# Series circuit problem

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1. May 11, 2015

### saulwizard1

1. The problem statement, all variables and given/known data
Find the current for the next circuit. Consider the values given:
V1= 2 V, V2=3V, V3= 5V, R5= 4Ω Y R4= 2Ω

2. Relevant equations
RT=R1+R2+R3+Rn
I=V/RT
P=VI=V^2/R
3. 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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2. May 11, 2015

### Staff: Mentor

I suggest that you review Kirchhoff's laws. KVL would be very useful here.

3. May 12, 2015

### saulwizard1

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: May 12, 2015
4. May 12, 2015

### Staff: Mentor

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.

5. May 12, 2015

### saulwizard1

VT=2-3+5=4V
it's correct?

6. May 12, 2015

### Staff: Mentor

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"?

7. May 12, 2015

### saulwizard1

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?

8. May 12, 2015

### Staff: Mentor

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)

9. May 12, 2015

### saulwizard1

Yes, I didn´t see the sign
So it would be VT=2+3+5=10V

10. May 12, 2015

### Staff: Mentor

Okay, you should now be able to write out KVL for the loop, including the drops associated with the resistances.

11. May 12, 2015

### saulwizard1

10V-(6 ohms*I)=0
10V=6 ohms*I
I=10V/6 Ohms
I=1.66 A

12. May 12, 2015

### Staff: Mentor

Yup. That's better!

13. May 12, 2015

### saulwizard1

Ok, thank you so much for the help.