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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# An RLC circuit such as that of Fig. 31-7 has ${\mathbit{R}}{\mathbf{=}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{Ω}}}$, ${\mathbit{C}}{\mathbf{=}}{\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{μF}}}$, ${\mathbit{L}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{H}}}$, and ${{\mathbit{\epsilon }}}_{{\mathbf{m}}}{\mathbf{=}}{\mathbf{30}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{\text{V}}}$. (a) At what angular frequency ${{\mathbit{\omega }}}_{{\mathbf{d}}}$ will the current amplitude have its maximum value, as in the resonance curves of Fig. 31-16? (b) What is this maximum value? At what (c) lower angular frequency ${{\mathbit{\omega }}}_{\mathbf{d}\mathbf{1}}$ and (d) higher angular frequency ${{\mathbit{\omega }}}_{\mathbf{d}\mathbf{2}}$ will the current amplitude be half this maximum value? (e) For the resonance curve for this circuit, what is the fractional half-width $\mathbf{\left(}{\mathbf{\omega }}_{\mathbf{d}\mathbf{2}}\mathbf{-}{\mathbf{\omega }}_{\mathbf{d}\mathbf{1}}\mathbf{\right)}\mathbf{/}\mathbf{\omega }\mathbf{}\mathbf{}$?

1. Angular frequency ${\omega }_{d}$ corresponding to maximum current amplitude is $224\text{rad/s}$.
2. Maximum value of amplitude of current is $6.0\text{A}$.
3. Lower angular frequency ${\omega }_{d1}$ corresponding to half of the maximum current amplitude is $219\text{rad/s}$.
4. Higher angular frequency ${\omega }_{d2}$ corresponding to half of the maximum current amplitude is $228\text{rad/s}$.
5. Fractional half width is $0.040$.
See the step by step solution

## Step 1: The given data

1. Amplitude of emf, ${\epsilon }_{\mathrm{m}}=30.0\text{V}$
2. Resistance, $R=5.0\text{Ω}$
3. Capacitance, $C=20.0\text{μF}$
4. Inductance, $L=1.0\text{H}$

## Step 2: Understanding the concept of frequency of LC circuit

When driving angular frequency is equal to the natural angular frequency of the circuit, then the current amplitude is maximum because of Z=R. This condition is satisfied only when the capacitive reactance is exactly matched with inductive reactance, and it is called the condition of resonance. So, by substituting the given values in the corresponding formulae, we can find the required quantities.

The capacitive reactance of the capacitor,

${{\mathbf{X}}}_{{\mathbf{C}}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{\omega }}_{\mathbf{d}}\mathbf{C}}$ ...(1)

The inductive reactance of the inductor,

${{\mathbit{X}}}_{{\mathbit{L}}}{\mathbf{=}}{{\mathbit{\omega }}}_{{\mathbit{d}}}{\mathbit{L}}$ ...(2)

The impedance of the LCR circuit for the driving frequency,

${\mathbit{Z}}{\mathbf{=}}\sqrt{{\mathbit{R}}^{\mathbf{2}}\mathbf{+}{\left({\mathbit{X}}_{\mathbit{L}}\mathbf{-}{\mathbit{X}}_{\mathbit{c}}\right)}^{\mathbf{2}}}$ ...(3)

The current equation using Ohm’s law,

${\mathbit{I}}{\mathbf{=}}\frac{{\mathbit{\epsilon }}_{\mathbit{m}}}{\mathbit{Z}}$ ...(4)

Here, R is the resistance of the resistor, C is the capacitance of the capacitor, L is the inductance of the inductor and ${{\mathbit{\omega }}}_{{\mathbit{d}}}$ is the driving angular frequency.

## Step 3: a) Calculation of the angular frequency

For a given resistance R and emf ${\epsilon }_{m}$, I is maximum when .role="math" localid="1663000356720" ${\chi }_{L}-{\chi }_{C}=0$.

This implies ${X}_{L}={X}_{C}$, thus using equations (1) and (2) we get the angular frequency ${\omega }_{d}$ corresponding to maximum current amplitude as follows:

${\omega }_{d}L=\frac{1}{{\omega }_{d}C}\phantom{\rule{0ex}{0ex}}{\omega }_{d}^{2}=\frac{1}{LC}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{LC}}$

For the given values, we have

${\omega }_{d}=\frac{1}{\sqrt{\left(1.0\text{H}\right)×\left(20.0×{10}^{-6}\text{F}\right)}}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{3}}{\sqrt{20.0}}\text{rad/s}\phantom{\rule{0ex}{0ex}}=223.6\text{rad/s} \phantom{\rule{0ex}{0ex}}\approx 224\text{rad/s}\phantom{\rule{0ex}{0ex}}$

Hence, the angular frequency is $224\text{rad/s}$.

## Step 4: b) Calculation of the maximum current

For the condition of resonance, we have $Z=R$,

So the maximum value of amplitude of current is given using equation (4) as follows:

$I=\frac{30.0\text{V}}{5.0\text{Ω}}\phantom{\rule{0ex}{0ex}}=6.0\text{A}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the current is $6.0\text{A}$.

