Question

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

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

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.

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.

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: \[R_S = R_1 + R_2\]

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: \[\frac{1}{R_P} = \frac{1}{1\Omega} + \frac{1}{Rx}\]

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) \(1\Omega\).

Key concepts

Equivalent Resistance

Equivalent resistance is a way of simplifying a complex network of resistors into a single resistor that has the same effect. It allows us to analyze and solve circuit problems more efficiently. By replacing a set of resistors with their equivalent resistance, we make calculations manageable without changing the overall electrical behavior.

To find the equivalent resistance, we often need to identify whether the resistors are in series or parallel. The challenge is to reduce them step-by-step until you have a single equivalent resistance value. Once simplified, calculating the total current and voltage around the circuit becomes straightforward. Throughout this process, Kirchhoff’s laws can be used to reinforce the understanding of circuit behavior.

Parallel and Series Resistor Calculation

In circuit analysis, understanding how to calculate resistors in series and parallel is essential. For resistors in series, the equivalent resistance is the sum of all individual resistances. It is calculated using the formula:

• \[ R_S = R_1 + R_2 + R_3 + ext{...} \]

Meanwhile, resistors in parallel have a more complex formula. It involves the reciprocal of the sum of the reciprocals of the individual resistances:

• \[ \frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ext{...} \]

In solving problems, it's crucial to substitute known values into these formulas. In the infinite circuit problem, the resistors formed both series and parallel connections. By understanding and applying these rules, one can simplify even intricate networks.

Infinite Series Circuit

An infinite series circuit contains an indefinite repetition of circuit elements, making the direct application of normal circuit formulas complex. In such a scenario, recognizing patterns and symmetries is key. For instance, an infinite set of resistors could be treated as a repetitive unit, representing a constant equivalent resistance.

• Each part of the repeating sequence has consistent equivalent resistance because it is essentially a replica of the larger structure.
• By identifying the equivalent resistance of one repeating unit, you can infer the resistance for larger sections or the entire circuit.

The application of this understanding allows you to set up equations that clearly denote the consistent resistance values. In our exercise, analyzing such a pattern helped in deriving the equivalent resistance by effectively collapsing the infinite circuit into a manageable equivalent resistance calculation.