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Chapter 18: Problem 56

The resistance of a bulb filaments is \(100 \Omega\) at a temperature of \(100{ }^{\circ} \mathrm{C}\). If its temperature coefficient of resistance be \(0.005 l{ }^{\circ} \mathrm{C}^{-1}\), its resistance will become \(200 \Omega\) at a temperature of (a) \(200^{\circ} \mathrm{C}\) (b) \(300^{\circ} \mathrm{C}\) (c) \(400^{\circ} \mathrm{C}\) (d) \(500^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The resistance becomes \(200 \Omega\) at \(300^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the formula

The formula to find the resistance at a new temperature is \( R_t = R_0 (1 + \alpha (t - t_0)) \), where \( R_t \) is the resistance at temperature \( t \), \( R_0 \) is the original resistance, \( \alpha \) is the temperature coefficient of resistance, \( t \) is the new temperature, and \( t_0 \) is the initial temperature.
02

Identify known values

We know the initial resistance \( R_0 = 100 \Omega \), the temperature coefficient \( \alpha = 0.005 \; ^{\circ}C^{-1} \), the initial temperature \( t_0 = 100 \; ^{\circ}C \), and the final resistance \( R_t = 200 \Omega \). We need to find the new temperature \( t \).
03

Rearrange the formula to find the temperature

Start by rearranging the formula: \( R_t = R_0 (1 + \alpha (t - t_0)) \). Solving for \( t \) gives \( t = \frac{R_t}{R_0 \alpha} + t_0 - \frac{1}{\alpha} \).
04

Substitute the known values into the formula

Using the given values, substitute them into the rearranged formula: \( t = \frac{200}{100 \times 0.005} + 100 - \frac{1}{0.005} \).
05

Simplify the expression

Calculate each part of the expression: \( \frac{200}{100 \times 0.005} = 400 \), and \( \frac{1}{0.005} = 200 \). Thus the expression becomes \( t = 400 + 100 - 200 \).
06

Calculate the final temperature

Finish the calculation: \( t = 400 + 100 - 200 = 300 \; ^{\circ}C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance Calculation
To grasp how resistor values change, it's essential to start with the basics of resistance calculations. Resistance measures how much a material opposes the flow of electric current. With a device like a bulb filament, resistance is significant since it regulates electrical flow, directly impacting performance. In the exercise, we see initial resistance given as 100 惟, and we're asked to determine what changes to another resistance value, 200 惟, specify in terms of temperature. This involves computations showing how resistance shifts relative to a change in temperature. Understanding this will help you know how devices perform under different thermal conditions. It鈥檚 akin to calibrating your expectations or configurations for varying environmental temperatures.
Temperature Effects on Resistance
Temperature can significantly influence the resistance of a material. As temperature rises, the atomic structure of a material vibrates more intensely. This increased vibration clashes with moving electrons, boosting the resistance a conductor offers. For metals, like the tungsten in a light bulb filament, this relationship between temperature and resistance is usually linear within practical ranges. Meaning, resistance grows consistently as temperature rises. The exercise illustrates this with a step-by-step method to discover the temperature for a specified resistance, emphasizing how any change in temperature directly affects electrical resistance values. Key points to remember:
  • Increased temperature generally means increased resistance in metallic conductors.
  • Understanding this can help design and use applications that consider how temperature affects electrical characteristics.
Resistance Formula
The calculation of resistance at different temperature levels requires using a specific mathematical formula. This formula incorporates a term called the 'temperature coefficient of resistance,' denoted by \( \alpha \), which tells us how much the resistance changes per degree of temperature change.The formula is expressed as:\[ R_t = R_0 (1 + \alpha (t - t_0)) \]where:
  • \( R_t \) is the resistance at new temperature \( t \)
  • \( R_0 \) is the original resistance
  • \( \alpha \) is the temperature coefficient of resistance
  • \( t \) is the new temperature
  • \( t_0 \) is the initial temperature
This formula is pivotal in many engineering and physics problems concerning how electrical devices behave in different thermal conditions. It allows one to predict changes and ensure the device operates within safe and efficient limits. By manipulating this formula, we effectively forecast how resistance will vary as the temperature undergoes alterations, crucial for maintaining optimal performance and safety.

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Most popular questions from this chapter

\(n\) cells of emfs \(E_{1}, E_{2}, E_{3}, \ldots, E_{n}\) and internal resistance \(r_{1}, r_{2}, r_{3}, \ldots, r_{n}\) are connected in series to form a closed circuit with zero external resistance. For each cell the ratio of emf to internal resistance is \(K\), where \(K\) is a constant; then current in the circuit is (a) \((1 / K)\) (b) \(K\) (c) \(K^{2}\) (d) \(\left(1 / K^{2}\right)\)

Two different conductors have same resistance at \(0^{\circ} \mathrm{C}\). It is found that the resistance of the first conductor at \(t_{1}^{\circ} \mathrm{C} .\) It is equal to the resistance of the second conductor at \(t_{2}{ }^{\circ} \mathrm{C}\). The ratio of the temperature coefficients of resistance of the conductors, \(\alpha_{1} / \alpha_{2}\) is (a) \(\frac{t_{1}}{t_{2}}\) (b) \(\frac{t_{2}-t_{1}}{t_{2}}\) (c) \(\frac{t_{2}-t_{1}}{t_{1}}\) (d) \(\frac{t_{2}}{t_{1}}\)

The wire of the electric fire element glows red hot but the copper wire in the leads from the plug remains cool, although the same current is passing through each. This is because the element wire (a) is longer than the copper wire (b) has much higher resistance than the copper wire (c) is coiled but the copper wire is not (d) has a much lower resistance than the copper wire

The electrical conductivity of the metal decreases with temperature because (a) the energy of the electrons increases with temperature (b) a metal expands on heating (c) the atoms of the metal vibrate more at higher temperature (d) metals have low specific heat

In a meter bridge with standard resistance of \(5 \Omega\) in the left gap, the ratio of balancing lengths of meter bridge wire is \(2: 3 .\) The unknown resistance is (a) \(15 \Omega\) (b) \(10 \Omega\) (c) \(3.3 \Omega\) (d) \(7.5 \Omega\)
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