are in series.

iv)
I=V/R
I=10/180
I=1/18 A

v)Using law of super position
By shorting V2 I obtain I=0.02A
By shorting V1 I obtain I= 1/18A
I = 0.02+1/18 = 0.076A

Please correct me where I am going wrong. Your help is appreciated. Thanks!
in parts 3 and 4, what is the voltage across the 60 and 40 ohm resistors?
recall ohms law, what does that tell you about the current through the resistors.
 
donpacino said:
in parts 3 and 4, what is the voltage across the 60 and 40 ohm resistors?
recall ohms law, what does that tell you about the current through the resistors.

donpacino could you elaborate a little more. Would the voltage across the 60 and 40 ohm be the total voltage - voltage across 80 ohm?.
 
  • #10
johndough999 said:
donpacino could you elaborate a little more. Would the voltage across the 60 and 40 ohm be the total voltage - voltage across 80 ohm?.
in parts 3 and 4 you said you were setting V1 to zero... If you set V1 to zero, what is the voltage across the 60 and 40 ohm resistors
 
  • #11
donpacino said:
in parts 3 and 4 you said you were setting V1 to zero... If you set V1 to zero, what is the voltage across the 60 and 40 ohm resistors

the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor
 
  • #12
johndough999 said:
the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor
nope. V1 is short circuited, which means the voltage across it is zero
 
  • #13
johndough999 said:
the voltage across 40 and 60 ohm would be the supply voltage V2 - voltage across 80 ohm resistor
well, technically that is correct, what is the voltage across the 80 ohm resistor?
 
  • #14
So if V1 is short circuited and the voltage across it zero, voltage across 60 and 40 ohms is zero therefore voltage across 80 ohm resistor is equal to V2?
 
  • #15
johndough999 said:
So if V1 is short circuited and the voltage across it zero, voltage across 60 and 40 ohms is zero therefore voltage across 80 ohm resistor is equal to V2?
yes. so knowing that, look at your answer to parts 3 and 4 again
 
  • #16
iii) 80 0hms

iv) I=V/R
I=10/80
I=1/8 A

v) 0.02A? since current in the centre branch when V1 is short circuited is 0A.
 
  • #17
johndough999 said:
iii) 80 0hms

iv) I=V/R
I=10/80
I=1/8 A

v) 0.02A? since current in the centre branch when V1 is short circuited is 0A.

for part V yes.

I also just re-read your origonal questions and realized I missed something for parts 2 and 4. You know the current through the source V1 due to V1, however the will be current flowing through the source V1 that is due to the source V2. the same is true for part 4.
 
  • #18
donpacino said:
for part V yes.

I also just re-read your origonal questions and realized I missed something for parts 2 and 4. You know the current through the source V1 due to V1, however the will be current flowing through the source V1 that is due to the source V2. the same is true for part 4.

Does that mean the answers are incorrect?
 
  • #19
johndough999 said:
Does that mean the answers are incorrect?
Yes, you have half of the answer for parts two and four. You then need to use superposition to find the correct answer (hint, you already did most of the work)
 
  • #20
donpacino said:
Yes, you have half of the answer for parts two and four. You then need to use superposition to find the correct answer (hint, you already did most of the work)

Why can i not just simply use ohms law to find the current for part 2 and 4?
 
  • #21
johndough999 said:
Why can i not just simply use ohms law to find the current for part 2 and 4?
You have to include the current from both sources
 
  • #22
donpacino said:
You have to include the current from both sources

The same way as for the last part, where I find the current in the center branch?
 
  • #23
johndough999 said:
The same way as for the last part, where I find the current in the center branch?
Its the current due to V1 + the current due to V2. You had it correct
 
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  • #24
donpacino said:
Its the current due to V1 + the current due to V2. You had it correct
Thanks for the help!
 
  • #25
no problem
 

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