## Step 5: c) Calculation of the lower angular frequency

To find lower angular frequency ${\omega }_{d1}$ corresponding to half of the maximum current amplitude we have the following equation using equation (3) in equation (4) as follows:

$\frac{I}{2}=\frac{{\epsilon }_{m}}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\frac{{\epsilon }_{m}}{2R}=\frac{{\epsilon }_{m}}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\frac{1}{2R}=\frac{1}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}}\phantom{\rule{0ex}{0ex}}\frac{1}{4{R}^{2}}=\frac{1}{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}\phantom{\rule{0ex}{0ex}}$

Solving further

$4{R}^{2}={R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}\phantom{\rule{0ex}{0ex}}3{R}^{2}={\left({X}_{L}-{X}_{C}\right)}^{2}\phantom{\rule{0ex}{0ex}}$

Taking the square root on both sides of the above equation, we get that

localid="1663002854238" $±\sqrt{3}·R={X}_{L}-{X}_{C}\phantom{\rule{0ex}{0ex}}=\left({\omega }_{d}L-\frac{1}{{\omega }_{d}C}\right)\text{(from equations (1) and (2))}\phantom{\rule{0ex}{0ex}}$

On simplifying the above equation further, we get a quadratic equation in ${\omega }_{d}$ as follows:

$\left(LC\right){\omega }_{d}^{2}±\sqrt{3}·\left(RC\right){\omega }_{d}-1=0$

The two roots of this equation will give lower angular frequency ${\omega }_{d1}$ and higher angular frequency ${\omega }_{d2}$ corresponding to half of the maximum current amplitude.

On solving the quadratic equation, we get,

${\omega }_{d}=\frac{-\left(±\sqrt{3}·RC\right)±\sqrt{3{\left(RC\right)}^{2}+4LC}}{2LC}$

As negative frequency is not possible, so we can neglect negative sign in the second term. Thus, we get the angular frequency equation as:

${\omega }_{d}=\frac{-\left(±\sqrt{3}·RC\right)+\sqrt{3{\left(RC\right)}^{2}+4LC}}{2LC}$

Now considering the positive sign, i.e. $+\sqrt{3}·RC$, we can get the lower angular frequency ${\omega }_{d1}.$ from the above equation as follows:

${\omega }_{d1}=\frac{-\left(\sqrt{3}·RC\right)+\sqrt{3{\left(RC\right)}^{2}+4LC}}{2LC}$

We have, the above values as:

$RC=5.0\text{Ω}×\left(20.0×{10}^{-6}\right)\text{F}\phantom{\rule{0ex}{0ex}}={10}^{-4}\text{ΩF}\phantom{\rule{0ex}{0ex}}$

And

$LC=1.0\text{H}×\left(20.0×{10}^{-6}\right)\text{F}\phantom{\rule{0ex}{0ex}}=0.2×{10}^{-4}\text{HF}\phantom{\rule{0ex}{0ex}}$

Now, substituting the above values, we can get the lower angular frequency as follows:

${\omega }_{d1}=\frac{-\left(\sqrt{3}×{10}^{-4}\text{Ω F}\right)+\sqrt{3×{10}^{-8}{\left(\text{ΩF}\right)}^{2}+4×\left(0.2×{10}^{-4}\right)\text{HF}}}{2×\left(0.2×{10}^{-4}\right)\text{HF}}\phantom{\rule{0ex}{0ex}}=219\text{rad/s}$

Hence, the value of the frequency is $219\text{rad/s}$.

## Step 6: d) Calculation of the higher angular frequency

Similarly, considering the negative sign, i.e. $-\sqrt{3}·RC$, from equation, we can get the higher angular frequency ${\omega }_{d2}$ for the above values as follows:

${\omega }_{d2}=\frac{\left(+\sqrt{3}·RC\right)+\sqrt{3{\left(RC\right)}^{2}+4LC}}{2LC}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{3}×{10}^{-4} \text{ΩF}\right)+\sqrt{3×{10}^{-8}{\left(\text{ΩF}\right)}^{2}+4×\left(0.2×{10}^{-4}\right)\text{HF}}}{2×\left(0.2×{10}^{-4}\right)\text{HF}}\phantom{\rule{0ex}{0ex}}=228\text{rad/s}\phantom{\rule{0ex}{0ex}}$

Hence, the value of the frequency is $228\text{rad/s}$.

## Step 7: e) Calculation of the fractional half-width

Resonance curve for this circuit will have the following fractional width as follows:

$\frac{\left({\omega }_{d2}-{\omega }_{d1}\right)}{\omega }=\frac{228.0\text{rad/s}-219.0\text{rad/s}}{224.0\text{rad/s}}\phantom{\rule{0ex}{0ex}}=0.040\phantom{\rule{0ex}{0ex}}$

Hence, the value of the fractional width is $0.040$.

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