Kirchhoff's Laws

In most cases when we build a circuit, we already know which batteries and resistors we are using. What we don't know are the currents that will flow through the wires when the circuit is set up: these currents are unknown variables. In this circuit, for instance, there are three unknowns: the current \(I_1\) through the battery, and then the currents \(I_2\) and \(I_3\) which pass through the two batteries.

To solve for three unknown variables, we need three independent equations. We can get these equations using Kirchhoff's Laws:

Using a combination of the junction and loop rules, we can come up with the equations we need to solve for the unknown currents.
These two resistors are said to be "in parallel" with each other and with the battery. Find the current through both resistors and through the battery.
We have three unknown currents — \(I_1\), \(I_2\), and \(I_3\) — so we need three equations from Kirchhoff's Laws:
  • Junction Rule: To use the junction rule we need a junction: let's pick the one at the top. We see that current \(I_1\) goes into that junction, and the other currents go out of it. Therefore we get our first equation $$\color{black}{I_1=I_2+I_3}$$ How about the bottom junction? There, currents \(I_2\) and \(I_3\) flow in and \(I_1\) flows out, so we get the equation $$I_2+I_3=I_1$$ It's the same equation! That doesn't count. Typically you will find that at least one of the junction rule equations is redundant, and doesn't count as an independent equation. We stil need two more equations, so we turn to the
  • Loop Rule:
    There are three loops our bug could take in this circuit, as shown here. the loop on the left, the loop on the right, and a big loop around the perimeter of the circuit. We only need two more equations, so we can choose any two of these loops...but here are all three: $$\color{red}{+6-2I_2=0} \qquad \color{purple}{+6-3I_3=0} \qquad \color{blue}{-2I_2+3I_3=0}$$
Once we have our equations, we can solve for the currents. The first two loop equations are easy, and tell us that $$\color{red}{\boxed{I_2=3\u{A}}} \qquad \color{purple}{\boxed{I_3=2\u{A}}}$$ Notice that, because \(I_1\) doesn't pass through any resistors, it doesn't appear in any loop rule. We need the junction rule to find it: $$\color{black}{I_1=I_2+I_3=3\u{A}+2\u{A}\implies \boxed{I_1=5\u{A}}}$$

Find all the currents in this circuit.
Starting with the junction rule, we have $$\color{red}{I_1=I_2+I_3}$$ We can choose two of the three possible loop rules. I'll choose these two: $$\color{blue}{-3I_2+5I_3=0}$$ $$\color{black}{4-2I_1-3I_2=0}$$ Unlike in the previous example, all of these equations involve multiple variables, so we will have to solve them simultaneously. If you have experience with linear algebra you might break out a matrix right about now. For the rest of us, the trick is to eliminate all but one of the variables, solve for it, and then use that to find the others. For example, $$\newcommand\eq[3]{\color{#1}{#2}&\color{#1}{=}&\color{#1}{#3}} \begin{eqnarray} \eq{blue}{3I_2}{5I_3}\implies \color{blue}{I_2={5\over3}I_3}\\ \eq{red}{I_1}{\color{blue}{{5\over3}I_3}\color{red}{+I_3={8\over3}I_3}}\\ \eq{black}{4-2\color{red}{{8\over3}I_3}\color{black}{-3}\color{blue}{{5\over3}I_3}}{0}=\color{black}{4-{31\over3}I_3}\\ \end{eqnarray}$$ We rewrote the first two equations to solve for \(I_1\) and \(I_2\) in terms of \(I_3\), and then plugged them into the third equation, which is now entirely in terms of \(I_3\). We can now solve for \(I_3\), and substitute it into the first two equations to get the other variables. $$\begin{eqnarray} \eq{black}{4-{31\over3}I_3}{0} \implies I_3={12\over31}\u{A} =0.39\u{A}\\ \eq{red}{I_1}{{8\over3}\left({12\over31}\u{A}\right)}={32\over31}\u{A} =1.03\u{A}\\ \eq{blue}{I_2}{{5\over3}\left({12\over31}\u{A}\right)}={20\over31}\u{A} =0.64\u{A}\\ \end{eqnarray}$$ (Note: I did the calculation in fractions, and as you can see they can get pretty unwieldy in these problems. Please feel free to use decimals in your own calculations.)

Find the currents through the three batteries. Assume that current can flow backwards through our ideal batteries with no ill effects.

The first thing we need to do is to label the currents. In our previous examples it was obvious which way the currents would flow, but here it's not so clear. That's ok, though. We can still define a direction for each current, as shown in the diagram to the left. If we get the direction wrong, then the current will turn out to be negative.

Going counterclockwise around the right loop gives us the equation $$9-8-1I_3=0$$ which we can solve right away to find \(I_3=1\u{A}\).

Going clockwise around the left loop gives us the equation $$2-8-2I_1=0 \implies 2I_1=-6$$ which means \(I_1=-3\u{A}\). That means that a current of 3 amps will flow downward through the 2V battery.

Lastly, if we look at the junction at the top, $$I_1+I_2+I_3=0$$ This would have no solution if all the currents were positive, but of course we know that \(I_1\) is negative. Solving this for \(I_2\): $$I_2=-I_1-I_3=-(-3\u{A})-(1\u{A})=2\u{A}$$ so current will flow up the 8V battery.