Voltage in a 2 battery series circuit

AI Thread Summary
In a series circuit with two batteries, the correct approach to find the current involves applying Kirchhoff's Loop Rule, which states that the sum of potential drops must equal the sum of potential rises. When calculating the total voltage, one must subtract the opposing battery voltage instead of adding them, as they influence the overall potential difference experienced by the circuit. The initial misunderstanding led to an incorrect current calculation of 0.77 Amps instead of the correct 0.38 Amps. The confusion often arises from misinterpreting the direction of voltage changes in the circuit. Understanding the arrangement of the batteries and their effects on voltage is crucial for accurate calculations.
Dart82
Messages
56
Reaction score
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:
 

Attachments

  • circuit.GIF
    circuit.GIF
    4.6 KB · Views: 2,215
Physics news on Phys.org
Mark the poisitive side of the cells in your image.
Does that give you a clue?
 
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.
 
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
 
i see; i mistakenly counted the 2nd battery as a rise instead of a drop. Thanks again!
 
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
 
Which part do you find confusing?
 
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.
 
Resistance in series adds like this:

R1=32
R2=20

Rsum=32+20
 
Back
Top