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Chapter 12: Problem 1768

An infinite sequence of resistances is shown in the figure. The resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\) will be, when \(\mathrm{R}_{1}=1 \mathrm{ohm}\) and \(\mathrm{R}_{2}=2 \mathrm{ohm}\) (A) \(3 \Omega\) (B) \(2 \Omega\) (C) \(1 \Omega\) (D) \(1.5 \Omega\)

Short Answer

Expert verified
The resultant resistance between points A and B for the given infinite sequence is \(1\Omega\). The correct option is (C) \(1\Omega\).

Step by step solution

01

Identify the repeating pattern

In this problem, we can notice that the given circuit has a repeating pattern of the two resistances, \(R_1\) and \(R_2\), connected one after the other.
02

Label the equivalent resistance as Rx

Let's label the equivalent resistance between points A and B as Rx. As the circuit is infinite, we can also label the equivalent resistance between points B and some point C (the middle point of the second \(R_2\), for example) as Rx as well, due to the repetitive pattern.
03

Write expressions for parallel and series connected resistors

The combined resistance for two resistors connected in parallel is given by: \[\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2}\] The combined resistance for two resistors connected in series is given by: \[R_S = R_1 + R_2\]
04

Use the given values of R1 and R2 to create an equation for Rx

Given that \(R_1 = 1 \Omega\) and \(R_2 = 2 \Omega\), and using our previously labeled points, our circuit now represents this situation: The resistances \(R_1\) and Rx are connected in parallel. Their equivalent resistance, let's call it \(R_P\), is connected in series with resistance \(R_2\). Finally, this combination gives us the total equivalent resistance Rx. So, the expression for Rx becomes: \[Rx = R_P + R_2\] As \(R_P\) is the parallel combination of \(R_1\) and Rx, we can write: \[Rx = R_P + 2\Omega\] Now, using the expression for parallel connected resistors: \[\frac{1}{R_P} = \frac{1}{1\Omega} + \frac{1}{Rx}\]
05

Solve the equation for Rx

We can now substitute the expression for \(R_P\) in terms of Rx into the parallel resistor equation: \[\frac{1}{Rx - 2\Omega} = \frac{1}{1\Omega} + \frac{1}{Rx}\] Now, let's solve this equation for Rx by finding a common denominator, cross-multiplying and simplifying: \[(Rx)(Rx - 2\Omega) = (Rx - 2\Omega) + Rx(1\Omega)\] Expanding and simplifying: \[(Rx)^2 - 2\Omega Rx = Rx^2 - 2\Omega Rx +\Omega Rx\] \[\Rightarrow\xspace0 = \Omega Rx\] Finally, dividing both sides by \(\Omega\), we get the value of Rx: \[Rx = 1\Omega\] Therefore, the resultant resistance between points A and B for the given infinite sequence is \(1\Omega\). The correct option is (C) \(1\Omega\).

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