Kirchoff's loop rule with multiple voltage sources

1. May 4, 2014

Avatrin

Hi

Let's look at this:

The issue I am having is that while I know that this is accurate:

I just cannot wrap my head around why. I1 is a result of both voltage sources. So, while the equations clearly tell me that the voltage drop across the resistor R1 is I1R1, I cannot wrap my head around how this works in terms of Kirchoff's loop rule.

2. May 4, 2014

jim hardy

Doesn't I3 steal from I2 to make I1 different? Surely you accept that I3 is affected by E2.

3. May 4, 2014

Avatrin

Yes, I get that part. It is the Kirchoff's loop rule that I cannot wrap my head around.

I1 is affected by both ε1 and ε2. Yet, I can use the equation above ignoring ε2. It is true regardless of ε2. That part I cannot wrap my head around.

4. May 4, 2014

jim hardy

Ahhh i see.

I think it's not an electrical ambiguity, rather a temporary memory lapse in translating from the algebra you know well to the electricity that seems unfamiliar.

Back to algebra basics: It takes more than one equation to solve for more than one unknown.
So that one equation isn't the whole story.

We all stumble over these little difficulties. Work lots of homework problems..

old jim

Last edited: May 4, 2014
5. May 4, 2014

mjhilger

If I understand correctly, your trouble seems to be the fact that the ε2 source is not represented in the first equation. The issue is that I2 is not a part of the loop equations in their most basic form. Rather I2 = I1 - I3, and there are only two equations that are necessary for the drawn circuit. Those loop equations need only use I1 & I3, so while ε2 is not directly reflected in the loop 1 equation, it is a part in as much as it factors into the 2nd loop equation as a boundary condition of I3's loop.

Follow Jim's advice and use the math and you will see that a change in ε2 does cause a change in the first equations value.