An infinite ladder network of resistances is constructed with 1 ohm and 2 ohm resistances, as shown

Q: Why does removing one unit not change the total resistance?

Infinity minus one is still infinity! The infinite ladder is self-similar. If you look at the circuit starting from the second node, it looks exactly identical to the circuit starting from the first node. Hence, their equivalent resistances must be exactly the same.

Q: Why do we reject the negative root of the quadratic equation?

Resistance is a measure of opposition to current flow, which is a physical property of the material. In standard passive circuits, resistance cannot be negative. Therefore, we only accept the positive root.

Q: How do we know the current divides equally at the first junction?

At the first junction, the current has two paths: the $2\ \Omega$ resistor and the rest of the infinite ladder. We calculated the equivalent resistance of the entire infinite tail to be exactly $2\ \Omega$. Since both parallel paths have the same resistance ($2\ \Omega$), the current splits equally.

Q: What if the ladder was finite, say 100 units long?

If the ladder is finite, we cannot use the self-similarity trick. We would have to calculate the equivalent resistance from the end, working backwards step-by-step, or use difference equations. However, for a large number of units, the resistance converges very closely to the infinite ladder's value.

Q: Can this method be used for capacitors or inductors?

Absolutely! The concept of self-similarity applies to any infinite repeating network. You just need to use the appropriate series and parallel combination rules for capacitors or inductors to set up your equation.

Question

An infinite ladder network of resistances is constructed with $1\ \Omega$ and $2\ \Omega$ resistances, as shown in figure.
The 6 V battery between $A$ and $B$ has negligible internal resistance.
(a) Show that the effective resistance between $A$ and $B$ is $2\ \Omega$.
(b) What is the current that passes through the $2\ \Omega$ resistance nearest to the battery?

Sigma Percentile · Accepted Answer

### The Infinite Illusion

Imagine standing between two mirrors, looking at an endless corridor of reflections. If you step forward and pass the first reflection, the corridor still looks exactly the same, stretching into infinity. This profound concept of self-similarity is the secret weapon we will use to conquer this infinite ladder network.

In mathematics and physics, infinity is a bizarre and beautiful beast. Infinity minus one is still infinity. This means that if we take our infinite ladder of resistors and chop off the very first repeating unit—the first $1\ \Omega$ series resistor and the first $2\ \Omega$ parallel resistor—the remaining tail of the circuit is indistinguishable from the original ladder.

### The Power of Self-Similarity

Let's assign a variable to the unknown. Let the total equivalent resistance of the entire infinite ladder across terminals A and B be $x$. Because of the self-similarity we just discussed, the resistance of the infinite tail, starting from the second node, must also be exactly $x$.

This is a moment of pure elegance. We can replace that entire daunting, endless tail of resistors with a single, simple resistor of value $x$. Our infinite circuit has suddenly collapsed into a finite, manageable circuit with just three components: the first $1\ \Omega$ resistor, the first $2\ \Omega$ resistor, and our equivalent resistor $x$.

Now, let's analyze this simplified circuit. The $2\ \Omega$ resistor and our equivalent resistor $x$ are connected in parallel. We know that for two resistors in parallel, the equivalent resistance is the product divided by the sum. So, the resistance of this parallel pair is $\frac{2x}{2+x}$.

This parallel combination is in series with the initial $1\ \Omega$ resistor. Therefore, the total resistance of the entire circuit is $1 + \frac{2x}{2+x}$. But wait! We already defined the total resistance of the entire circuit as $x$. This gives us our master equation:

$$x = 1 + \frac{2x}{2+x}$$

### The Quadratic Resolution

We have transformed a complex physics problem into a straightforward algebra problem. Let's solve it. Multiplying the entire equation by $(2+x)$ to clear the denominator, we get:

$$x(2+x) = (2+x) + 2x$$

Expanding this yields $x^2 + 2x = 3x + 2$. Bringing all terms to one side gives us a neat quadratic equation:

$$x^2 - x - 2 = 0$$

Factoring this quadratic equation is simple. We look for two numbers that multiply to $-2$ and add to $-1$. Those numbers are $-2$ and $1$. So, we can write $(x-2)(x+1) = 0$. This gives us two possible mathematical solutions: $x = 2$ and $x = -1$.

Here, we must put our physics hats back on. Resistance is a physical property that opposes the flow of current; it cannot be negative in this context. Therefore, we discard the negative root. The effective resistance of the infinite ladder is exactly $2\ \Omega$.

### Unveiling the Current

With the total resistance found, the second part of the problem is a breeze. We have a $6	ext{ V}$ battery connected across the terminals A and B. The total resistance of the circuit is $2\ \Omega$. According to Ohm's Law, the total current $i$ drawn from the battery is:

$$i = \frac{V}{R} = \frac{6}{2} = 3	ext{ A}$$

This main current of $3	ext{ A}$ flows through the first $1\ \Omega$ resistor and then reaches a junction. Here, it must split into two paths: the first $2\ \Omega$ resistor, and the rest of the circuit. But remember, we found that the equivalent resistance of the rest of the circuit is also exactly $2\ \Omega$!

Because the resistance of both parallel paths is identical ($2\ \Omega$ each), the current will divide perfectly equally. The current through the first $2\ \Omega$ resistor is simply half of the main current:

$$i_1 = \frac{3}{2} = 1.5	ext{ A}$$

And there we have it! A problem that looked infinitely complex was unraveled using the elegant logic of self-similarity and basic circuit principles.

Visualized Solution

$Infinite Ladder Network$

$Concept of Infinity$

$Equivalent Circuit$

$Parallel Combination$

$Total Resistance Equation$

$Solving the Equation$

$Effective Resistance$

$Current Calculation Setup$

$Main Current$

$Current Division$

$Final Answer$

$The Way Forward$

The Sigma Insight: Combination of Resistors

The Infinite Illusion

The Power of Self-Similarity

The Quadratic Resolution

Unveiling the Current

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Infinite Ladder Network

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Concept of Infinity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Combination

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Resistance Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effective Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current Calculation Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Main Current

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Current Division

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward