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Chapter 12: Problem 1800

A wire of resistor \(R\) is bent into a circular ring a circular ring of radius \(\mathrm{r}\) Equivalent resistance between two points \(\mathrm{X}\) and \(\mathrm{Y}\) on its circumference, when angle xoy is \(\alpha\), can be given by (A) $\left\\{(\mathrm{R} \alpha) /\left(4 \pi^{2}\right)\right\\}(2 \pi-\alpha)$ (B) \((\mathrm{R} / 2 \pi)(2 \pi-\alpha)\) (C) \(\mathrm{R}(2 \pi-\alpha)\) (D) \((4 \pi / \mathrm{R} \alpha)(2 \pi-\alpha)\)

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: The correct answer is (B) \(\frac{R}{2 \pi}(2 \pi - \alpha)\).

Step by step solution

01

Calculate the length of each segment

To find the length of each segment between points X and Y, first, we need to determine the circumference of the entire circle, which can be given by: C (Circumference) = 2πr Now, since angle XOY = α, we can use the fact that the ratio of the arc length to the total circumference equals the ratio of the angle (in radians) to 2π. Therefore, the length of segments XY and YX can be calculated as: Length of XY = (α/2π) * C = (α/2π) * 2πr = αr Length of YX = (2π - α)/2π * C = ((2π - α)/2π) * 2πr = (2π - α)r
02

Calculate the resistance of each segment

Now, using the formula R = ρ * (Length/Area), where ρ is the resistivity of the wire, we can find the resistance of each segment. Since we are given only the total resistance R of the wire and we know that resistivity and wire area will remain constant throughout the wire, we can say: Resistance of XY = (α/2π) * R Resistance of YX = ((2π - α)/2π) * R
03

Determine the equivalent resistance between points X and Y

Now, we can use the concept of resistances in parallel to find the equivalent resistance REQ between the two points X and Y. The formula is: 1/REQ = 1/RXY + 1/RYX Now replacing RXY and RYX with calculated resistance values from step 2, we get: 1/REQ = 1/((α/2π) * R) + 1/(((2π - α)/2π) * R) Simplifying: REQ = \(\frac{R \alpha(2 \pi - \alpha)}{2\pi [(\alpha)(2 \pi - \alpha)]}\) After cancellation: REQ = \(\frac{R(2 \pi - \alpha)}{2\pi}\) Comparing to the given options, we see that this expression for equivalent resistance matches option (B). So, the correct answer is (B) \(\frac{R}{2 \pi}(2 \pi - \alpha)\).

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Most popular questions from this chapter

Three identical resistors connected in series with a battery, together dissipate \(10 \mathrm{~W}\) of power. What will be the power dissipated, if the same resistors are connected in paralle1 across the same battery? (A) \(60 \mathrm{~W}\) (B) \(30 \mathrm{~W}\) (C) \(90 \mathrm{~W}\) (D) \(120 \mathrm{~W}\)

The resistance of the wire made of silver at \(27^{\circ} \mathrm{C}\) temperature is equal to \(2.1 \Omega\) while at \(100^{\circ} \mathrm{C}\) it is \(2.7 \Omega\) calculate the temperature coefficient of the resistivity of silver. Take the reference temperature equal to \(20^{\circ} \mathrm{C}\) (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (B) \(0.402 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (C) \(40.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) (D) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)

Two wires of equal diameters of resistivity's \(\rho_{1}\) and \(\rho_{2}\) are joined in series. The equivalent resistivity of the combination is.... (A) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (B) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$ (C) $\left\\{\left(\rho_{1} \ell_{2}+\rho_{2} \ell_{1}\right) /\left(\ell_{1}+\ell_{2}\right)\right\\}$ (D) $\left\\{\left(\rho_{1} \ell_{1}+\rho_{2} \ell_{2}\right) /\left(\ell_{1}-\ell_{2}\right)\right\\}$

In the circuit shown in fig the potential difference across \(3 \Omega\) is. (A) \(2 \mathrm{~V}\) (B) \(4 \mathrm{~V}\) (C) \(8 \mathrm{~V}\) (D) \(16 \mathrm{~V}\)

In a wheat stone's bridge, three reststance \(P, Q\) and \(R\) connected in three arm a and the fourth arm is formed by two resistances \(\mathrm{S}_{1}\) and \(\mathrm{S}_{2}\) connected in parallel The condition for bridge to be balanced will be. (A) $(\mathrm{P} / \mathrm{Q})=\left\\{\mathrm{R} /\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\}$ (B) $(\mathrm{P} / \mathrm{Q})=\left\\{(2 \mathrm{R}) /\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\}$ (C) $(\mathrm{P} / \mathrm{Q})=\left[\left\\{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\} /\left\\{\mathrm{S}_{1} \mathrm{~S}_{2}\right\\}\right]$ (D) $(\mathrm{P} / \mathrm{Q})=\left[\left\\{\mathrm{R}\left(\mathrm{S}_{1}+\mathrm{S}_{2}\right)\right\\} /\left\\{2 \mathrm{~S}_{1} \mathrm{~S}_{2}\right\\}\right]$
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