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

AI Thread Summary
The discussion focuses on calculating the current in a series circuit with given voltage sources and resistances. Initially, there were errors in voltage polarity signs, leading to incorrect calculations. After reviewing Kirchhoff's laws and correcting the signs, the total voltage was recalculated to 10V. The final current was determined to be 1.66A using the correct total resistance of 6 ohms. The importance of accurately identifying voltage polarities in circuit analysis is emphasized throughout the conversation.
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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