Example: Resistors in parallel

Understanding the situation

In this examples, we'll consider the simplest example of a parallel circuit: one battery and two resistors connected in parallel.

Two circuit elements are in parallel means that they are connected so that whatever the potential is at the top of one of them is also the potential at the top of the other; similarly, the bottoms of the two elements are also at the same potential. (same voltage change)

Let's see how this works by solving a problem to determine the potential drops and currents in a pair of parallel resistors.

Presenting a sample problem

Two resistors of resistance $R_1$ and $R_2$ are connected to each other and to a battery as shown in the diagram(s) at the right. If the battery maintains a voltage difference of $V_0$ across its terminals, find the current in and voltage drop across each resistor. Assume the battery and resistors are ideal and the connecting wires are resistanceless.

Solving this problem

We've drawn the diagram in two forms: the semi-realistic one on the left and the more standard symbolic form on the right. 

To answer the questions in the problem we have our five tools to bring to bear: Kirchhoff's principles, the 0-resistance-wire heuristic, and our definition of what a battery is.

1. The total amount of current flowing into any volume in an electrical network equals the amount flowing out. (K1)
2. Across any single resistor, the potential drop across the resistor is proportional to the current through the resistor. The proportionality constant is the resistance: $ΔV = IR$. (K2)
3. Following around any loop in an electrical network the potential has to come back to the same value (sum of drops = sum of rises). (K3)
4. An element (e.g., a wire) having a 0 resistance is at a uniform potential: There is no potential drop since $R = 0$. (Heuristic)
5. A battery is a device that maintains a fixed potential difference across its terminals. (Definition of a battery)

To approach this problem, let's first get a sense of what's happening ("tell the story of the problem"). The battery maintains a potential difference across its terminals by principle 5 so it's like it's trying to push current out of its high end. Once it's connected up into a loop, it can create a flow — an electric current. In steady state (which takes about a nanosecond to establish in a circuit like this that would fit on a table). As the current $I$ runs around the circuit, as it passes through the wires, nothing happens to the potential (it doesn't change) by principle 4. When the current reaches the node joining the two resistors it must split. Principle 1 tells us how. Some will run through one of the resistors, some through the other. We don't yet know how much, but we might expect more might run through the small resistance ("follow the path of least resistance"). When the current gets through the resistors, it will rejoin and go back through the battery. For the currents that run through each resistors, there is a drop whose amount is governed by principle 2.  

Let's play this out in detail in terms of the symbols. Call the current through the battery $I$. When it reaches the junction it splits. We don't know how, but let's give them names. Let the current in $R_1$ be $I_1$ and the current in $R_2$ be $I_2$. Principle 1 tells us that

$$I= I_1 + I_2$$

and therefore that when they get back together at the bottom of the resistors, we have $I$ again.

A useful second step is often to "map the potential". We only have one battery, so we can pick the low end of the battery as the 0 of our potential.

Since the battery maintains a fixed potential difference across its terminals, we then know the high end. We can then follow wires in both directions, using our heuristic to specify the potential (since there is no change in a resistanceless wire). When we get to a resistor, there will be a change as we cross it. 

In our series example, the current was easy to deal with since there was just one loop. Here, that's no longer the case as the current splits. We'll see that the voltage is actually easier to deal with in this case. If we follow the voltages along the resistanceless wires from the top and bottom of the battery, we see that we know the voltages everywhere. There is no need to introduce a new name for an unknown voltage.

What else do we know? We've used principles 1, 4, and 5 in setting up our diagrams, and principle 3 is automatically satisfied. (That's what principle 4 does for us in a single loop network. It won't always do that for us.) Clearly, we have to satisfy principle 2: Ohm's law. Since this is ONLY true for a single resistor, we have to apply it once for each resistor. It give the following two equations:

$$V_0 = I_1R_1$$

$$V_0 = I_2R_2$$

In this case, since we know $V_0$ and $R_1$ and $R_2$, these equations give us the currents in the resistors. Since they add up to give us the $I$ in the battery we then know all the currents and all the voltage drops.

$$I = I_1 + I_2 = \frac{V_0}{R_1} + \frac{V_0}{R_2}$$

Making sense of the result

In general, it's not enough to solve a problem; we need to look at our answer and see that it makes sense. Here, that's especially important since resistors in parallel are going to be components of more complex problems we will analyze. Let's step back and see what's happened. Here are the results:

$$I_1 = V_0 \frac{1}{R_1}$$

$$I_2 = V_0 \frac{1}{R_2}$$

$$I = I_1 + I_2 = V_0 \bigg(\frac{1}{R_1} + \frac{1}{R_2}\bigg)$$

Since the thing multiplying $V_0$ is just a constant, we can write that equation so our pair of resistors looks just the same as if we have a single resistor connected across the battery with resistance $R^{parallel}_{effective}$ where $I = V_0/R^{parallel}_{effective}$.  This suggests the useful result:

To the outside circuit, a pair of resistors in series behaves the same as a single resistor with an inverse resistance equal to the sum of the inverse resistances: 
$$\frac{1}{R^{effective}_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}.$$

he resistors each experience the same potential drop, so they divide the current according to Ohm's law, by $I = \Delta V/R$. The larger resistance gets the smaller current, the current taking the path of least resistance proportionately.

It's interesting to compare the series and parallel results for resistors with a similar analysis for springs and capacitors. We can do this whenever we have a linear relation $y = Ax$ where in some cases the $x$ adds and the$y$ is constant and in other cases the $y$ adds and the x is constant. Anything like this gives similar relations for how "$A$" gets added to make an $A_{effective}$.

Joe Redish 2/21/16