Calculate Current Through R3: Negative Possible?

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The discussion centers on calculating the current through resistor R3 and whether a negative current is a valid outcome. A negative current indicates that the actual flow is opposite to the assumed direction. Participants emphasize the importance of correctly applying loop equations, noting that the potential should decrease in the direction of current flow. The correct equations for the loops are outlined, with one participant confirming their understanding of the voltage changes involved. Accurate calculations and proper direction indication are essential for resolving the current through R3.
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Homework Statement



Calculate the current through R3.

Homework Equations



V=IR

The Attempt at a Solution



my attempt is attached. My question revolves around my answer...is a negative current a legitimate answer? What does it mean?
 

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It is a legitimate answer provided you show a direction arrow (which you have done). It means that the current is not traveling in the direction indicated by your direction arrow. It is going in the opposite direction.
 
Looking at your loop equations, they seem a bit suspect. If you are summing the changes in voltage across each element along a loop, equating the sum to zero, you should get something different.
 
Sastronaut said:

Homework Statement



Calculate the current through R3.

Homework Equations



V=IR

The Attempt at a Solution



my attempt is attached. My question revolves around my answer...is a negative current a legitimate answer? What does it mean?

Your equations 1 and 2 are wrong. Note, that the potential decreases in the direction of the current.

ehild
 
What should my equation for loop 1 be?
 
You start at the negative pole of the battery, and follow the change of potential in the direction, indicated by the green arrow. The potential increases by 5 V traversing the battery. Then it decreases by I3R3 along R3. Then again, it decreases by R1I1, when you return to your initial position. The net change of potential is zero: 5-I3R3-I1R1=0.
Do the same for the other loop. Note that going in direction opposite to the current, the potential would increase.

ehild

attachment.php?attachmentid=61707&d=1378954564.jpg
 
okay I think I am following you so for loop 2 the equation is then -10V+I3R3+I2R2=0?
 
Sastronaut said:
okay I think I am following you so for loop 2 the equation is then -10V+I3R3+I2R2=0?

Yes.

ehild
 
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