Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

In an oscillating LC circuit, $L = 3.00 mH$ and $C = 2.70 μF$ . At $t = 0$ the charge on the capacitor is zero and the current is 2.00 A. (a) What is the maximum charge that will appear on the capacitor? (b) At what earliest time $t > 0$ is the rate at which energy is stored in the capacitor greatest, and (c) What is that greatest rate?

The maximum charge that will appear on the capacitor is $1.8 \times 10^{- 4} C$ .
The rate at which energy stored is the greatest in the capacitor is $7.07 \times 10^{- 5} s$ .
The greatest rate is 66.7 W.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The inductance $L = 3.00 mH or 0.003 H$ and capacitance $C = 2.70 μF or 2.70 \times 10^{- 6} F$ are present in the oscillating LC circuit.
At $t = 0$ , the charge on the capacitor, $q = 0$ and $i = 2.00 A$ .

Step 2: Understanding the concept of charge and energy of LC circuit

By using the concept of charge, which is the function of time, we first find the current, which is also a function of time. Using the initial condition, we can find the maximum charge that will appear on the capacitor. Further, by using the equation of energy stored in the capacitor, we can find the time rate at which energy stored in the capacitor is the greatest.

Formulae:

The current equation related to the charge rate, $i = \frac{d q}{d t}$ (i)

The angular frequency of an LC oscillation, $ω = \frac{1}{\sqrt{L C}}$ (ii)

The electrical energy stored by the capacitor, $U_{E} = \frac{q^{2}}{2 C}$ (iii)

The charge is a function of time, $q = Q \sin ω t$ (iv)

Where Q is the maximum charge on the capacitor and $ω$ is the angular frequency of oscillation.

The angular frequency of an oscillation, $ω = \frac{2 π}{T}$ (v)

Step 3: a) Calculation of the maximum charge on the capacitor

Substituting equation (iv) in equation (i), we get the current equation as follows:

$\begin{array}{rcl} i & = & \frac{d (Q \sin ω t)}{d t} \\ = & ω Q \cos ω t \end{array}$

At time t = 0 then,

$\cos 0 = 1$

And

$i = ω Q$

So the charge Q is given by substituting the value of equation (ii) in the above equation as follows:

$\begin{array}{rcl} Q & = & i \sqrt{L C} \\ = & 2.00 \times \sqrt{3.00 \times 10^{- 3} \times 2.70 \times 10^{- 6}} \\ = & 2.00 \times 9 \times 10^{- 5} \\ = & 1.8 \times 10^{- 4} C \end{array}$

Hence, the value of the charge is $1.8 \times 10^{- 4} C$ .

Step 4: b) Calculation of the time at which rate of the energy stored is greatest

By substituting the value of charge from equation (iv) in equation (iii), we get the energy equation as follows:

$U_{E} = \frac{Q^{2} \sin^{2} ω t}{2 C}$

And its rate of change is now given as:

$\frac{d U_{E}}{d t} = \frac{d}{d t} (\frac{Q^{2} \sin^{2} ω t}{2 C})$

But by using the trigonometric identity, we can write,

$\sin^{2} a = \frac{1 - \cos 2 a}{2}$

And

$\cos a \sin a = \frac{1}{2} \sin 2 a$

Thus, the energy rate equation becomes:

$\begin{array}{rcl} \frac{d U_{E}}{d t} & = & \frac{Q^{2} ω \sin ω t \cos ω t}{C} \\ = & \frac{Q^{2} ω \sin 2 ω t}{2 C} . . . . . . . . . . . . . . . . . . . . . . . . . . . (a) \end{array}$

The greatest rate of change occurs when $\sin 2 ω t = 1,$ i.e.,

$\begin{array}{rcl} 2 ω t & = & \frac{π}{2} \\ t & = & \frac{π}{4 ω} \\ = & \frac{π}{4} \sqrt{L C} (∵ from equation (ii)) \\ = & \frac{π}{4} \sqrt{(3 \times 10^{- 3}) (2.70 \times 10^{- 6})} \\ = & \frac{π}{4} \times 9 \times 10^{- 5} \\ = & 7.07 \times 10^{- 5} s \end{array}$

Hence, the value of time is $7.07 \times 10^{- 5} s$ .

Step 5: c) Calculation of that greatest energy rate

From equation (a), we can get the greatest rate of energy stored for $\sin 2 ω t = 1$ as follows:

$\begin{array}{rcl} \frac{d U_{E}}{d t} & = & \frac{Q^{2} ω}{2 C} \\ = & \frac{2 π Q^{2}}{2 T C} (∵ ω = \frac{2 π}{T} f r o m e q u a t i o n (v)) \\ = & \frac{π Q^{2}}{T C} . . . . . . . . . . . . . . . . . . . . (b) \end{array}$

Now, using equation (ii) in equation (v), the period of oscillations can be given as follows:

$\begin{array}{rcl} T & = & 2 π \sqrt{L C} \\ = & 2 π \sqrt{3.00 \times 10^{- 3} \times 2.70 \times 10^{- 6}} \\ = & 5.655 \times 10^{- 4} s \end{array}$

Now, substituting the above value in equation (b) with other given data, we can get the greatest rate value as follows:

$\begin{array}{rcl} \frac{d U_{E}}{d t} & = & \frac{π {(1.80 \times 10^{- 4})}^{2}}{5.655 \times 10^{- 4} \times 2.70 \times 10^{- 6}} \\ = & 66.7 W \end{array}$

From this result, we can conclude that the energy in the capacitor is indeed increasing at $t = \frac{T}{8}$ .

Hence, its greatest rate is 66.7 W.

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of charge and energy of LC circuit

Step 3: a) Calculation of the maximum charge on the capacitor

Step 4: b) Calculation of the time at which rate of the energy stored is greatest

Step 5: c) Calculation of that greatest energy rate

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Fundamentals Of Physics

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Step 3: a) Calculation of the maximum charge on the capacitor

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Step 5: c) Calculation of that greatest energy rate

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