Class Content > Electric currents > Kirchoff's principles

Prerequisites:

• Quantifying electric current
• Resistive electric flow: Ohm's law
• Kirchhoff's principles

Kirchhoff's principles summarize some of the key ideas that control the flow of electric current through networks of connected electrical elements. In this webpage we'll go through a couple of examples to illustrate how this works -- and as a result, derive the rules for how resistances behave when connected in series and parallel. Note that the principles describe the stead state of a network -- what happens when the currents have been established and everything has settled down to a steady flow. They don't describe the start up (first nanosecond or so) or what happens when things are changing in time.

Briefly the principles are:

1. Flow rule -- In a steady-state electrical network, the current flowing into any volume equals the amount flowing out of the volume.
2. Ohm's law -- In any resistor, the current flowing through that resistor is related to the potential drop across the resistor by Ohm's law: ΔV = IR.
3. Loop rule -- If you follow around any loop in a steady-state electrical network the sum of the rises of the potential and the drops of the potential will be equal.

We add a useful heuristic:

4. Resistanceless conductor rule -- A resistanceless conductor (e.g., a wire or anything that has R = 0) has 0 potential drop along it, even when it is carrying current.

The approach

The idea in analyzing an electrical network is to figure out what is the current through each element and what is the voltage drop across each element. Often there will by many things you don't know. The common approach of figuring things out one at a time in sequence usually will not work. You have to name all your unknowns and then write equations relating them.

All these unknowns can be tied together in an intricate way and this is what makes these problems challenging. Some of our principles (1,2,4) are local -- apply at a particular point or element, others are global -- tie together the whole network (2,3). The analysis of a network often requires you to switch perspectives -- writing an equation about a single element and then writing one that relates multiple elements.

Example 1: Two resistors in series

The simplest non-trivial example is a single battery with two resistors connected sequentially (in series). This is shown in the diagram at the right. To analyze this, we could define 6 separate currents and the corresponding voltage drops -- in the battery, two resistors, and three wires. Doing this kind of mindless algorithm makes for impossible equations. The best idea is to have some "stakes in the ground" -- intuitions you know you can trust -- and rely on those.
Let's analyze this a step at a time. First, we identify the elements. Let's assume that the battery provides a voltage difference V0 and the resistors have resistances RA and RB (not the same). We can choose any single point in our network as 0 voltage (NOT more than one point can be chosen freely, though). If we have only a single battery, it's convenient to choose the low end of the battery as 0. This gives the diagram at the right.
Now let's think about and name the currents through each element. The flow role tells us that since we only have one loop, we must have the same current everywhere (like the rope model). So we can name the current through each piece of the circuit, I.  (Imagine a box covering any element. There is only one wire going in and one wire going out so the current going in must equal the current going out.)
Now let's map the voltages. We have chosen the low end of the battery to be 0 potential, so the high end is V0. Since we are treating the wires as resistanceless, every point connected to the 0 end of the battery is at 0 potential -- until we reach a resistor. Every point connected to the V0 end is at potential V0 -- until we reach a resistor. We therefore know what the potential is everywhere -- except between the two resistors. Give the potential at that point the name V1.

Now our diagram is fully labeled. We are ready to write our equations. The trick here is now going to be trying to keep track of which of these similarly looking symbols are our knowns and which are our unknowns! Of course, it doesn't really matter! (Honest!) What our rules give are equations that are relations among the variables -- whether you know them or not. In some problems you might know one set of things, in another, a different set.

In this case, we'll assume we know that voltage difference provided by the battery (V₀) and the magnitude of each resistance (R_A and R_B). What we want to find the the current, I, and the voltage between the resistors, V₁.

What do we have to work with? We have Ohm's law in each resistor (a local principle), and the loop rule (a global one). But we only have 2 unknowns and these will be three equations. Let's see if the Ohm's laws will do for us. Since the currents are the same, here's what the Ohm's law for each resistor gives:

V₀-V₁ = IR_A

V₁-0 = IR_B

There are two equations for two unknowns so they should suffice. We can use the second equation to substitute for V₁ in the first equation to get (we don't have to keep the "-0"!)

