Question

An infinite number of resistors are connected in parallel. If \(R_{1}=10 \Omega, R_{2}=10^{2} \Omega, R_{3}=10^{3} \Omega,\) and so on, show that \(R_{e q}=9 \Omega\).

Short answer

Answer: The equivalent resistance of the infinite parallel resistors circuit is 9Ω.

Step-by-step solution

Write the formula for resistors in parallel

The formula for the equivalent resistance of resistors in parallel is given by: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \)

Substitute the given values

Given that \(R_1 = 10\Omega\), \(R_2 = 10^2\Omega\), \(R_3 = 10^3\Omega\), and so on, we can rewrite the formula as: \( \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \cdots \)

Recognize the geometric series

The given series is a geometric series with a first term \(a = \frac{1}{10}\) and a common ratio \(r = \frac{1}{10}\). To find the sum of an infinite geometric series, we can use the formula: \( S = \frac{a}{1 - r} \)

Calculate the sum of the infinite geometric series

Applying the formula for the sum of an infinite geometric series to our problem, we have: \( S = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9} \)

Find the equivalent resistance

We know that the sum of the geometric series is equal to the reciprocal of the equivalent resistance: \( \frac{1}{R_{eq}} = S = \frac{1}{9} \) To find \(R_{eq}\), take the reciprocal of the sum: \( R_{eq} = \frac{1}{\frac{1}{9}} = 9\Omega \) Therefore, the equivalent resistance of the infinite parallel resistors circuit is 9Ω.

Key concepts

Equivalent Resistance

When resistors are connected in parallel, the total or "equivalent" resistance follows a specific rule. The formula for resistors in parallel is:\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots \)This means the total resistance decreases as you add more resistors in parallel.
The effect broadens the current path, allowing more current to flow compared to a single resistor in use. Now normally, with just a few resistors, you might calculate this sum directly. However, when dealing with an infinite number of resistors, the calculation becomes a little trickier.
In this example, each new resistor's resistance value increases by a power of 10. Let's dive into how this forms a geometric series, helping us calculate the equivalent resistance even for infinite scenarios.

Geometric Series

Recognizing the pattern in the resistance values helps us to see that they form a geometric series. A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant, known as the "common ratio."
In our case:

• The first term \(a\) is \(\frac{1}{10}\) (from \(R_1 = 10\Omega\)).
• The common ratio \(r\) is \(\frac{1}{10}\), as each subsequent term's denominator increases tenfold.

Let's consider how these resistances are used in the parallel equation:\( \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{10^2} + \frac{1}{10^3} + \ldots \).
This series is geometric due to the consistent factor \(\frac{1}{10}\) between terms. This geometric relationship sets us up to use another formula to handle the infinity of terms.

Infinite Series

In solving for an infinite series' sum, especially a geometric one, you will typically use the formula:\[ S = \frac{a}{1 - r} \]where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio.
For our particular series, we input:

• \(a = \frac{1}{10}\)
• \(r = \frac{1}{10}\)

Plug these values into the formula:\[ S = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9} \]This calculated sum \(S\) represents the total of the inverses of the infinitely many resistors' values.
The reciprocal of \(S\), which is \(9\), gives us the equivalent resistance \(R_{eq},\) confirming our conclusion that \(R_{eq} = 9 \Omega.\)
This demonstrates how the properties of an infinite series can simplify what looks like an overwhelmingly large problem, yielding a finite and manageable solution.