/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 A \(160-\mu \mathrm{F}\) capacit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 87

A \(160-\mu \mathrm{F}\) capacitor charged to \(450 \mathrm{V}\) is discharged through a \(31.2-\mathrm{k} \Omega\) resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is \(2.50 \mathrm{g}\) and its specific heat is \(1.67 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant?

Short Answer

Expert verified
The RC time constant of the circuit is approximately 4.992 seconds. The energy dissipated in the resistor is approximately 16.2 Joules, causing a temperature increase of about 3.9掳C. The new resistance after the temperature increase is approximately 31.15 k惟, an insignificant change of about 0.16%.

Step by step solution

01

Find the time constant (RC)

To find the time constant for this RC circuit, use the formula: \(RC\), where R is the resistor's value in ohms, and C is the capacitor's value in farads. Given values: \(C = 160\mu F = 160 \times 10^{-6} F\) and \(R = 31.2 k\Omega = 31.2 \times 10^3\Omega\). Calculate the time constant \(RC\): \(RC = R \times C = (31.2 \times 10^3\Omega) \times (160 \times 10^{-6} F)\)
02

Calculate the time constant

Perform the calculation: \(RC = (31.2 \times 10^3\Omega) \times (160 \times 10^{-6} F) \approx 4.992s\) The time constant is approximately 4.992 seconds.
03

Calculate the energy dissipated in the resistor

The energy dissipated in the resistor can be calculated using the formula: \(E = \frac{1}{2} CV^2\), where E is the energy, C is the capacitance in farads, and V is the applied voltage in volts. Given values: \(C = 160 \times 10^{-6} F\) and \(V = 450V\). Calculate the energy: \(E = \frac{1}{2} (160 \times 10^{-6}) (450)^2\)
04

Calculate the energy

Perform the calculation: \(E = \frac{1}{2} (160 \times 10^{-6}) (450)^2 \approx 16.2J\) The energy dissipated in the resistor is approximately 16.2 Joules.
05

Calculate the temperature increase

Using the formula \(Q = mc\Delta T\), where Q is the energy in Joules, m is the mass in kg, c is specific heat in J/(kg路掳C), and \(\Delta T\) is the temperature increase in 掳C, we can calculate the temperature increase. Given values: \(Q = 16.2J\), \(m = 2.50g = 0.0025kg\), and \(c = 1.67 \times 10^3 J/(kg\cdot掳C)\). Rearrange the formula to isolate \(\Delta T\): \(\Delta T = \frac{Q}{mc}\)
06

Calculate the temperature increase

Plug in the given values and perform the calculation: \(\Delta T = \frac{16.2}{(0.0025)(1.67 \times 10^3)} \approx 3.9^{\circ}C\) The temperature increase of the resistor is approximately 3.9掳C.
07

Calculate the new resistance

To find the new resistance, we need to consider that the resistivity of pure carbon at room temperature (20掳C) is \(5.0 \times 10^{-3} \Omega m\), and the temperature coefficient, \(\alpha\), is approximately \(-(5.0 \times 10^{-4}) K^{-1}\). The dependence of resistance on temperature is given by the formula \(R_{new} = R_0(1 + \alpha\Delta T)\), where \(R_{new}\) is the new resistance, \(R_0\) is the initial resistance, and \(\Delta T\) is the temperature increase. Given values: \(R_0 = 31.2 k\Omega\), \(\alpha = -(5.0 \times 10^{-4})K^{-1}\), and \(\Delta T = 3.9^{\circ}C\). Calculate the new resistance: \(R_{new} = R_0(1 + \alpha\Delta T) = (31.2 \times 10^3\Omega)(1 - (5.0 \times 10^{-4} K^{-1})(3.9))\)
08

Calculate the new resistance

Perform the calculation: \(R_{new} = (31.2 \times 10^3\Omega)(1 - (5.0 \times 10^{-4} K^{-1})(3.9)) \approx 31.15 k\Omega\) The new resistance is approximately 31.15 k惟.
09

Determine the significance of the change in resistance

Given that the initial resistance was 31.2 k惟, and the new resistance after the temperature increase is approximately 31.15 k惟, the change in resistance is relatively small. Since the difference is only 50惟, which is around a 0.16% change, it appears that this small change in resistance is not significant for most applications.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Time Constant
The time constant in an RC circuit is a crucial concept, representing the time it takes for the capacitor to charge or discharge to about 63.2% of its full charge. In formulas, it's denoted by \( \tau \) and calculated using the equation \( \tau = RC \). Here, \( R \) is the resistance in ohms, and \( C \) is the capacitance in farads.

