2 Batteries 4 Resisters

Thread Starter

mer584

Joined Feb 23, 2008
6
PROBLEM: For the circuit (see attach) find the potnetial difference between points a and b. Each resistor has R=75ohms and each battery is 1.5V.


MY ATTEMPT/CONFUSION: I'm getting really confused applying Kirchhoffs Rules because there are only two points and we want to find the difference between them, but they are technically the same on both sides.

So I know that because this is a series I will be constant, so I have added all of the R values (75 x 4) to find that V/R where v= (1.5+1.5) =3V yeilds
3/300 = .01 Amperes

then at each resister to find the potential its going to be IxR =.01 x 75 = .75V

I'm quite confused, this problem also seems really simple.
 

recca02

Joined Apr 2, 2007
1,212
find the current,
find the voltage drop on each resistance,
take any point as reference, algebraically add voltage drop and battery voltage.
that is voltage drop on two resistance and the battery voltage encountered while traversing from point a-b.
 

Thread Starter

mer584

Joined Feb 23, 2008
6
so that was what I started to do...correctly I hope?

I found a voltage drop of .75V at each resistor (based on I=.01) since they are all the same.
Does this mean from point a to b I will have .75+ .75 + 1.5 ?
 

Thread Starter

mer584

Joined Feb 23, 2008
6
I guess I don't really understand how the sign works over the battery. I know for the current you just assign it. I thought if it was going over the negative to the positive portion of the battery there was a rise in voltage. Perhaps i have this backwards? That also makes the potential 0...and I'm not sure I understand why that would be.
 

recca02

Joined Apr 2, 2007
1,212
Traverse a loop in direction of current and take voltages dropped by it (IR) as positive and voltages dropped by currents in opposite direction as -ve.
When u encounter a voltage source while traveling the loop take it as -ve if it supports the current direction else take +ve.
Then equate all these to zero. (all these rules i have made from my experience, books may/may not help much but practice certainly does)
................................By Recca02 ;)
This is something I do when confused. It usually works for me.

The reason why it becomes zero in the above case is rather easy its because the points are located such that the potential at those points is same..u yourself observed this similarity in position.:rolleyes:.
 
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