# Example: A complex network

#### Prerequisites

- Kirchhoff's principles
- Example: Resistors in series
- Example: Resistors in parallel
- Example: Batteries in series and parallel

## Understanding the situation

In our previous three examples, we didn't actually have to use Kirchhoff's loop rule at all. The networks were simple enough that using our "resistanceless conductor heuristic" (that says a resistanceless conductor has the same potential on it everywhere, even if it is carrying a current) allowed us to map the values of the potential everywhere. That won't always be the case. Often, we'll do our mapping and our access to various points in the network will be blocked from all directions by resistors. There will be places where we won't know the potential. The solution here, as in many other analogous examples in physics, is to give names (symbols -- typically a *V* with a subscript) to the values of the potentials in those places and then write equations that involve them. If we get enough independent equations we can solve for our new unknowns. Let's consider a problem that requires such an approach.

While this model looks kind of random, it's actually a step in the direction of modeling the properties of a cell membrane. The top represents the outside of the membrane, the bottom the inside, and the cross-link the possibility of motion of ions across the membrane. A more realistic model includes capacitors and repeats the basic unit multiple times.

## Presenting a sample problem

The network shown at the right has two batteries (each 1.5 V) and three resistors (each 3.0 Ω). Assuming that the batteries and resistors are all ideal circuit elements and the wires conducting them can be treated as resistanceless, find the potential drop across and the current through each of the four elements.

## Solving this problem

A key feature of this network is that there are two **panes**. That is, if we consider just the shape of the connections, we can find two independent loops. Of course our parallel circuits had that too, but there we had a particular current that we could think of as "more important" that splits. We did that and it worked. We could do that here too, but we'll introduce a different technique that works when the connections are lots more complex: **loop currents. **For each independent pane we introduce a loop of current as shown in the figure at the right. Although we are given the numerical values of the resistors and batteries, we'll work with symbols until the end, just for practice. We'll call the resistances $R$ and the potential rises produced by each battery as $V_0$. Keep in mind that these are givens — not unknowns.

We choose one loop of current in each pane. How this works is that in the left pane, battery A and resistor 1 are in series -- only current $I_1$ goes through them. In the right pane, battery B and resistor 2 are in series — only current $I_2$ goes through them. But for resistor $R_3$, it has current $I_1$ going down through it and current $I_2$ going up through it. So the current through that resistor is $I_1 - I_2$ (or its negative — we'll see how the signs work when we do Ohm's law). These currents are the unknowns we have to find (along with the voltage drops).

Now let's see what our "resistanceless conductor heuristic" buys us. Remember, this says that a resistanceless conductor is at a constant potential, even if there is a current flowing through it. Let's do this a step at a time, since it's a bit tricky and there are a number of issues going on at once.

Let's start with the diagram on the left below. The first thing we have to do is choose a point in the network that we are going to consider our 0 potential. Let's choose the bottom of the battery A (and therefore the whole bottom wire) as our 0. That puts a 0 at the positive terminal of battery B which seems a bit peculiar, since batteries raise the potential, but this gives us a good chance to see how to work with negative signs in these networks. The top of battery A is $V_0$ above its negative side, so that tells us the potential all up the left side of the diagram.

Next, we know that battery B produces a rise of $V_0$ from its negative to its positive terminal. But its positive terminal is at 0! No problem; this just means that we have to have its negative terminal at a potential of $-V_0$. By our resistanceless conductor heuristic, we know the whole right side of the network is at that same potential. This is shown in the middle diagram.

Once those are done, the only place we don't know the potential is at the three-way junction between the resistors. If we knew the current in either of the top resistors we could find it by Ohm's law, but we don't know the currents. Are we stuck? Not at all. Whenever we don't know the value of something we need in a physics problem, we just give it a name and add it to our list of unknowns (so far, $I_1$ and $I_2$). Let's call if $V_1$ as shown in the diagram on the right.

Our third step is to apply Kirchhoff's resistance principle (Ohm's law) to each of the resistors. Since we don't actually know either the currents or the voltage drops, the approach we use is to write all the relationships we can and see if we have enough equations to find all our unknowns.

