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Chapter 12: Problem 1746

A circuit with an infinite no of resistance is shown in fig. the resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\), when $\mathrm{R}_{1}=1 \Omega\( and \)\mathrm{R}_{2}=2 \Omega$ will be (A) \(4 \Omega\) (B) \(1 \Omega\) (C) \(2 \Omega\) (D) \(3 \Omega\)

Short Answer

Expert verified
The equivalent resistance between points A and B in the infinite circuit with resistors R₁ = 1 Ω and R₂ = 2 Ω is \(R_{eq} = 1 \Omega\), which corresponds to answer (B).

Step by step solution

01

Observe the repeating pattern in the circuit

Examine the infinite circuit and notice that there is a repeating pattern starting from point A. Part of the circuit consists of one R₂ resistor connected in parallel with another branch that has one R₁ resistor in series with one R₂ resistor connected in series with another recursion of the same pattern. In other words, the circuit has a "self-similar" nature that we can make use of in our calculations.
02

Replace infinite part with an equivalent resistor

As previously mentioned, there is a self-similar pattern within the circuit. We can identify this pattern and replace it with an equivalent resistor, say R_eq. The branch that contains R₂ parallel with the recursive infinite circuit will reduce to another equivalent resistor, which will also be equivalent to R_eq since the pattern repeats from this point as well.
03

Apply the Parallel and Series Formulas for Equivalent Resistance

To determine the equivalent resistance, R_eq, we'll use the parallel rule for resistors in parallel and the series rule for resistors in series. The formula for resistors in parallel is: \[\frac{1}{R_{eq}}= \frac{1}{R_{1}} + \frac{1}{R_{2}}\] The formula for resistors in series is: \[R_{eq} = R_{1} + R_{2}\]
04

Set up the recursive equation

Using our observation from Step 2, we can replace the part in the recursive infinite circuit with R_eq. We have the following setup: R_eq = R₁ + R₂ in parallel with another R_eq. Applying the parallel formula, this equation will simplify as: \(R_eq = R_1 + \frac{R_2 \times R_{eq}}{R_2 + R_{eq}}\)
05

Solve for R_eq

Now we can solve the recursive equation for R_eq. Substitute R₁ = 1 Ω and R₂ = 2 Ω into the equation and solve for R_eq: \(R_{eq} = 1 + \frac{2 \times R_{eq}}{2 + R_{eq}}\) Rearrange this equation to isolate R_eq and simplify: \(R_{eq}(2 + R_{eq}) = 1(R_{eq}+2)\) \(2R_{eq} + R_{eq}^2 = R_{eq} + 2\) \(R_{eq}^2 + R_{eq} - 2 = 0\) This is a quadratic equation. Using the quadratic formula or factoring, we find two solutions for R_eq: \(R_{eq}=1, -2\) However, only the positive value, 1 Ω, makes physical sense for resistance. So the equivalent resistance between A and B is 1 Ω, which corresponds to answer (B).

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Most popular questions from this chapter

The resistance of a copper coil is \(4.64 \Omega\) at \(40^{\circ} \mathrm{C}\) and \(5.6 \Omega\) at \(100^{\circ} \mathrm{C}\) Its resistance at $0^{\circ} \mathrm{C}$ will be (A) \(5 \Omega\) (B) \(4 \Omega\) (C) \(3 \Omega\) (D) \(2 \Omega\)

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

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Two resistors when connected in parallel have an equivalent of \(2 \Omega\) and when in series of \(9 \Omega\) The values of the two resistors are. (A) \(2 \Omega\) and \(9 \Omega\) (B) \(3 \Omega\) and \(6 \Omega\) (C) \(4 \Omega\) and \(5 \Omega\) (D) \(2 \Omega\) and \(7 \Omega\)

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) \(1.5 \Omega\)
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