Solution
Let the resistance of each voltmeter be $R_V$ and the internal resistance of each battery be $r$. Let the potential at the $n$-th node (top wire) be $U_n$ and the bottom wire be ground ($0$). Then the reading of the $n$-th voltmeter is $V_n = U_n$.
1. Recurrence Relation:
Apply Kirchhoff’s Current Law (KCL) at the $n$-th node ($n \ge 1$). The sum of currents leaving the node is zero:
$$ I_{left} + I_{down} + I_{right} = 0 $$
The current leaving to the left (towards node $n-1$) flows against the potential rise of the battery $\mathcal{E}$.
$$ \frac{V_n – (V_{n-1} – \mathcal{E})}{r} + \frac{V_n}{R_V} + \frac{V_n – (V_{n+1} + \mathcal{E})}{r} = 0 $$
Note on Polarity: For the voltage readings to decay ($V_{n+1} < V_n$) in a passive-like manner, the batteries must be opposing the current flow from left to right, or arranging potentials such that they cancel in the recurrence. Let's arrange the terms:
$$ \frac{V_n - V_{n-1} + \mathcal{E}}{r} + \frac{V_n}{R_V} + \frac{V_n - V_{n+1} - \mathcal{E}}{r} = 0 $$
Notice that the constant $\mathcal{E}$ terms cancel out ($\frac{\mathcal{E}}{r} - \frac{\mathcal{E}}{r} = 0$). This leaves a homogeneous difference equation for voltage:
$$ \frac{2V_n - V_{n-1} - V_{n+1}}{r} + \frac{V_n}{R_V} = 0 $$
Multiplying by $r$:
$$ 2V_n - V_{n-1} - V_{n+1} + \frac{r}{R_V} V_n = 0 $$
$$ V_{n+1} + V_{n-1} = \left( 2 + \frac{r}{R_V} \right) V_n $$
Dividing by $V_n$ and using the given geometric ratio $\frac{V_{n-1}}{V_n} = \eta$ and $\frac{V_{n+1}}{V_n} = \frac{1}{\eta}$:
$$ \frac{1}{\eta} + \eta = 2 + \frac{r}{R_V} $$
$$ \frac{r}{R_V} = \eta + \frac{1}{\eta} - 2 = \frac{(\eta - 1)^2}{\eta} $$
2. Boundary Condition at the First Node ($n=0$):
The first voltmeter ($V_0$) is at the start of the semi-infinite chain. There is no circuit to its left.
Apply KCL at node 0:
$$ I_{down} + I_{right} = 0 $$
$$ \frac{V_0}{R_V} + \frac{V_0 – (V_1 + \mathcal{E})}{r} = 0 $$
(Assuming the battery polarity pushes current to the left, acting as a source for the first node).
Rearranging:
$$ \frac{V_0}{R_V} = \frac{\mathcal{E} + V_1 – V_0}{r} $$
Substitute $V_1 = V_0 / \eta$:
$$ \frac{r}{R_V} V_0 = \mathcal{E} + V_0 \left( \frac{1}{\eta} – 1 \right) $$
Now substitute the expression for $\frac{r}{R_V}$ derived from the recurrence:
$$ \left( \frac{(\eta – 1)^2}{\eta} \right) V_0 = \mathcal{E} – V_0 \left( \frac{\eta – 1}{\eta} \right) $$
$$ \mathcal{E} = V_0 \left[ \frac{(\eta – 1)^2}{\eta} + \frac{\eta – 1}{\eta} \right] $$
$$ \mathcal{E} = \frac{V_0}{\eta} \left[ (\eta – 1)^2 + (\eta – 1) \right] $$
$$ \mathcal{E} = \frac{V_0}{\eta} (\eta – 1) [ (\eta – 1) + 1 ] $$
$$ \mathcal{E} = \frac{V_0}{\eta} (\eta – 1) [ \eta ] $$
$$ \mathcal{E} = V_0 (\eta – 1) $$
3. Calculation:
Given $V_0 = 15 \text{ V}$ and $\eta = 1.1$.
$$ \mathcal{E} = 15 \times (1.1 – 1) $$
$$ \mathcal{E} = 15 \times 0.1 $$
$$ \mathcal{E} = 1.5 \text{ V} $$
Answer: The EMF of the battery is $1.5 \text{ V}$.