Question

Find the equivalent resistance (in $\Omega$) between the terminals A and B as shown on the diagram below. Put R = 120, r = 80 and neglect the resistance of leads.

Answer 1

145.83

Answer 2

Let $R_{eq}$ be the equivalent resistance of the infinite ladder network between the terminals A and B. The network consists of repeating units, each composed of a horizontal resistor $r$ and a vertical resistor $R$. Since the network is infinite, the equivalent resistance of the network starting from the point after the first unit is also $R_{eq}$.

Consider the first unit of the network. It consists of a resistor $r$ in the upper branch, connected from terminal A to an intermediate node, let's call it C. From node C, a resistor $R$ is connected to the lower branch. The lower branch is a continuous wire, and terminal B is located on this lower branch. The infinite network continues to the right of node C and the corresponding point on the lower branch.

Let's consider the equivalent resistance of the infinite network starting from node C and the lower branch. Due to the infinite nature and repeating structure, this equivalent resistance is also $R_{eq}$. So, from node C, we have a resistor $R$ connected to the lower branch, in parallel with the equivalent resistance of the rest of the infinite network to the right, which is $R_{eq}$. The equivalent resistance from C to the lower branch is the parallel combination of $R$ and $R_{eq}$:

$R_{C, lower} = R \parallel R_{eq} = \frac{R \cdot R_{eq}}{R + R_{eq}}$

The total equivalent resistance between A and B, $R_{eq}$, is the sum of the resistance of the first horizontal resistor $r$ (between A and C) and the equivalent resistance from C to the lower branch.

$R_{eq} = r + R_{C, lower}$ $R_{eq} = r + \frac{R \cdot R_{eq}}{R + R_{eq}}$

Now, we solve this equation for $R_{eq}$:

$R_{eq}(R + R_{eq}) = r(R + R_{eq}) + R \cdot R_{eq}$ $R \cdot R_{eq} + R_{eq}^2 = rR + rR_{eq} + R \cdot R_{eq}$ $R_{eq}^2 - rR_{eq} - rR = 0$

This is a quadratic equation in $R_{eq}$. Using the quadratic formula, we get:

$R_{eq} = \frac{-(-r) \pm \sqrt{(-r)^2 - 4(1)(-rR)}}{2(1)}$ $R_{eq} = \frac{r \pm \sqrt{r^2 + 4rR}}{2}$

Since resistance must be positive, we take the positive root:

$R_{eq} = \frac{r + \sqrt{r^2 + 4rR}}{2}$

We are given $R = 120 \, \Omega$ and $r = 80 \, \Omega$. Substituting these values:

$R_{eq} = \frac{80 + \sqrt{80^2 + 4 \cdot 80 \cdot 120}}{2}$ $R_{eq} = \frac{80 + \sqrt{6400 + 38400}}{2}$ $R_{eq} = \frac{80 + \sqrt{44800}}{2}$

Let's simplify the square root:

$44800 = 448 \times 100 = 64 \times 7 \times 100 = 8^2 \times 7 \times 10^2 = (8 \times 10)^2 \times 7 = 80^2 \times 7$ $\sqrt{44800} = \sqrt{80^2 \times 7} = 80 \sqrt{7}$

Substitute this back into the expression for $R_{eq}$:

$R_{eq} = \frac{80 + 80 \sqrt{7}}{2}$ $R_{eq} = \frac{80(1 + \sqrt{7})}{2}$ $R_{eq} = 40(1 + \sqrt{7})$

To find the numerical value, we use the approximation $\sqrt{7} \approx 2.64575$.

$R_{eq} \approx 40(1 + 2.64575)$ $R_{eq} \approx 40(3.64575)$ $R_{eq} \approx 145.83$