Solution to Question 7
1. Identifying the Repeating Unit
The circuit is an infinite ladder network. Since the ladder is infinite, adding or removing one repeating section does not change the equivalent resistance. Let the equivalent resistance of the entire infinite network looking into the input terminals (left side) be $R_{eq}$.
We can visualize the circuit as a single “L-section” (one series resistor $R$ and one shunt resistor $R$) connected in series with the rest of the infinite ladder. The rest of the ladder, looking to the right of the first shunt resistor, is identical to the original ladder and thus has resistance $R_{eq}$.
2. Setting up the Equation
The circuit simplifies to:
- A resistor $R$ in series with a parallel combination.
- The parallel combination consists of the vertical resistor $R$ and the rest of the ladder $R_{eq}$.
Mathematically, this is expressed as:
$$ R_{eq} = R + (R \parallel R_{eq}) $$ $$ R_{eq} = R + \frac{R \cdot R_{eq}}{R + R_{eq}} $$3. Solving the Quadratic Equation
Multiply through by $(R + R_{eq})$ to clear the denominator:
$$ R_{eq}(R + R_{eq}) = R(R + R_{eq}) + R \cdot R_{eq} $$ $$ R \cdot R_{eq} + R_{eq}^2 = R^2 + R \cdot R_{eq} + R \cdot R_{eq} $$Subtract $R \cdot R_{eq}$ from both sides:
$$ R_{eq}^2 = R^2 + R \cdot R_{eq} $$ $$ R_{eq}^2 – R \cdot R_{eq} – R^2 = 0 $$This is a quadratic equation in $R_{eq}$ of the form $ax^2 + bx + c = 0$, where $a=1$, $b=-R$, and $c=-R^2$. Using the quadratic formula:
$$ R_{eq} = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$ $$ R_{eq} = \frac{R \pm \sqrt{(-R)^2 – 4(1)(-R^2)}}{2} $$ $$ R_{eq} = \frac{R \pm \sqrt{R^2 + 4R^2}}{2} $$ $$ R_{eq} = \frac{R \pm \sqrt{5R^2}}{2} = \frac{R(1 \pm \sqrt{5})}{2} $$Since resistance cannot be negative, we discard the solution with the minus sign ($\sqrt{5} > 1$, so $1-\sqrt{5}$ is negative).
Answer
The equivalent resistance is:
$$ R_{eq} = \frac{R(1 + \sqrt{5})}{2} $$