Current in Series Circuit: 2V, 3V, 5V, 4Ω, 2Ω

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Discussion Overview

The discussion revolves around calculating the current in a series circuit with given voltage sources and resistances. Participants explore the application of Kirchhoff's laws, particularly Kirchhoff's voltage law (KVL), to analyze the circuit and address issues related to voltage signs and polarities.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant attempts to find the current by summing the voltages and dividing by the resistance, indicating limited background knowledge.
  • Several participants suggest reviewing Kirchhoff's laws, emphasizing the importance of KVL for solving the circuit.
  • There are discussions about potential sign problems in the voltage calculations, with participants questioning the signs of the voltages based on their polarities.
  • One participant proposes a calculation resulting in a total voltage of 4V, which is later challenged regarding the correctness of the signs used.
  • Another participant suggests that all voltage sources should have the same sign if they have the same polarity, raising questions about how potential rises and drops are summed.
  • After addressing the sign issues, a participant recalculates the total voltage to be 10V and proceeds to apply KVL to find the current, resulting in a value of 1.66 A.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the signs of the voltages and the correct application of KVL. While some calculations are refined, there is no consensus on the initial interpretations of the voltage signs.

Contextual Notes

Participants highlight the importance of correctly identifying the polarities of voltage sources and the potential rises and drops when applying KVL. There are unresolved questions about the consistency in summing potential rises versus drops.

saulwizard1
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Homework Statement


Find the current for the next circuit. Consider the values given:
V1= 2 V, V2=3V, V3= 5V, R5= 4Ω Y R4= 2Ω

Homework Equations


RT=R1+R2+R3+Rn
I=V/RT
P=VI=V^2/R

The Attempt at a Solution


My attempt of solutions is to sum the first two voltages and divide it between the first resistance. My background context is that I only know very few about this topic.
 

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I suggest that you review Kirchhoff's laws. KVL would be very useful here.
 
gneill said:
I suggest that you review Kirchhoff's laws. KVL would be very useful here.
According to the Kirchoff's law of voltages:
RT=6 ohms
VT=?
-2+3-2I+5-4I=0
6+-6I=0
6I=6
I=6/6=1A
is the result right? I have some doubts in the signs of the voltages.
 
Last edited:
Definitely sign problems. Pay attention to the polarities of the voltage sources as you do your "KVL walk" around the loop in the direction of the current.
 
gneill said:
Definitely sign problems. Pay attention to the polarities of the voltage sources as you do your "KVL walk" around the loop in the direction of the current.
VT=2-3+5=4V
it's correct?
 
saulwizard1 said:
VT=2-3+5=4V
it's correct?
No. Check the polarities of the voltage sources as you go around the loop. Suppose you "walk" clockwise around the loop starting at the bottom of V1. Is there a potential rise or drop as you pass through each source? Do they have different polarity orientations with respect to your "walk"?
 
gneill said:
No. Check the polarities of the voltage sources as you go around the loop. Suppose you "walk" clockwise around the loop starting at the bottom of V1. Is there a potential rise or drop as you pass through each source? Do they have different polarity orientations with respect to your "walk"?
The first one is=-2
The second has the same polarity, and it has a rise of potential
For the third, it has the same polarity and it has a rise of potential.
that's correct?
 
Why is the first one -2? If all the polarities are the same, shouldn't they all have the same sign? Are you summing potential rises or potential drops? (Some prefer to sum drops as positive, rises as negative, which is perfectly fine as long as one is consistent about it)
 
gneill said:
Why is the first one -2? If all the polarities are the same, shouldn't they all have the same sign? Are you summing potential rises or potential drops? (Some prefer to sum drops as positive, rises as negative, which is perfectly fine as long as one is consistent about it)

Yes, I didn´t see the sign
So it would be VT=2+3+5=10V
 
  • #10
saulwizard1 said:
Yes, I didn´t see the sign
So it would be VT=2+3+5=10V
Okay, you should now be able to write out KVL for the loop, including the drops associated with the resistances.
 
  • #11
gneill said:
Okay, you should now be able to write out KVL for the loop, including the drops associated with the resistances.
10V-(6 ohms*I)=0
10V=6 ohms*I
I=10V/6 Ohms
I=1.66 A
 
  • #12
Yup. That's better!
 
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  • #13
gneill said:
Yup. That's better!
Ok, thank you so much for the help.
 

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