Voltage in a 2 battery series circuit

In summary: Kirchhoff's rules state that the sum of potential drops must equal the sum of potential rises. So, in this problem, the voltage drops (from the battery in series with R3) must equal the voltage rises (from the battery in series with R4).
  • #1
Dart82
57
0

Homework Statement

What is the magnitude and direction of the current in the following circuit?
[URL=http://imageshack.us][PLAIN]http://img234.imageshack.us/img234/7587/circuituk2.gif[/URL][/PLAIN]
resistance for series: Req=R1+R2+R3+R4...etc
Kirchhoff's rules --> Loop Rule: sum of potential drops must equal sum of potential rises
V=IR



The Attempt at a Solution


1st added all the voltage drops: I(5+27+12+8) = I52
2nd added all the Voltage rises: 30 + 10 = 40V

3rd solved for V when V=IR
40 volts/52 ohms= I
*since resistors are in series they experience the same current right?

my answer .77 Amps clockwise [which is wrong] the right answer is .38 Amps clockwise

My question is this: Why am i not supposed to add the voltages (30+10) together when solving this problem?? If i solve for this problem in the following way, 30V-10V = I(52ohms) then i get .38 Amps clockwise (which is right) but i don't understand why i have to subtract the voltages!:confused:
 

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  • #2
Mark the poisitive side of the cells in your image.
Does that give you a clue?
 
  • #3
Yes. It seems the only logical explanation i can think of is that when 2 batteries "oppose" each others flow; then one battery will reduce the voltage of the other.
 
  • #4
Yes, the equivalent voltage will either be a sum or (in this case) difference of the 2 voltages. Going in a clockwise loop from A, a particle will go from low potential to high potential (+30V) and then go from high potential to low potential (-10V). If the batteries were arranged + to - in series, you could add the voltages because there would be a rise followed by another rise
 
  • #5
i see; i mistakenly counted the 2nd battery as a rise instead of a drop. Thanks again!
 
  • #6
i'm staring at the same problem and yet don't get how to solve it ? can anyone offer me any help?! I might have this problem on my final exam this coming monday may 10th , 2010. help please
 
  • #7
Which part do you find confusing?
 
  • #8
Phrak said:
Which part do you find confusing?

the kirchhoffs rule as a whole. I tried solving this problem in the following way :
R= 1/32 + 1/20 = 0.083 ohms
I= V/R = 30/0.083= 360 amps
which I know is wrong so what can I do, I spent the last hour watching a professor from MIt working a similar problem yet he did not show it with numbers or real answers.
 
  • #9
Resistance in series adds like this:

R1=32
R2=20

Rsum=32+20
 

1. What is the total voltage in a 2 battery series circuit?

The total voltage in a 2 battery series circuit is the sum of the individual voltages of each battery. For example, if each battery has a voltage of 1.5 volts, the total voltage in the circuit would be 3 volts.

2. How does the voltage change in a 2 battery series circuit?

In a series circuit, the voltage remains constant throughout the circuit. This means that the voltage in a 2 battery series circuit will not change and will be the same as the voltage of each individual battery.

3. Can the total voltage in a 2 battery series circuit be greater than the voltage of one battery?

No, the total voltage in a 2 battery series circuit cannot be greater than the voltage of one battery. This is because the batteries are connected in a series, meaning that the voltage of each battery adds up to create the total voltage in the circuit.

4. How does the resistance affect the voltage in a 2 battery series circuit?

In a series circuit, the voltage is divided between the components based on their resistance. This means that the higher the resistance, the more voltage will be dropped across that component. Therefore, if the resistance is increased in a 2 battery series circuit, the voltage across the batteries will decrease.

5. What happens to the voltage in a 2 battery series circuit if one battery is removed?

If one battery is removed from a 2 battery series circuit, the total voltage in the circuit will be reduced by the voltage of the removed battery. This is because the circuit is now broken, and the remaining battery cannot provide the full voltage on its own.

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