Help w/Current Through Batteries

[STEF2098]

In the figure below, the resistances are R1 = 1.1 , R2 = 1.5 , and the ideal batteries have emfs 1 = 2.0 V, and 2 = 3 = 5.5 V.



http://www.webassign.net/hrw/hrw7_27-40.gif




I feel like I've tried everything to solve this problem; I just don't get it!

 

[collinsmark]

Use Kirchhoff's circuit laws to solve this. There are more than one way to do this problem, even when using Kirchhoff's laws. But no matter which way you go about it, you will end up with 2 equations and 2 unknowns. Solve the simultaneous equations and things will fall into place after that.

[Edit: Alternately, you might be able to use Thévenin/Norton equivalent transformations to grind your way to an answer. But I still suggest using Kirchhoff's laws if you have the choice.]

 

[STEF2098]

Wouldn't you have three unknowns? The current through R2, the current through R1 in the left loop, and the current through R1 in the right loop? That's what's been confusing me.

 

[vela]

Yeah, if you do it that way, you'll have three unknowns and three equations. I'm guessing you're having problems coming up with that third equation. Am I right? Show us what you've gotten so far.

 

[collinsmark]

Wouldn't you have three unknowns? The current through R2, the current through R1 in the left loop, and the current through R1 in the right loop? That's what's been confusing me.

Of course you get to define your current loops yourself.

But let's suppose that you defined your current loops to be i₁ on the left side of the circuit, and another current loop i₂ on the right side of the circuit. And, let's assume that you have defined both loops to go around clockwise. (These definitions are up to you, btw.)

Then by that definition, the current going through R₂ from the direction of top to bottom, is simply i₁ - i₂. Or if you are tracing the current from bottom to top, the current is i₂ - i₁. The important thing is that either way you can express the overall current through R₂ as a simple function of the two unknowns. There is no reason to have a third unknown.

Of course the particular resistor or resistors, who's current is a function of multiple unknowns completely depends on how you define your current loops at the beginning. And the details are a function of how you define the direction of the loops (clockwise or counterclockwise). The current loop definitions I used in the above paragraph is just one possible definition of many.

 

[STEF2098]

Okay, so, defining the current to be traveling in a clockwise direction:
LEFT LOOP:
E1-i1R1-(i2)R2-E2-i1ri=0

RIGHT LOOP:
-E3-i3R1+E2-(i2)R2-i3R1=0

I feel like a third variable is necessary. Because the R1's on either side aren't equal.
THere is one current going through the left, another through the right, and another through the middle.
I set a third equation to be i2+i1-i3=0.
But I wouldn't know where to go from there

 

[collinsmark]

Okay, so, defining the current to be traveling in a clockwise direction:
LEFT LOOP:
E1-i1R1-(i2)R2-E2-i1ri=0

Well, that's not the way I would have done it. But let's use your method. I should still point out though that there is no need to define i₂ at all, the way that you've defined it. i₂ isn't even a loop. It starts and ends, but doesn't loop back on itself. It is unnecessary.

But let's go ahead and use it anyway, even if it is unnecessary. But make sure you draw the direction of i₂ on your schematic. According to your second and third equation, i₂ points up.

So there is an error in your first equation! If you keep things in terms of i₂ which points up, and sum the voltage traced out by following i₁ around, the voltage across R₂ should get a positive sign, not negative (because you're going in the opposite direction of i₂).

$$E_1 - i_1 R_1 + i_2 R_2 -E_2 -i_1 R_1 = 0$$

RIGHT LOOP:
-E3-i3R1+E2-(i2)R2-i3R1=0

I feel like a third variable is necessary. Because the R1's on either side aren't equal.
THere is one current going through the left, another through the right, and another through the middle.
I set a third equation to be i2+i1-i3=0.
But I wouldn't know where to go from there

Using your method, solve for i₂ (i.e. i₂ = i₃ - i₁). Then anytime you see an i₂ in any of your first two equations, substitute the results in, which leaves only i₁ and i₃! That gives you two equations and two unknowns!

But before I finish, I want to point out that using a third variable, such as i₂ in this case, can make things more confusing than they need to be, thus making the solution more error prone.

Based on your loop definitions of i₁ and i₃, I suggest starting over and ignoring i₂ altogether. Trace through the loops in your circuit, and look at the equations that I came up with. Can you see how I did it? (Hint: it didn't involve a third variable or a third equation at all.)

Left Loop:
$$E_1 - i_1 R_1 - (i_1 - i_3)R_2 -E_2 -i_1 R_1 = 0$$

Right loop:
$$-E_3 - i_3 R_1 +E_2 -(i_3 - i_1) R_2 - i_3 R_1 = 0$$

 

[STEF2098]

Thanks, that actually helped alot!
And I looked over your method again, and it does make more sense.