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Expert-verified Found in: Page 863 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # An RL circuit consists of a $40.0\Omega$ 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 $188.5\Omega$ and $193\Omega$

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

See the step by step solution

## 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 $R=40.0\Omega$

The inductor in the RL circuit is $L=3.00mH\left(\frac{{10}^{-3}H}{1mH}\right)=3.00×{10}^{-3}H$

## Step 2: Finding Impedance(a)

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

${X}_{L}=2\pi fL$………..(1)

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

For each frequency, impedance can be expressed as,

$Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}$

………………(2)

At low frequency, f = 60 Hz, the value of ${X}_{C}=0\Omega$.

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

${X}_{L}=2×3.14×\left(60Hz\right)\left(3.00×{10}^{-3}H\right)\phantom{\rule{0ex}{0ex}}=1.13\Omega$

………………(3)

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

$Z=\sqrt{\left({\left(40\Omega \right)}^{2}+{\left(1.13\Omega -0\Omega \right)}^{2}\right)}\phantom{\rule{0ex}{0ex}}=40.02\Omega$

At high-frequency $f=10.0kHz\left(\frac{{10}^{3}Hz}{1kHz}\right)=1.00×{10}^{4}Hz$, substituting the given values in the above equation (1), we get

${X}_{L}^{\text{'}}=2×3.14×\left(1×{10}^{4}Hz\right)\left(3×{10}^{-3}H\right)\phantom{\rule{0ex}{0ex}}=188.5\Omega$

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

${Z}^{\text{'}}=\sqrt{{\left({\left(40\Omega \right)}^{2}+\left(188.5\Omega \right)-0\Omega \right)}^{2}}\phantom{\rule{0ex}{0ex}}=192.7\Omega$

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

## 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, $Z=531\Omega$, 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, $Z=190\Omega$, that is similar to the value obtained without the capacitor.

Thus, the capacitor has a smaller effect at high frequencies. ### Want to see more solutions like these? 