Series circuit answers correct?

AI Thread Summary
The discussion revolves around solving a series circuit problem involving three batteries and resistors. The initial calculations for currents and resistances were questioned, particularly the determination of I3 and the voltage across R3. It was clarified that the total voltage of 70V does not apply across R3, and instead, the voltage across R3 should be used to calculate the current. The correct current through R3 was confirmed to be 2A, derived from Ohm's law using the voltage across R3 and its resistance. Accurate circuit analysis requires careful attention to voltage drops across individual components.
confused12345
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Homework Statement


3 batteries: Vt= 40v+20v+10v = 70v
R1= unknown
R2= Unknown
R3= 10 ohm
I= unkown
The ones which I am looking for are:
I3=___
I1=___
V1= ___
R1=___
R2=___



Homework Equations


V=IR
I=V/R
R=V/I


The Attempt at a Solution


I3= 7a
I1= 7a
V1=70v
R1= 10 ohm
R2= 10 ohm
Diagram:

Is this correct?
 

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Hard to say. Strike that--impossible to say. Need a circuit diagram, please.
 
There seems to be some problems with your results. You'll have to give some details on how you arrived at your answers. How did you determine i3 to be 7A?
 
I used the equation I=V/R
70/10 ohm <--- since that was the only resistance already given?
 
I see. Well, presuming that the 70V you've quoted is the sum of the voltages shown for the battery and the voltages across R2 and R3, then that 70V is NOT across R3. It's the potential across R1, which you don't yet have a value for.

Your figure shows a voltage across R3. Why don't you use that voltage and that resistance to determine the current?
 
So if I use
I=V/R
=20/10
=2a
That is the correct current? I was using the totals to find I the first time.
 
confused12345 said:
So if I use
I=V/R
=20/10
=2a
That is the correct current? I was using the totals to find I the first time.

Yes, that is the correct current. It is a direct consequence of Ohm's law applied to R3 where you are given both the value of the resistance and the voltage across it.
 
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