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Chapter 12: Problem 1776

The tungsten filament of bulb has resistance equal to \(18 \Omega\) at \(27^{\circ} \mathrm{C}\) temperature \(0.25 \mathrm{~A}\) of current flows, when \(45 \mathrm{~V}\) is connected to it If $\alpha=4.5 \times 10^{-3} \mathrm{~K}^{-1}$ for a tungsten then find the temperature of the filament. (A) \(2160 \mathrm{~K}\) (B) \(1800 \mathrm{~K}\) (C) \(2070 \mathrm{~K}\) (D) \(2300 \mathrm{~K}\)

Short Answer

Expert verified
The temperature of the tungsten filament is \(2300 \mathrm{K}\), which corresponds to answer choice \((D)\).

Step by step solution

01

Write the given information

We are given the following information: - The resistance of the tungsten filament at 27℃ (initial temperature) is 18 ohms - The current flowing through the filament is 0.25 A - The voltage connected to it is 45 V - The temperature coefficient of resistance for tungsten is \(4.5 \times 10^{-3} \mathrm{K}^{-1}\)
02

Calculate the power dissipated in the filament

Using Ohm's Law, the power dissipation in the filament can be calculated as: \[P = IV\] Where \(P\) is the power, \(I\) is the current, and \(V\) is the voltage. Substitute the given values: \[P = (0.25 \mathrm{~A})(45 \mathrm{~V})\] \[P = 11.25 \mathrm{~W}\]
03

Determine temperature change

Now, we need to find the temperature change from the initial temperature 27℃. Divide the formula for resistance by \(R_0\): \[\frac{R_t}{R_0} = 1 + \alpha \Delta T\] Since we are looking for the final temperature, we can use Ohm's Law to find \(R_t\) by substituting \(V = 45 \mathrm{V}\) and \(I = 0.25 \mathrm{A}\): \[R_t = \frac{V}{I} = \frac{45 \mathrm{~V}}{0.25 \mathrm{~A}} = 180 \Omega\] Now substitute the given values: \[\frac{180 \Omega}{18 \Omega} = 1 + (4.5 \times 10^{-3} \mathrm{K}^{-1}) \Delta T\]
04

Solve for the temperature change

Now, solve for \(\Delta T\): \[10 = 1 + (4.5 \times 10^{-3} \mathrm{K}^{-1}) \Delta T\] \[9 = (4.5 \times 10^{-3} \mathrm{K}^{-1}) \Delta T\] \[\Delta T = \frac{9}{(4.5 \times 10^{-3} \mathrm{K}^{-1})} = 2000 \mathrm{K}\]
05

Calculate the final temperature of the filament

Now we can calculate the final temperature by adding the initial temperature to the temperature change: \[T = T_0 + \Delta T\] \[T = 27 ℃ + 2000 \mathrm{K}\] Convert the initial temperature to Kelvin by adding 273 to 27℃: \[T = (273 + 27) \mathrm{K} + 2000 \mathrm{K} = 2300 \mathrm{K}\] Therefore, the temperature of the tungsten filament is 2300 K, which corresponds to answer choice (D).

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Most popular questions from this chapter

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The voltmeter reading will be. (A) \(1.96 \mathrm{~V}\) (B) \(1.8 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(0.96 \mathrm{~V}\)

The length of a potentiometer wire is \(600 \mathrm{~cm}\) and it carries a current of \(40 \mathrm{~m} \mathrm{~A}\) for cell of emf \(2 \mathrm{~V}\) and internal resistance \(10 \Omega\), the null point is found to be at $500 \mathrm{~cm}$ on connecting a voltmeter across the cell, the balancing length is decreased by \(10 \mathrm{~cm}\) The resistance of the voltmeter is (A) \(500 \Omega\) (B) \(290 \Omega\) (C) \(49 \Omega\) (D) \(20 \Omega\)

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