If we label the pds across successive vertical resistors "$V_n$", "$V_{n+1}$" and so on we can write the Kirchoff current law at a junction as$$\frac{V_n-V_{n+1}}{R_A}=\frac{V_{n+1}}{R_B}+\frac{V_{n+1}-V_{n+2}}{R_A}$$
in which $R_A$ and $R_B$ are the values of a horizontal and vertical resistor respectively. If you're happy with the assumption that $\frac{V_{n+2}}{V_{n+1}}=\frac{V_{n+1}}{V_{n}}\ [=\alpha\ \,\text{say}]\ $ throughout the ladder, then you have a quadratic equation for $\alpha$. Knowing $\alpha$ you can find the input current for a given input pd, and hence the input resistance.

Possibly rather less convincing than the 'no change with an extra section' method, but at least it's rather different! And it does give the same answer!

A few numerical investigationsâ€¦ Suppose the right hand end of the ladder is a vertical resistor (2 ohm) and the last section of the ladder is a horizontal resistor (1 ohm) connected to that vertical resistor. Then the resistance, $R_1$, of the last section seen from the left is 3 ohm. The resistance, $R_2$, of the last two sections together is, seen from the left, 11/5 ohm, then we have 43/21 ohm, 171/85 ohm. These resistances, I found, fit the pattern$$R_{m}\text{\ohm}=2+\frac{3}{4^{m}-1}.$$Clearly the convergence to 2 ohm will be very rapid!

**Addition to post**
Simplify notation by calling the horizontal resistance "*X*" and the vertical resistance "$\beta X$". Call the resistance of a ladder of *m* sections (one horizontal resistor joined on the right to a vertical resistor down to a 'ground' rail), "$r_{m}X$". Then, assuming that adding one more section on to the left of an infinite ladder will make no difference to its resistance we have$$r_{\infty}X=X+\frac{\beta X.r_{\infty}X}{\beta X+r_{\infty}X}\ \ \ \text{giving}\ \ \ \ \ r_{\infty}=\frac{1}{2}\left(1+\sqrt{1+4\beta}\right)$$

$\beta=2$ gives $r_{\infty}=2$, as we know. Certain other values of $\beta$ will also give rational values of $r_{\infty}$, for example, $\beta \ =$12, 6, 2, 3/4, 5/16, 9/64 give respectively, $r_{\infty}=\ $4, 3, 2, 3/2, 5/4, 9/8. No doubt in each of these cases we can find a general formula for $r_m$ as we did for when $\beta=2$, by inspecting the bit left over when we subtract $r_\infty$ from $r_1$, $r_2$ and so on.

For most values of $\beta$, this procedure won't stand a chance of working, because $r_{\infty}$ will be irrational, while $r_1$, $r_2$ and so on are clearly rational if $\beta$ is rational! Instead, for the general case, we use binomial series... Expanding the expression for $r_\infty$ given above we get
$$r_\infty=1+\beta-\beta^2+2\beta^3-5\beta^4+14\beta^5-42\beta^6+132\beta^7-429\beta^8+...$$
But clearly, $r_1=1+\beta$.

So $r_1$ agrees with $r_\infty$ up to the second term!

We find without much difficulty that$$r_2=1+\frac{\beta(1+\beta)}{1+2\beta}$$
Expanding$\frac{1}{1+2\beta}$ binomially and multiplying out, we get$$r_2=1+\beta-\beta^2 +2\beta^3 -4\beta^4+...$$
So we now have agreement with $r_\infty$ up to the fourth term!

Continuingâ€¦$$r_3=1+\frac{\beta(1+3\beta +\beta^2)}{1+4\beta +3\beta^2}$$and this gives us$$r_3=1+\beta-\beta^2+2\beta^3-5\beta^4+14\beta^5-41\beta^6+â€¦$$
So we now have agreement with $r_\infty$ up to the sixth term!

This may not have been very elegant, but I'm certainly now convinced! [Presumably $\beta$ is restricted to being less than 1 for this method to be valid.] Thank you for this interesting problem.

I am not
confident about this approach. I am confused about the inclusion of
the internal resistor in this calculation.

Whether or not to include the internal resistor in the calculation with the switch open depends on what $V$ is, which is not clear from the wording. Your calculation assumes $V$ is the battery emf, $\epsilon $. If in fact that is the case, they should have said $\epsilon$ = 5 volts. Then your calculation with the switch open that includes the internal resistance would be correct.

But in my opinion, if $V$ is the battery emf $\epsilon$ the problem becomes trivial. So I suspect $V$ is intended to be the battery terminal voltage and not the emf, as I show in the figure below. If that's the case then the voltage $V$ is across R1 and R2 in series not R1, R2, and r in series.

If $V$ is the battery terminal voltage, you can determine the $\epsilon$ by simply adding the voltage across $r$ due to $I$ to $V$. Then you can use the value of $\epsilon$ directly in doing the calculations with the switch closed.

Hope this helps.

## Best Answer

Yes and you've showed that using KCL and then current division. But, as a comment points out, the result follows simply from your problem statement.

(1) the resistors are connected in parallel which

meansthey have the same voltage across(2) there is a voltage source across the parallel resistors and no other components in the circuit so the voltage across either resistor is $V$, the terminal voltage of the voltage source

(3) the voltage produced by the voltage source is

constantThat's really all there is to it. Since the terminal voltage of the voltage source is constant, adding yet another resistor in parallel does not change $V$; $V$ is independent of the total current.

By Ohm's law, the current through the resistor is the voltage across divided by the resistance.

Thus,

$$I_n = \frac{V}{R_n},\qquad V\; \mathrm{constant}$$