It's a bit tricky to keep track of signs in these sort of problems and they matter a lot. The rule is that the voltage drops across a resistance in the direction of the current flow. That is, the voltage is higher on the end of the resistor where the current is entering. This may seem a problem since we don't know the value — or even the direction! — of some of the currents, and we don't know the value — or even the sign — of some of the potentials. The great thing about solving these problems with algebra is THAT DOESN'T MATTER! We assume a particular direction for the current and if we get it wrong, our value for it will come out negative. Similarly for the potentials. This is one reason why it's really useful to work with algebraic symbols.

Here are the equations for Ohm's law across resistors 1, 2, and 3. The tricky one is #3 where there are two currents. I've assumed $V_1$ is positive so the current flows downward through $R_3$ (that is, $I_1$is greater than $I_2$). Check to be sure you understand how to get each equation.

$$I_1R = V_0 - V_1$$

$$I_2R = V_1 - (-V_0)$$

$$(I_1 - I_2)R = V_1$$

The first equation comes from applying Ohm's law to $R_1$. The current through it is $I_1$ and is going left to right. So we assume $V_0$ is higher than $V_1$ and the drop across the resistor is therefore $V_0-V_1$.

The second equation comes from applying Ohm's law to $R_2$. The current through it is $I_2$ and is also going left to right (along the top). So we assume $V_1$ is higher than $(-V_0)$ and the drop across the resistor is therefore $V_1-(-V_0)$. Note that $V_0$ is positive, so $-V_0$ is negative, but that doesn't matter: the drop is the initial (assumed higher) potential minus the final (assumed lower) potential going in the direction we assume the current is going. And we put in whatever the symbol is for the higher and lower potentials. Since the negative of a negative is a positive, the drop can be written $V_1+V_0$.

The third equation comes from applying Ohm's law to $R_3$. This is a bit tricky since there are two currents flowing through it, $I_1$ going down and $I_2$ going up. Since we think the current might flow down, we'll assume $V_1$ is positive (higher than zero). It doesn't matter since the math will give us a negative sign if we choose wrong. Since we have chosen the positive direction of flow as down, the current through $R_3$ will be $I_1 - I_2$. The potential drop is just $V_1 - 0 = V_1$.

So our three equations come down to just these:

$$I_1R = V_0 - V_1$$

$$I_2R = V_1 - V_0$$

$$(I_1 - I_2)R = V_1$$

Our givens are $R$ and $V_0$, and our unknowns are $I_1$, $I_2$, and $V_1$. We can solve first for $V_1$ by substituting the $IR$ terms from the first two equations into the third giving

$$V_0-V_1 - (V_1 +V_0) = V_1$$

The V0's cancel and the $V_1$s add giving $3V_1 = 0$. The only value that works for this is $V_1 = 0$, giving $I_1 = I_2 = V0/R$.

This means that the current through both batteries and $R_1$ and $R_2$ is $V0/R =$ (1.5 V)/(3.0 Omega) = 0.5 Amps. The potential drop across the batteries are both 1.5 Volts, of course (given), and since $V_1$ is 0, the potential drops across each resistor is just $V_0$ or 1.5 Volts. The current through $R_3$ is 0 (since $I_1 = I_2$) and therefore the drop across it is also 0.

## Making sense of the answer

Well, that's the answer, but it seems rather strange. Our final step in a problem should always be looking at the answers we got and seeing if it makes sense. Why did we get no current down the middle bar of the network? Could we have seen that right away?

In fact we could have. If we think of each battery as providing a push to drive a current, each is pushing the same amount and trying to push the current through the middle resistor in opposite directions. Since the resistance each battery's attempted current sees is the same, their currents will be the same and therefore cancel in the middle by symmetry.

To see if you understand what's going on, consider where you could move the resistors so the result would not change, and where you could move them so that the result would no longer be zero in the middle. If the resistors or batteries were of different values would you still get zero down the middle? What about if you flipped one of the batteries?

Joe Redish 3/16/16

Last Modified: May 3, 2019