Potential Differences and Currents in a Diamond Circuit

[. Arctic.]

Homework Statement

In the figure ε = 15.1 V, R1 = 1560 Ω, R2 = 2130 Ω, and R3 = 4210 Ω. What are the potential differences (in V) (a) VA - VB, (b) VB - VC, (c) VC - VD, and (d) VA - VC? The figure consists of single ideal batter connecting to a diamond shape circuit ABCD. Going clockwise starting with point A, line AB has resistor 1, line AD has resistor 2, line CD also has a resistor 1, and line CA also has a resistor two. A line between CB is between points C and B, and it contains resistor 3.

ε = 12V, R₁ = 2000Ω, R₂ = 3000Ω, R₃ = 4000Ω

Homework Equations

ε - i₁R₂ - i₂R₂ = 0 (1)
ε - (i₁ - i₂)R₃ = 0 (2)

The Attempt at a Solution

What I did was move around equation (2), so that I would end up with

i₂ = i₁ - (ε/R₃), and then I would plug that back into equation one, which would help me find the current. The thing is that it doesn't match up with the answers in the book, so I would like to know how to reach the correct answer. i 1-3, in that order, are 0.002625 A, 0.00225 A, and 0.000375 A. Help would be much appreciated. Thanks.

 

[phinds]

In the figure

Uh ... what figure would that be ?

 

[. Arctic.]

Figure Associated With Problem

Oops. Forgot to upload the drawing. Sorry. It's up now.

 

[haruspex]

ε - i₁R₂ - i₂R₂ = 0 (1)
ε - (i₁ - i₂)R₃ = 0 (2)

You haven't defined i1 and i2. No matter how I try to define them I don't understand how you get those equations.

 

[. Arctic.]

Where the Equations Came From

I'm following along with the solution manual on this one. For this part, the book said,

"The symmetry of the problem allows us to use i2 as the current in both of the R2
resistors and i1 for the R1 resistors. We see from the junction rule that i3 = i1 – i2. There
are only two independent loop rule equations:

where in the latter equation, a zigzag path through the bridge has been taken. Solving, we
find i1 = 0.002625 A, i2 = 0.00225 A and i3 = i1 – i2 = 0.000375 A. Therefore, VA – VB =
i1R1 = 5.25 V."

I'm just not getting where they are getting their currents from. I've tried looking at the R_eq, and using that to find the the current by using, i = (V/R_eq), but it seems as though that didn't work. Of course, I've also used the R for when the current passes through resistors 1, 2, and 3.

 

[SammyS]

Homework Statement

In the figure ε = 15.1 V, R1 = 1560 Ω, R2 = 2130 Ω, and R3 = 4210 Ω. What are the potential differences (in V) (a) VA - VB, (b) VB - VC, (c) VC - VD, and (d) VA - VC? The figure consists of single ideal batter connecting to a diamond shape circuit ABCD. Going clockwise starting with point A, line AB has resistor 1, line AD has resistor 2, line CD also has a resistor 1, and line CA also has a resistor two. A line between CB is between points C and B, and it contains resistor 3.

ε = 12V, R₁ = 2000Ω, R₂ = 3000Ω, R₃ = 4000Ω

...

It appears that you have two completely different ses of values for the voltage and the resistances.

Which set are you using for this problem?

 

[haruspex]

"The symmetry of the problem allows us to use i2 as the current in both of the R2
resistors and i1 for the R1 resistors.

OK, so in (1) i₁*R₂ was a typo for i₁*R₁; but I still don't see how you get equation (2).

 

[. Arctic.]

Equation (2) comes from the loop rule. I know that the current running through R₁ is i₁ and the current running through R₂ is i₂. These two currents add up to the main current i₃. The equation would be i₃ = i₁ + i₂. At the moment the currents i₁ and i₂ run through R₃, they add up together to become the main current i₃. At least, that's what I believe my understanding of the situation is. Correct me if I'm wrong.

 

[haruspex]

the currents i₁ and i₂ run through R₃

No. Think about the currents flowing into and out of the node at the left of the diamond.