Consider the circuit shown, with a 3V battery and three resistors. A 0.5A current flows through the circuit, and I want to know the potential difference between the points i (for initial) and f (final) shown. How do I go about it?

- The battery maintains a potential difference of 3V. Suppose the negative end of the battery has a potential of \(V_i\); then the positive end of the battery has a potential of \(V_i+3\u{V}\).
- When a current flows through a resistor, it drops through a
potential difference of
$$\Dl V=IR=(0.5\u{A})(3\u{\Omega})=1.5\u{V}.$$
The potential
*drops*from \(V_i+3\u{V}\) to \(V_i+3\u{V}-1.5\u{V}\).

The tally method is a way of standardizing this calculation. Imagine a small bug (represented here by a purple dot) which travels along the wires of a circuit, keeping track of changes in the potential as it goes. To find the potential difference between two points on a circuit:

- Start the bug at the initial position, and begin moving towards the final position along any path.
- If the bug crosses a
*battery*from negative to positive, add the battery's emf \(\cal E\) to the tally. (We'll call this "moving up" the battery.) If the bug "moves down" the battery, subtract the emf from the tally. - If the bug moves through a resistor \(R\) in the same direction as the current \(I\) (we can call this "downstream" or "with the current"), subtract \(IR\) from the tally. If it moves upstream, add \(IR\) to the tally.
- When the bug reaches the end position, the tally is the potential difference \(\Dl V\) between the points. The final potential \(V_f\) is the tally added to the initial potential: \(V_f=V_i+\Delta V\).

Interactive 11.2.2

In the circuit shown, the point in the upper-left corner has a
potential has 3V. What is the potential at the lower-right corner?

(A) We start at \(V=3\u{V}\). If we go clockwise around the loop, we go
*down* the 1Ω resistor, which is a drop of
\((0.25\u{A})(1\u{\Omega})=0.25\u{V}\). We then go *down* the
3Ω resistor for an additional drop of 0.75V, for a total drop of
1V. Thus the potential at the bottom-right corner is 3V-1V=2V.
Show this on the circuit.

(B) We can also go counterclockwise around the loop. We go
*down* the battery, so the potential *drops* by 2V. We
then go *up* the 4Ω resistor, which adds
\((0.25\u{A})(4\u{\Omega})=1\u{V}\) to the tally. The potential at the
bottom corner is
$$3\u{V}-2\u{V}+1\u{V}=\boxed{2\u{V}}$$
which is the same as before.
Show this on the circuit.

Start your bug in the upper-right corner, and go clockwise. The
bug goes down both resistors and then up the battery before
returning to its starting point. The tally is
$$-(2\u{\Omega})I-(3\u{\Omega})I+4\u{V}$$
Because the bug completed a loop, this tally must be equal to
zero:
$$\begin{eqnarray}
-2I-3I+4&=&0\\
\implies 4&=&5I\\
\implies I&=&\boxed{0.8\u{A}}\\
\end{eqnarray}$$