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

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SUMMARY

The discussion focuses on calculating the current in a series circuit with voltage sources of 2V, 3V, and 5V, and resistances of 4Ω and 2Ω. The total resistance (RT) is determined to be 6Ω, leading to a final current (I) calculation of 1.66A using Kirchhoff's Voltage Law (KVL). Participants emphasized the importance of correctly identifying the polarities of voltage sources during the KVL analysis, which directly affects the accuracy of the results. The correct application of KVL and attention to voltage signs are crucial for solving circuit problems accurately.

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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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Likes saulwizard1
  • #13
gneill said:
Yup. That's better!
Ok, thank you so much for the help.
 

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