Question

Two circuits (as shown in the following figure) are called Circuit A and Circuit B. The equivalent resistance of Circuit A is x and that of Circuit B is y (between 1 and 2). The value of $y-x$ is $\eta R$. Then the value of $\eta$ is _____.

Answer 1

0.0593

Answer 2

To find the value of $\eta$, we first need to calculate the equivalent resistances of Circuit A ($x$) and Circuit B ($y$). Both are infinite ladder networks, which can be solved by assuming the equivalent resistance of the infinite part of the circuit is the same as the total equivalent resistance.

Circuit A Analysis:

Circuit A is a ladder network where each repeating unit consists of a $2R$ resistor in the horizontal branch and an $R$ resistor in the vertical branch, connecting the top rail to the bottom rail.

Let $x$ be the equivalent resistance of Circuit A between terminals 1 and 2. If we consider the first unit cell (a $2R$ horizontal resistor and an $R$ vertical resistor), the rest of the infinite ladder network (from the node after the first $2R$ resistor to the bottom rail) will also have an equivalent resistance of $x$.

The equivalent resistance $x$ can be expressed as: The $2R$ resistor is in series with the parallel combination of the vertical $R$ resistor and the equivalent resistance $x$ of the rest of the ladder. $x = 2R + \frac{R \cdot x}{R + x}$ To solve for $x$, multiply both sides by $(R+x)$: $x(R+x) = 2R(R+x) + Rx$ $Rx + x^2 = 2R^2 + 2Rx + Rx$ $Rx + x^2 = 2R^2 + 3Rx$ Rearrange into a quadratic equation: $x^2 - 2Rx - 2R^2 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-2R$, $c=-2R^2$: $x = \frac{-(-2R) \pm \sqrt{(-2R)^2 - 4(1)(-2R^2)}}{2(1)}$ $x = \frac{2R \pm \sqrt{4R^2 + 8R^2}}{2}$ $x = \frac{2R \pm \sqrt{12R^2}}{2}$ $x = \frac{2R \pm 2R\sqrt{3}}{2}$ Since resistance must be positive, we take the positive root: $x = R(1 + \sqrt{3})$

Circuit B Analysis:

Circuit B is a different type of ladder network. Each repeating unit consists of a $2R$ resistor in the top horizontal branch, an $R$ resistor in the bottom horizontal branch, and an $R$ resistor connecting the two horizontal branches.

Let $y$ be the equivalent resistance of Circuit B between terminals 1 and 2. If we consider the first unit cell (a $2R$ top horizontal, an $R$ vertical, and an $R$ bottom horizontal resistor), the rest of the infinite ladder network (from the nodes after this first unit) will also have an equivalent resistance of $y$.

The equivalent resistance $y$ can be expressed as: The $2R$ resistor (top) is in series with a parallel combination. This parallel combination consists of the vertical $R$ resistor and the series combination of the bottom $R$ resistor and the equivalent resistance $y$ of the rest of the ladder. $y = 2R + \frac{R \cdot (R+y)}{R + (R+y)}$ $y = 2R + \frac{R^2 + Ry}{2R + y}$ To solve for $y$, multiply both sides by $(2R+y)$: $y(2R+y) = 2R(2R+y) + R^2 + Ry$ $2Ry + y^2 = 4R^2 + 2Ry + R^2 + Ry$ $2Ry + y^2 = 5R^2 + 3Ry$ Rearrange into a quadratic equation: $y^2 - Ry - 5R^2 = 0$ Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-R$, $c=-5R^2$: $y = \frac{-(-R) \pm \sqrt{(-R)^2 - 4(1)(-5R^2)}}{2(1)}$ $y = \frac{R \pm \sqrt{R^2 + 20R^2}}{2}$ $y = \frac{R \pm \sqrt{21R^2}}{2}$ $y = \frac{R \pm R\sqrt{21}}{2}$ Since resistance must be positive, we take the positive root: $y = \frac{R(1 + \sqrt{21})}{2}$

Calculate $\eta$:

We are given that $y-x = \eta R$. $y-x = \frac{R(1 + \sqrt{21})}{2} - R(1 + \sqrt{3})$ Factor out $R$: $y-x = R \left( \frac{1 + \sqrt{21}}{2} - (1 + \sqrt{3}) \right)$ $y-x = R \left( \frac{1 + \sqrt{21} - 2(1 + \sqrt{3})}{2} \right)$ $y-x = R \left( \frac{1 + \sqrt{21} - 2 - 2\sqrt{3}}{2} \right)$ $y-x = R \left( \frac{\sqrt{21} - 2\sqrt{3} - 1}{2} \right)$ Comparing this with $y-x = \eta R$, we get: $\eta = \frac{\sqrt{21} - 2\sqrt{3} - 1}{2}$

The value of $\sqrt{21} \approx 4.5826$ and $\sqrt{3} \approx 1.732$. $\eta = \frac{4.5826 - 2(1.732) - 1}{2}$ $\eta = \frac{4.5826 - 3.464 - 1}{2}$ $\eta = \frac{0.1186}{2}$ $\eta = 0.0593$

The question asks for the value of $\eta$.

The final answer is $\boxed{0.0593}$.

Explanation of the solution:

1. Identify the repeating unit: For both infinite ladder circuits, identify the basic repeating unit.
2. Formulate the self-similarity equation: Assume the equivalent resistance of the infinite ladder is $R_{eq}$. Replace the infinite part of the circuit after the first repeating unit with $R_{eq}$ itself.
3. Solve the quadratic equation: The resistance equation will typically reduce to a quadratic equation. Solve for the positive root, as resistance cannot be negative.
4. Calculate the difference: Subtract the equivalent resistance of Circuit A ($x$) from that of Circuit B ($y$) and express the result in the form $\eta R$ to find $\eta$.

Answer: The value of $\eta$ is $\frac{\sqrt{21} - 2\sqrt{3} - 1}{2}$. Numerically, $\eta \approx 0.0593$.