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Chapter 12: Problem 1745

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) \(2 \Omega\)

Short Answer

Expert verified
(B) \(4 \Omega\)

Step by step solution

01

Write down the temperature coefficients of resistance formula

We need to use the following formula that relates the resistance (R) at a given temperature (T) to the resistance at a reference temperature (R₀) and the temperature coefficient of resistance (α): \( R(T) = R_0 (1 + \alpha T) \)
02

Set up two equations from given values

Since we are given the resistance at temperatures 40°C and 100°C, we can set up two equations from the formula mentioned in Step 1. For T = 40°C and R(40) = 4.64Ω: \( 4.64 = R_0 (1 + \alpha \cdot 40) \) For T = 100°C and R(100) = 5.6Ω: \( 5.6 = R_0 (1 + \alpha \cdot 100) \)
03

Solve for α

Rearrange the first equation to get α in terms of R₀: \( \alpha = \frac{4.64}{R_0} - \frac{1}{40} \) Now substitute the expression for \(\alpha\) in the second equation: \( 5.6 = R_0 \left(1 + 100 \left(\frac{4.64}{R_0} - \frac{1}{40}\right)\right) \)
04

Solve for R₀

Now we need to solve for R₀. We can first simplify the equation obtained in step 3: \( 5.6 = R_0 \left(1 + 464 - \frac{100R_0}{40}\right) \) Solve for R₀: \( R_0 = -7.51 \) However, this is not possible as the resistance should be positive. We may have made a mistake in our calculations. Let's go back to step 3.
05

(Revised): Eliminate R₀ from equations and solve for α

Instead of expressing α in terms of R₀, we can eliminate R₀ from the two given equations. Divide the second equation by the first equation: \( \frac{5.6}{4.64} = \frac{R_0 (1 + \alpha * 100)}{R_0 (1 + \alpha * 40)} \) Cancel out R₀ and simplify: \( 1.2069 = \frac{1 + 100\alpha}{1+40\alpha} \) To solve for α, cross multiply and solve the linear equation: \( \alpha = 0.004225 \)
06

(Revised): Solve for R₀

Now, use the first equation to solve for R₀: \( 4.64 = R_0 (1 + 0.004225 * 40) \) \( R_0 = 4.0002 \) Now we have the resistance R₀ at 0°C. The closest option to our calculated value is: Answer: (B) \(4 \Omega\)

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Most popular questions from this chapter

What maximum power can be obtained from a battery of \(\mathrm{emf} \mathrm{E}\) and internal resistance \(\mathrm{r}\) connected with an external resistance \(\mathrm{R}\) ? (A) \(\left(E^{2} / 4 r\right)\) (B) \(\left(E^{2} / 3 r\right)\) (C) \(\left(E^{2} / 2 r\right)\) (D) \(\left(\mathrm{E}^{2} / \mathrm{r}\right)\)

In the circuit shown, the current sources are of negligible internal resistances. What is the potential difference between the points \(\mathrm{B}\) and \(\mathrm{A}\) ? (A) \(-4.0 \mathrm{~V}\) (B) \(4.0 \mathrm{~V}\) (C) \(-8.0 \mathrm{~V}\) (D) \(8.0 \mathrm{~V}\)

Figure, shows a network of eight resistors numbered 1 To 8 , each equal to $2 \Omega\(, connected to a \)3 \mathrm{~V}$ battery of negligible internal resistance The current \(\mathrm{I}\) in the circuit is .... (A) \(0.25 \mathrm{~A}\) (B) \(0.5 \mathrm{~A}\) (C) \(0.75 \mathrm{~A}\) (D) \(1.0 \mathrm{~A}\)

The resistivity of a potentiometer wire is $40 \times 10^{-8} \mathrm{ohm} \mathrm{m}\( and its area of cross-section is \)8 \times 10^{-6} \mathrm{~m}^{2}\(. If \)0.2$ amp current is flowing through the wire, the potential gradient will be. (A) \(10^{-2}\) Volt \(/ \mathrm{m}\) (B) \(10^{-1}\) Volt \(/ \mathrm{m}\) (C) \(3.2 \times 10^{-2}\) Volt \(/ \mathrm{m}\) (D) 1 Volt \(/ \mathrm{m}\)

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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