Question

Find equivalent Resistance blw A and B.

$R_{AB}$=$R_{eq}$ = ?

A R c R/2 R/4 R/8 R/16 R/32 R/64 R/128

B R D R/2 R/4 R/8 R/16 $\frac{R}{32}$ R/64 $\frac{R}{128}$ $\infty$/ladder

Answer 1

$R \left( \frac{1 + \sqrt{17}}{2} \right)$

Answer 2

The problem describes an infinite ladder network where the resistances in each subsequent section are scaled by a factor of 1/2. Let the equivalent resistance between points A and B be $R_{eq}$.

The first "unit" of the ladder consists of:

1. A resistor of value R in the upper branch (between A and C). Let's call this $R_{AC}$.
2. A resistor of value R in the lower branch (between B and D). Let's call this $R_{BD}$.
3. A vertical resistor of value R connecting the upper and lower branches (between C and D). Let's call this $R_{CD1}$.

The network to the right of points C and D is an infinite ladder. All the resistances in this part of the network are half of the corresponding resistances in the original ladder. For example, the next horizontal resistors are R/2, the next vertical resistor is R/2, and so on. Therefore, if the equivalent resistance of the entire ladder from A to B is $R_{eq}$, then the equivalent resistance of the infinite ladder starting from C and D will be $R_{eq}/2$. Let's call this $R_{CD\_ladder}$.

Now, we can redraw the circuit by replacing the infinite ladder to the right of C and D with its equivalent resistance $R_{CD\_ladder} = R_{eq}/2$.

The circuit between points C and D now has two resistors in parallel:

1. The vertical resistor $R_{CD1} = R$.
2. The equivalent resistance of the rest of the ladder $R_{CD\_ladder} = R_{eq}/2$.

The equivalent resistance between C and D, due to these parallel components, is: $R_{CD\_parallel} = R_{CD1} \parallel R_{CD\_ladder}$ $R_{CD\_parallel} = \frac{R \times (R_{eq}/2)}{R + (R_{eq}/2)}$ $R_{CD\_parallel} = \frac{R \cdot R_{eq}}{2R + R_{eq}}$

Now, the entire circuit from A to B can be viewed as three resistors in series:

1. $R_{AC}$ (from A to C)
2. $R_{CD\_parallel}$ (from C to D)
3. $R_{BD}$ (from D to B)

So, the equivalent resistance $R_{eq}$ is: $R_{eq} = R_{AC} + R_{CD\_parallel} + R_{BD}$ Substitute the values: $R_{AC} = R$, $R_{BD} = R$. $R_{eq} = R + \frac{R \cdot R_{eq}}{2R + R_{eq}} + R$ $R_{eq} = 2R + \frac{R \cdot R_{eq}}{2R + R_{eq}}$

Now, we solve this equation for $R_{eq}$. Let $X = R_{eq}$. $X = 2R + \frac{RX}{2R + X}$ Multiply both sides by $(2R + X)$: $X(2R + X) = 2R(2R + X) + RX$ $2RX + X^2 = 4R^2 + 2RX + RX$ $2RX + X^2 = 4R^2 + 3RX$ Rearrange the terms to form a quadratic equation: $X^2 - RX - 4R^2 = 0$

Using the quadratic formula $X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-R$, $c=-4R^2$: $X = \frac{-(-R) \pm \sqrt{(-R)^2 - 4(1)(-4R^2)}}{2(1)}$ $X = \frac{R \pm \sqrt{R^2 + 16R^2}}{2}$ $X = \frac{R \pm \sqrt{17R^2}}{2}$ $X = \frac{R \pm R\sqrt{17}}{2}$ Since resistance must be a positive value, we take the positive root: $R_{eq} = \frac{R + R\sqrt{17}}{2}$ $R_{eq} = R \left( \frac{1 + \sqrt{17}}{2} \right)$

The final answer is $R \left( \frac{1 + \sqrt{17}}{2} \right)$.

Explanation of the solution:

1. Identify the repeating pattern: The network is an infinite ladder where each subsequent section's resistances are scaled by 1/2.
2. Define equivalent resistance: Let the equivalent resistance of the entire infinite ladder be $R_{eq}$.
3. Utilize infinite ladder property: Due to the scaling, the equivalent resistance of the ladder starting from the second set of nodes (C and D) is $R_{eq}/2$.
4. Simplify the first section: The vertical resistor (R) in the first section is in parallel with the equivalent resistance of the rest of the ladder ($R_{eq}/2$). Calculate this parallel combination.
5. Formulate the equation: The two horizontal resistors (R each) in the first section are in series with the calculated parallel combination. Set up an equation for $R_{eq}$.
6. Solve the quadratic equation: Solve the resulting quadratic equation for $R_{eq}$, choosing the positive root as resistance cannot be negative.