Irodov 3.176: A Current with No Voltage – Amit & Deepali Sinha Foundation

Problem Irodov 3.176

$N$ sources with different emfs are joined in series into a closed ring, every one with its positive terminal facing the same way round the ring. The emf of each source is proportional to its internal resistance, $\mathcal{E} = \alpha R$ , with $\alpha$ a given constant. The connecting wires have negligible resistance. Find:

the current in the ring
the potential difference between two points $A$ and $B$ of the ring that divide it into $n$ links on one side and $N - n$ on the other

The ring of sources, drawn with eight. Going clockwise, every cell is met at its short (-) plate first; A and B divide the ring into five links and three. — The ring of sources, drawn with eight. Going clockwise, every cell is met at its short ( $-$ ) plate first; $A$ and $B$ divide the ring into five links and three.

Given

$N$ sources in series, all aiding
$\mathcal{E}_i = \alpha R_i$
wire resistance negligible
$n$ links from $A$ to $B$

Find

(a)

I

· (b)

\varphi_A - \varphi_B

Solution

Step 1(a) The current round the ring

Each source is a link of emf $\mathcal{E}_i$ in series with its own internal resistance $R_i$ , as drawn below. All the sources face the same way round the ring, so all their emfs drive the current in the same sense, clockwise in the figure, and one current $I$ flows through every link.

One link of the ring: the emf E_i = α R_i and the internal resistance R_i in series, with the current I in the clockwise sense. — One link of the ring: the emf $\mathcal{E}_i = \alpha R_i$ and the internal resistance $R_i$ in series, with the current $I$ in the clockwise sense.

The loop rule round the whole ring:

\sum_{i=1}^{N} \mathcal{E}_i = I\sum_{i=1}^{N} R_i

Substituting $\mathcal{E}_i = \alpha R_i$ on the left:

\alpha\sum_{i=1}^{N} R_i = I\sum_{i=1}^{N} R_i

The sum of the resistances cancels:

I = \alpha

(1)

Step 2(b) Each link has no voltage across it

Walk from $A$ to $B$ through the $n$ links on one side, in the sense of the current. In link $i$ the potential rises by $\mathcal{E}_i$ across the emf and falls by $IR_i$ across the internal resistance:

\varphi_B - \varphi_A = \sum_{i=1}^{n}\left(\mathcal{E}_i - IR_i\right) = \sum_{i=1}^{n}\left(\alpha R_i - \alpha R_i\right)

Every term vanishes on its own:

\varphi_A - \varphi_B = 0

(2)

The same holds for any $n$ , so every point of the ring is at one potential.

Check

Units: $\alpha = \mathcal{E}/R$ is volts per ohm, a current, so $I = \alpha$ is dimensionally right. A special case: with $N$ identical sources of emf $\mathcal{E}$ and resistance $R$ , the current is $N\mathcal{E}/(NR) = \mathcal{E}/R$ , which is $\alpha$ , and the terminal voltage of each is $\mathcal{E} - IR = 0$ , as found.

Remark

A current flows with no potential difference anywhere along the ring: the electrostatic field is zero everywhere, inside the sources as well as in the wires, and each charge is pushed only by the non-electrostatic forces inside the sources. The proportionality $\mathcal{E}_i = \alpha R_i$ is exactly what makes every source short-circuit itself, $\mathcal{E}_i = IR_i$ , with the one common current. Break the proportionality and points of the ring take different potentials.