/*! 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 63 The charge flowing through a res... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 63

The charge flowing through a resistance \(R\) varies with time \(t\) as \(Q=a t-b t^{2} .\) The total heat produced in \(R\) from \(t=0\) to the time when value of \(Q\) becomes again zero is (A) \(\frac{a^{3} R}{6 b}\) (B) \(\frac{a^{3} R}{3 b}\) (C) \(\frac{a^{3} R}{2 b}\) (D) \(\frac{a^{3} R}{b}\)

Short Answer

Expert verified
The total heat produced in the resistance from \(t=0\) to the time when the value of \(Q\) becomes again zero is \(\frac{a^{3} R}{6 b}\).

Step by step solution

01

Calculate the current

First, we need to find the current flowing through the resistance. Knowing that \(i(t)=\frac{dQ}{dt}\), we differentiate the given equation of \(Q(t)\) with respect to time: \(i(t) = \frac{dQ}{dt} = \frac{d(at - bt^2)}{dt} = a - 2bt\)
02

Calculate the power

Now, we will find the instantaneous power being dissipated through the resistance using the formula \(P = i^2R\): \(P(t) = i^2(t)R = (a - 2bt)^2R\)
03

Determine the time when the value of Q is zero again

To find when the charge becomes zero again, we have to solve the equation \(Q(t) = 0\) for \(t\): \(at - bt^2 = 0\) Which gives, \(t(at - bt) = 0\) The two solutions are \(t=0\) and \(t = \frac{a}{b}\). Since we already know that \(t=0\) is the starting time, the second solution, \(t=\frac{a}{b}\), is the time when the charge becomes zero again.
04

Calculate the total heat produced

To find the total heat produced in the resistance, we integrate the instantaneous power over the time interval \(t=0\) to \(t=\frac{a}{b}\): \[ H = \int_0^\frac{a}{b} P(t)dt = R \int_0^\frac{a}{b}(a - 2bt)^2dt \]
05

Evaluate the integral

Evaluate the integral to get the total heat produced: \[ H = R \int_0^\frac{a}{b}(a^2 - 4abt + 4b^2t^2)dt = R \left[ a^2t - \frac{4}{3}abt^2 + \frac{4}{3}b^2t^3 \right]_0^\frac{a}{b} \]
06

Plug in the values

Substitute the values (\(t = \frac{a}{b}\)) into the integral and simplify: \[ H = R \left( a^2 \cdot \frac{a}{b} - \frac{4}{3}ab \cdot \left(\frac{a}{b}\right)^2 + \frac{4}{3}b^2 \cdot \left(\frac{a}{b}\right)^3 \right) \] After simplification, we get \[ H = \frac{a^3 R}{6b} \] So the correct answer is (A).

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!

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

An ammeter is obtained by shunting a \(30 \Omega\) galvanometer with a \(30 \Omega\) resistance. What additional shunt should be connected across it to double the range? (A) \(15 \Omega\) (B) \(10 \Omega\) (C) \(5 \Omega\) (D) None of these

The average bulk resistivity of the human body (apart from the surface resistance of the skin) is about \(5 \Omega m\). The conducting path between the hands can be represented approximately as a cylinder \(1.6 \mathrm{~m}\) long and \(0.1 \mathrm{~m}\) diameter. The skin resistance may be made negligible by soaking the hands in salt water. A lethal shock current needed is \(100 \mathrm{~mA}\). Note that a small amount of potential difference could be fatal if the skin is damp. What potential difference is needed between the hands for a lethal shock current? (A) \(100 \mathrm{~V}\) (B) \(10 \mathrm{~V}\) (C) \(120 \mathrm{~V}\) (D) \(150 \mathrm{~V}\)

A potentiometer is a device used for measuring EMF and internal resistance of a cell. It consists of two circuits, one is main circuit in which there is a cell of given emf \(\varepsilon^{\prime}\) and given resistance \(R\) which is connected across a wire of length \(100 \mathrm{~cm}\) and having resistance \(r\) and another circuit having unknown EMF \(\varepsilon\) and galvanometer. For a given potentiometer, if \(\varepsilon^{\prime}=30 \mathrm{~V}, r=1 \Omega\), and resistance \(R\) varies with time \(t\) given by \(R=2 t\). The jockey can move on wire with constant velocity \(10 \mathrm{~cm} / \mathrm{s}\) and switch \(S\) is closed at \(t=0\)If jockey starts moving from \(A\) at \(t=0\) and balancing point found at \(t=1 \mathrm{~s}\) then the value of \(\varepsilon\) is (A) \(1 \mathrm{~V}\) (B) \(2 \mathrm{~V}\) (C) \(3 \mathrm{C}\) (D) \(4 \mathrm{~V}\)

When \(5 \mathrm{~V}\) potential difference is applied across a wire of length \(0.1 \mathrm{~m}\), the drift speed of electrons is \(2.5 \times\) \(10^{-4} \mathrm{~ms}^{-1}\). If the electron density in the wire is \(8 \times\) \(10^{28} \mathrm{~m}^{-3}\), the resistivity of the material is close to (A) \(1.6 \times 10^{-7} \Omega \mathrm{m}\) (B) \(1.6 \times 10^{-6} \Omega \mathrm{m}\) (C) \(1.6 \times 10^{-5} \Omega \mathrm{m}\) (D) \(1.6 \times 10^{-8} \Omega \mathrm{m}\)

A galvanometer having a coil resistance of \(100 \Omega\) gives a full scale deflection, when a current of \(1 \mathrm{~mA}\) is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of \(10 \mathrm{~A}\), is (A) \(2 \Omega\) (B) \(0.1 \Omega\) (C) \(3 \Omega\) (D) \(0.01 \Omega\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Electrostatics

Read Explanation

Thermodynamics

Read Explanation

Dynamics

Read Explanation

Fluids

Read Explanation

Conservation of Energy and Momentum

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.