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Chapter 12: Problem 1777

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) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)

Short Answer

Expert verified
The temperature coefficient of resistivity of silver is approximately \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\).

Step by step solution

01

List the given information

We are provided with the following values: - Resistance at \(27^{\circ}\mathrm{C}\): \(R_{27} = 2.1 \Omega\) - Resistance at \(100^{\circ}\mathrm{C}\): \(R_{100} = 2.7 \Omega\) - Reference temperature: \(t_0 = 20^{\circ}\mathrm{C}\)
02

Set up two equations using the resistance formula

We can set up two equations using the temperature coefficient of resistivity formula: - Equation 1: \(R_{27} = R_0[1 + \alpha (27 - 20)]\) - Equation 2: \(R_{100} = R_0[1 + \alpha (100 - 20)]\)
03

Solve for R_0 using Equation 1

Using Equation 1, we can solve for the reference resistance \(R_0\): \(2.1 = R_0[1 + \alpha (7)]\)
04

Substitute R_0 from Equation 1 into Equation 2

Now we substitute the expression for \(R_0\) from Step 3 into Equation 2: \(2.7 = \frac{2.1}{1 + 7\alpha}[1 + 80\alpha]\)
05

Solve for alpha

Now, we need to solve the equation for \(\alpha\): \(\frac{2.7}{1 + 80\alpha} = \frac{2.1}{1 + 7\alpha}\) Cross-multiplying, we get: \(2.7 (1 + 7\alpha) = 2.1 (1 + 80\alpha)\) Expanding both sides: \(2.7 + 18.9\alpha = 2.1 + 168\alpha\) Now we can isolate \(\alpha\): \(150.9\alpha = 0.6\) \(\alpha = \frac{0.6}{150.9}\)
06

Calculate the value of alpha and choose the correct option

Now, we can calculate the value of \(\alpha\): \(\alpha \approx 0.00402^{\circ}\mathrm{C}^{-1}\) Now, we can see that the correct option is: (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\)

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Most popular questions from this chapter

A Potentiometer wire, \(10 \mathrm{~m}\) long, has a resistance of \(40 \Omega\). It is connected in series with a resistance box and a \(2 \mathrm{v}\) storage cell If the potential gradient along the wire is $0.1 \mathrm{~m} \mathrm{v} / \mathrm{cm}$, the resistance unplugged in the box is. (A) \(260 \Omega\) (B) \(760 \Omega\) (C) \(960 \Omega\) (D) \(1060 \Omega\)

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 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 \(7 \Omega\)

How would you arrange 48 cells each of e.m.f \(2 \mathrm{~V}\) and internal resistance \(1.5 \Omega\) so as to pass maximum current through the external resistance of \(2 \Omega\) ? (A) 2 cells in 24 groups (B) 4 cells in 12 groups (C) 8 cells in 6 groups (D) 3 cells in 16 groups

A bulb of \(300 \mathrm{~W}\) and \(220 \mathrm{~V}\) is connected with a source of \(110 \mathrm{~V}\). What is the \(\%\) decrease in power? (A) \(100 . \%\) (B) \(75 \%\) (C) \(70 \%\) (D) \(25 \%\)
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