For our circuit, given a \(160-\mu \mathrm{F}\) capacitor and a \(31.2-\mathrm{k}\Omega\) resistor, we calculate \(RC\) as follows:
  • Convert the units: \(C = 160 \times 10^{-6} F \) and \( R = 31.2 \times 10^3 \Omega\).
  • Calculate the time constant: \( RC = 31.2 \times 10^3 \Omega \times 160 \times 10^{-6} F \approx 4.992\) seconds.
A time constant of 4.992 seconds indicates how quickly or slowly the events occur in the circuit during charge or discharge phases.
Energy Dissipation
In an RC circuit, energy dissipation occurs when the electrical energy initially stored in the capacitor is transferred to the resistor, typically as heat. This process can be described using the energy formula: \( E = \frac{1}{2} CV^2 \).

To solve for the energy dissipated in our case:
  • Given: \( C = 160 \times 10^{-6} F \) and \( V = 450 V \).
  • Calculate energy: \( E = \frac{1}{2} \times 160 \times 10^{-6} \times 450^2 \approx 16.2 \) Joules.
This 16.2 Joules of energy is dissipated primarily as thermal energy, which impacts the temperature and physical properties of the resistor.
Temperature Increase
The dissipation of energy as heat can cause a temperature increase in the resistor. This increase is calculated using the formula \( Q = mc\Delta T \), where \( Q \) is the heat energy in joules, \( m \) is the mass in kilograms, \( c \) is the specific heat capacity in J/(kg路掳C), and \( \Delta T \) is the temperature change. The formula rearranges to \( \Delta T = \frac{Q}{mc} \).

For the given scenario:
  • Energy \( Q = 16.2 J \), mass \( m = 0.0025 kg \), specific heat \( c = 1.67 \times 10^3 J/(kg路掳C) \).
  • Calculate \( \Delta T: \Delta T = \frac{16.2}{(0.0025)\times(1.67 \times 10^3)} \approx 3.9^{\circ}C \).
This temperature increase of approximately 3.9掳C affects the resistor鈥檚 physical properties.
Resistor Properties
Resistors are fundamental components in circuits that impede the flow of electrical current. Their properties, such as resistance, are influenced by factors including temperature.

When a resistor conducts electricity, it converts some electrical energy into heat, influencing its resistance:
  • The nominal resistance is the resistance value under standard conditions.
  • As the temperature changes, so does the resistance, a phenomenon explained by the temperature coefficient \( \alpha \).
In our case, assuming the resistor is carbon, this change is vital for calculating the new resistance at an elevated temperature.
Carbon Resistance Change
Resistance in materials like carbon shifts with temperature variations. For carbon, this change can be estimated using the formula: \( R_{new} = R_0(1 + \alpha\Delta T) \), where \( R_0 \) is the initial resistance, \( \alpha \) is the temperature coefficient, and \( \Delta T \) is the temperature increase. Carbon has a negative temperature coefficient, meaning resistance slightly decreases with increasing temperature.

In our experiment:
  • Initial resistance \( R_0 = 31.2 k\Omega \).
  • Temperature coefficient \( \alpha = -(5.0 \times 10^{-4} K^{-1}) \) and \( \Delta T = 3.9^{\circ}C \).
  • Calculate new resistance: \( R_{new} = 31.2 \times 10^3(1 - (5.0 \times 10^{-4} \times 3.9)) \approx 31.15 k\Omega \).
This change is marginal, altering resistance by only around 0.16%, and is often negligible in practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 12.0-V emf automobile battery has a terminal voltage of \(16.0 \mathrm{V}\) when being charged by a current of 10.0 A. (a) What is the battery's internal resistance? (b) What power is dissipated inside the battery? (c) At what rate (in \(^{\circ} \mathrm{C} / \mathrm{min}\) ) will its temperature increase if its mass is \(20.0 \mathrm{kg}\) and it has a specific heat of \(0.300 \mathrm{kcal} / \mathrm{kg} \cdot^{\circ} \mathrm{C},\) assuming no heat escapes?

An ECG monitor must have an \(R C\) time constant less than \(1.00 \times 10^{2} \mu \mathrm{s}\) to be able to measure variations in voltage over small time intervals. (a) If the resistance of the circuit (due mostly to that of the patient's chest) is \(1.00 \mathrm{k} \Omega,\) what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)?

Your car's 30.0-W headlight and 2.40-kW starter are ordinarily connected in parallel in a 12.0 -V system. What power would one headlight and the starter consume if connected in series to a 12.0 -V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.)

(a) What is the internal resistance of a voltage source if its terminal potential drops by \(2.00 \mathrm{V}\) when the current supplied increases by 5.00 A? (b) Can the emf of the voltage source be found with the information supplied?

What voltage is involved in a \(1.44-\mathrm{kW}\) short circuit through a \(0.100-\Omega\) resistance?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Physics of Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Kinematics Physics

Read Explanation

Force

Read Explanation

Electromagnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.