Bridge Circuit (resistors, whoo)

1. Apr 29, 2009

GDGirl

1. The problem statement, all variables and given/known data
The resistor R in the above figure has a resistance of 21.7 Ω.
a) Find the current in the resistor R. (Enter a positive number for a current up and to the right, and negative for down and to the left.
b) Find the current in the 10-Ω resistor. (Enter a positive number for a downward current, negative for an upward current.)
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/10/03/P26_29.jpg [Broken]

2. Relevant equations
Current Leaving Battery= Voltage/total resistance
V=IR (Ohm's Law)

3. The attempt at a solution
Well, I tried to find the total resistance and ended up with 2.692 (I'm not sure this is right because of how the circuit is set up), then I divided the voltage by this, and got 2.228 as the current from the battery. Then I used this current to find the voltage at the point after the resistor R with Ohm's Law, which gives me 60.378V. I use the voltage drop between the points (which is -54.3788) and the resistance of the resistor to find the current (from Ohm's Law again), this gives me -2.007. This is wrong, as I expected. I'm not sure how to go about doing this. :/

Last edited by a moderator: May 4, 2017
2. Apr 29, 2009

davieddy

There is no quick fix for this.
It is a full blown exercise in applying Kirchoff's laws.

3. Apr 29, 2009

GDGirl

I didn't think about Kirchhoff's laws. :/ Dang. I suppose it's because my prof only ever used it in problems with multiple power sources. Alright, thanks

4. Apr 29, 2009

davieddy

I like to work with potentials at the junctions.
(Doing this trivially satisfies the law about potential drops around a closed
loop summing to zero).

Let the potential on the right be 0V and so on the left it is 6V.
Let the potential of the top junction be VA and the bottom be VB.

All the currents can be expressed easily in terms of VA and/or VB by
Ohm's Law. (e.g. the current through the 10 ohms is (VA-VB)/10)

Now apply "what flows in must flow out" to the top and bottom junctions,
and we get two simultaneous equations in VA and VB. Just solve them and
you have all you need.

David

Last edited: Apr 29, 2009
5. Apr 29, 2009

GDGirl

I think I understand how to do everything you're describing well enough- except the last bit. What do you mean by "apply 'what flows in must flow out' to the top and bottom junctions"? Like, I understand the what flows in part, obviously if you have a current leaving a battery, the same current should be going back into it, but I'm not sure what you mean about applying it to the junctions.

6. Apr 29, 2009

davieddy

Applying it to the top junction (A):

(6-VA)/R = (VA - VB)/10 + VA/2

Net current into something must be zero, or else charge is accumulating
in that something.

7. Apr 29, 2009

GDGirl

Alright, I think I've got it now, thanks.