Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

An RL circuit consists of a $40.0 Ω$ resistor and a 3.00 mH inductor. (a) Find its impedance Z at 60.0 Hz and 10.0 kHz . (b) Compare these values of Z with those found in Example 23.12 in which there was also a capacitor.

a) As a result, low and high-frequency impedance are respectively $188.5 Ω$ and $193 Ω$

b) At high frequencies, the capacitor has a smaller influence, while having a larger influence at lower frequencies.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Definition of circuit

A circuit is a closed path via which electricity can flow from one location to another. It could include transistors, resistors, and capacitors, among other electrical components.

Step 2: Given data

Resistance in the RL circuit is $R = 40.0 Ω$

The inductor in the RL circuit is $L = 3.00 m H (\frac{10^{- 3} H}{1 m H}) = 3.00 \times 10^{- 3} H$

Step 2: Finding Impedance(a)

The formula for the determination of impedance of inductance can be expressed as,

$X_{L} = 2 π f L$ ………..(1)

Here $X_{L}$ is inductive reactance, f is the frequency, and L is the inductance.

For each frequency, impedance can be expressed as,

$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}}$

………………(2)

At low frequency, f = 60 Hz, the value of $X_{C} = 0 Ω$ .

Substituting the given values in equation (1), we get

$X_{L} = 2 \times 3.14 \times (60 H z) (3.00 \times 10^{- 3} H) = 1.13 Ω$

………………(3)

Then, using the given data and value from equation (3), the impedance can be calculated using equation (2), such that,

$Z = \sqrt{({(40 Ω)}^{2} + {(1.13 Ω - 0 Ω)}^{2})} = 40.02 Ω$

At high-frequency $f = 10.0 k H z (\frac{10^{3} H z}{1 k H z}) = 1.00 \times 10^{4} H z$ , substituting the given values in the above equation (1), we get

$X_{L}^{'} = 2 \times 3.14 \times (1 \times 10^{4} H z) (3 \times 10^{- 3} H) = 188.5 Ω$

Then, using the given data and value from equation (3), the impedance can be calculated using equation (2), such that,

$Z^{'} = \sqrt{{({(40 Ω)}^{2} + (188.5 Ω) - 0 Ω)}^{2}} = 192.7 Ω$

Therefore, the impedance at low and high frequencies are $188.5 Ω$ and $193 Ω$ .

Step 3: Compare these values of Z(b)

Compare these values of Z with those found in the example 23.12 , in which there was also a capacitor.

As we can see, at f =60Hz , with a capacitor the value of impedance is, $Z = 531 Ω$ , that is about 13 times as high as without the capacitor.

The capacitor makes a large difference at low frequencies.

At f = 10 kHz , with a capacitor, the value of impedance is, $Z = 190 Ω$ , that is similar to the value obtained without the capacitor.

Thus, the capacitor has a smaller effect at high frequencies.

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College Physics (Urone)

Answers without the blur.

Short Answer

An RL circuit consists of a $40.0 Ω$ resistor and a 3.00 mH inductor. (a) Find its impedance Z at 60.0 Hz and 10.0 kHz . (b) Compare these values of Z with those found in Example 23.12 in which there was also a capacitor.

Step by Step Solution

Step 1: Definition of circuit

Step 2: Given data

Step 2: Finding Impedance(a)

Step 3: Compare these values of Z(b)

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College Physics (Urone)

Answers without the blur.

Short Answer

An RL circuit consists of a 40.0Ω resistor and a 3.00 mH inductor. (a) Find its impedance Z at 60.0 Hz and 10.0 kHz . (b) Compare these values of Z with those found in Example 23.12 in which there was also a capacitor.

Step by Step Solution

Step 1: Definition of circuit

Step 2: Given data

Step 2: Finding Impedance(a)

Step 3: Compare these values of Z(b)

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An RL circuit consists of a $40.0 Ω$ resistor and a 3.00 mH inductor. (a) Find its impedance Z at 60.0 Hz and 10.0 kHz . (b) Compare these values of Z with those found in Example 23.12 in which there was also a capacitor.