Why is the current calculated this way

Thread Starter

Malkawi99

Joined Oct 18, 2018
5
I can't comprehend how did this example reach equation (2.147). The first part which is -R2/(R1+R2)I is pretty easy and straightforward, it's just a current divider formula for I. The second part looks like a voltage divider formula, but the resistors are connected in parallel, how is voltage division applicable in this case?

 

WBahn

Joined Mar 31, 2012
29,979
While the first term is the current divider equation for I, notice that it is NOT what you might expect because it ignores the influence of V1. The current divider formula assumes that the two resistors are in parallel and these two are not. However, IF you turn the voltage source off (i.e., set its output voltage to zero so that it behaves like a short circuit), now they ARE in parallel. The leading minus sign comes from the fact that we want i1, which is defined in the opposite direction as the current from the current source.

As for the second term, it is NOT a voltage divider formula at all, in any way, shape, or form. Look at the units. It is a voltage divided by a resistance, which is a current. What if we turn the current source off (i.e., set its current output to zero so that it behaves like an open circuit)? Now that formula is the current i1.

So the first term is the current i1 due solely to the current source while the second term is the current i1 due solely to the voltage source.

As for where 2.147 comes from, the line above it gives the big hint -- you need to take all eight equations above it and manipulate them to get a solution in terms of the knowns and then manipulate that to separate that equation into terms involving each of the independent sources.

Take your best shot at doing that and post your work, we can then help you overcome whatever hurdles you come up against.
 

Thread Starter

Malkawi99

Joined Oct 18, 2018
5
Ok So I tried taking two routes, I first started by substituting i1=i2+i3 into V1= R1*i1, and tried to go from there. But it lead to a dead end. I then tried starting with i1=i2+i3, substituted a V equation and I into it, it also lead to a dead end. I am kind of stomped. Thanks for offering to help me out.

Here is my work, hopefully it was a small step in the right direction:
 

Thread Starter

Malkawi99

Joined Oct 18, 2018
5
I just realized that my second attempt is VERY close to the desired solution. But it seems that I am missing the R2/(R1+R2) term, which I can't comprehend its use in this circuit.
 

WBahn

Joined Mar 31, 2012
29,979
Best lesson you can learn is that units are your best friend. Constantly check your work at each step to see it is dimensionally consistent. You have a couple of places (in your top attempt) where your units stop working out -- which means that any and all work done from that point on is guaranteed to be wrong and is an utter waste of time. We all make stupid and silly math errors all the time, so get in the habit of always tracking your units throughout your work and you will spot a huge fraction of those mistakes immediately after making them instead of having to hunt for them after five or six pages of work finally reveals itself as being completely wrong.

Keep in mind that what you have is eight equations in eight unknowns. There are several systematic ways of solving them, such as Gaussian elimination. With equations this simple, however, it is usually easiest to make substitutions to eliminate first those variables that are easy to eliminate.

Also keep in mind that unless these eight equations are actually two (or more) separate and independent systems of equations, you won't be able to solve for any of the unknowns without using all eight of the equations. Since it appears that changing any of the circuit parameters will affect all eight unknowns, it is highly likely that all eight equations are coupled (and therefore not independent). Now, having said that, inspection reveals that Eqn 2.143 is independent of the others and this makes sense since KCL imposes this relationship regardless of whatever happens in the circuit. Similarly, Eqn 2.146 is imposed by KVL regardless of anything else. So we can set these aside and work with the other six.

Now go for the low hanging fruit. Use Eqn 2.139 through 2.142 to get rid of v0, v1, v2, and i3 in the other two equations (Eqn 2.144 and 2.145). You now have two equations in two unknowns, i1 and i2. Solve one equation for i2 and substitute that into the other equation and then solve that one for i1 (the only remaining unknown), collecting terms involving V and I separately.
 
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