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

### Kirchhoff's Junction Rule

*(This is sometimes called Kirchhoff's Current Law (KCL), or just the "junction rule".)*

### Kirchhoff's Loop Rule

*(This is sometimes called Kirchhoff's Voltage Law (KVL), or just the "loop rule".)*

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:***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: $$\color{red}{+6-2I_2=0} \qquad \color{purple}{+6-3I_3=0} \qquad \color{blue}{-2I_2+3I_3=0}$$

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.)

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.