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

• Engineering
• saulwizard1
In summary, the conversation discusses finding the current for a circuit with given values of voltage and resistance. The solution involves using Kirchhoff's laws and determining the polarity of the voltage sources while walking around the loop. Finally, the correct current is calculated to be 1.66 A.
saulwizard1

## 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

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.

#### Attachments

• 15-6.jpg
7.5 KB · Views: 443
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

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.

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

Yup. That's better!

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

## 1. What is the total voltage in a series circuit?

The total voltage in a series circuit is the sum of all the individual voltage drops. In this circuit, the total voltage would be 2V + 3V + 5V = 10V.

## 2. How do you calculate the current in a series circuit?

The current in a series circuit is the same at all points. To calculate the current, you can use Ohm's Law: I = V/R. In this circuit, the current would be 10V / (4Ω + 2Ω) = 1.67A.

## 3. What is the voltage drop across each resistor in a series circuit?

In a series circuit, the voltage drop across each resistor depends on its resistance. The voltage drop can be calculated using Ohm's Law: V = IR. In this circuit, the voltage drop across the 4Ω resistor would be 1.67A x 4Ω = 6.68V, and the voltage drop across the 2Ω resistor would be 1.67A x 2Ω = 3.34V.

## 4. How does the total resistance in a series circuit compare to the individual resistances?

In a series circuit, the total resistance is equal to the sum of the individual resistances. In this circuit, the total resistance would be 4Ω + 2Ω = 6Ω.

## 5. What would happen to the current if another 2V battery was added to the circuit?

If another 2V battery was added to the circuit, the total voltage would increase to 12V, and the current would also increase. This is because the total resistance in the circuit remains the same, but the total voltage has increased, following Ohm's Law: I = V/R.

• Engineering and Comp Sci Homework Help
Replies
1
Views
1K
• Engineering and Comp Sci Homework Help
Replies
26
Views
2K
• Engineering and Comp Sci Homework Help
Replies
12
Views
1K
• Engineering and Comp Sci Homework Help
Replies
5
Views
3K
• Engineering and Comp Sci Homework Help
Replies
3
Views
2K
• Engineering and Comp Sci Homework Help
Replies
7
Views
987
• Engineering and Comp Sci Homework Help
Replies
3
Views
999
• Engineering and Comp Sci Homework Help
Replies
17
Views
10K
• Engineering and Comp Sci Homework Help
Replies
19
Views
4K
• Engineering and Comp Sci Homework Help
Replies
4
Views
2K