V₀ - IR_B = IR_A

V₀ = IR_A + IR_B

V₀ = I(R_A + R_B)

I = V₀/(R_A + R_B)

We can now find the drop across each resistor:

Drop across R_A = V₀-V₁ = IR_A = V₀ R_A/(R_A + R_B)

Drop across R_B = V₁ = IR_B = V₀ R_B/(R_A + R_B)

This makes sense: the fraction of the total drop that happens across each resistor is the fraction of the total resistance that is in that resistor. And the effect of the resistor (as seen from the battery) is the same as a resistance that is the sum of the two:

R_series = R_A + R_B.

Let's check the flow rule.  As we go around the loop in the direction of the current, we find one rise -- in the battery, of an amount V₀; and we find two drops -- one across R_A and one across R_B of amounts that add up to

V₀ R_A/(R_A + R_B) + V ₀ R_B/(R_A + R_B) = V₀ (R_A + R_B)/(R_A + R_B) = V₀

so the sum of the rises equal the sum of the drops.

Example 2: Two resistors in parallel

A second fairly simple example is two resistors that are each connected directly to the battery. One representation of this looks like the figure at the right. Although it doesn't look like B is connected directly to the battery, since we are ignoring any potential drops along the wires, it is effectively. We call this parallel.
Let's assume we have the same devices we had before and label them the same way.
Now let's label currents. If a current I comes out of the battery, we won't  have a current I everywhere in the circuit. If we put a box around the T-junction under the resistor RA, we see that there is one current coming in and two coming out. We have to give each of these different labels. We do know that in any single leg -- where there is only one out and one in -- that the current has to be constant along that line by the flow rule. To make it easy to see what's what, we've called the current in battery, I, and colored it black. We've called the current in RA IA and colored it red. We've called the current in RB IA and colored it blue.
Now we need to map the voltage. We use principle 4 -- along any wire that has no resistance the voltage doesn't change. Note, therefore, that NOTHING happens at the split. The current splits, but the voltage, not being something that moves, does not. This shows us that the voltage drop across each of the resistors is equal to the full voltage drop across the battery. This is what be meant by saying that each resistor "is connected directly to the battery."

Now we can use our principles to write equations. Again, we'll assume we know the voltage across the battery and the resistances. In this case, then, our unknowns are the three currents. We can expect to need three equations.

The first is provided by the flow rule: into any place, the current going in equals the current going out.  This means we must have

I = I_A + I_B

Next, we can use Ohm's law across each of the resistors:

I_A = V₀/R_A

I_B = V₀/R_B

Since we know both V₀  and the two resistances, this gives us the solution for the two currents. Putting these together, it gives us the total current:

I = I_A + I_{B =} V₀/R_A + V₀/R_B = V₀ (1/R_A + 1/R_B)

If we put a box around both resistors and ask, what does the effect of the pair of resistors look like to the batter, we would see V₀ = IR_effective. A line or two of algebra shows that resistors in parallel produce a combined effect that looks like they add upside down!

1/R_effective = 1/R_A + 1/R_B

or flipping this over and simplifying,

R_effective = R_AR_B/(R_A + R_B)

We can check the loop rule on any of three different loops! The one including the battery and RA, the one including the battery and RB, and the one including only the two resistors. The first two are obvious since there is the rise from the battery and the drop across one resistor in each loop. Since the drop across each resistor is equal to the full rise in the battery, this works trivially. If we look at the rightmost loop not containing the battery, it's a little trickier. But if we follow the loop around, for one of the resistors we are going in the same direction as the current and we see a drop. But for the other resistor, we are going in the opposite direction of the current, so the drop across the resistor looks like a rise!

Joe Redish 3/1/12