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<h2 class="hd hd-2 unit-title">Adding Circuit Elements in Parallel</h2>
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Adding Circuit Elements in Parallel
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We have a circuit with a driving EMF in parallel with each of a resistor, with resistance [mathjaxinline]R[/mathjaxinline], an inductor with inductance [mathjaxinline]L[/mathjaxinline], and a capacitor with capacitance, [mathjaxinline]C[/mathjaxinline]. If we are driving the circuit with a voltage [mathjaxinline]V = V_0 e^{i \omega t}[/mathjaxinline], find the amplitude and phase of the current response, [mathjaxinline]I_0 e^{i\omega t}e^{-i \phi }[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]I_0[/mathjaxinline] = </p>
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<p style="display:inline">[mathjaxinline]\phi[/mathjaxinline] = </p>
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<h3 class="hd hd-2">W12PS1: Worked Example - Adding Circuit Elements in Parallel</h3>
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<h2 class="hd hd-2 unit-title">Low Pass Filter</h2>
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Low Pass Filter
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Consider an AC voltage source, [mathjaxinline]V_{in} = V_{in,0} e^{i \omega t}[/mathjaxinline], connected in series to a resistor with resistance [mathjaxinline]R[/mathjaxinline] and a capacitor with capacitance [mathjaxinline]C[/mathjaxinline] as shown in the figure below. The output voltage, [mathjaxinline]V_ C \equiv V_{out} = V_{out, 0} e^{i(\omega t - \delta )}[/mathjaxinline], is the voltage across the capacitor. In this problem, we shall solve for the ratio [mathjaxinline]V_{out, 0} / V_{in, 0}[/mathjaxinline] and identify the range of frequencies which are blocked by this filter. </p>
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<img src="/assets/courseware/v1/60cc6362e30f1ffad20b2fe2bc93d53a/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_Friday_W12d3_1-001.jpg" width="440"/>
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Remember that inverse trigonometric functions are entered in the edX as arcsin(), arccos(), arctan(), arccot(), etc. You may also enter [mathjaxinline]i[/mathjaxinline] in your answers here, where [mathjaxinline]i = \sqrt {-1}[/mathjaxinline]. </p>
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Discussions of all of these questions can be found in the video linked below. </p>
<p><b class="bfseries">(Part a)</b> What is the equivalent impedance of this circuit? Express your answer using R, C, V_0 for [mathjaxinline]V_{in,0}[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]Z_{eq} =[/mathjaxinline] </p>
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Impedance Modulus
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<p><b class="bfseries">(Part b)</b> What is the modulus of the equivalent impedance? Express your answer using R, C, V_0 for [mathjaxinline]V_{in,0}[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]|Z_{eq}| =[/mathjaxinline] </p>
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Current Amplitude
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<p><b class="bfseries">(Part c)</b> What is the amplitude of the current? Express your answer using R, C, V_0 for [mathjaxinline]V_{in,0}[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]I_0 =[/mathjaxinline] </p>
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Current Phase
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<p><b class="bfseries">(Part d)</b> What is the phase constant for the current? Express your answer using R, C, V_0 for [mathjaxinline]V_{in,0}[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\phi =[/mathjaxinline] </p>
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Capacitor Voltage Time Dependence
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<p><b class="bfseries">(Part e)</b> What is the voltage across the capacitor as a function of time? Please use [mathjaxinline]I_0[/mathjaxinline] and [mathjaxinline]\phi[/mathjaxinline], and do not expand them. Express your answer using C, t, I_0 for [mathjaxinline]I_0[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline], and phi for [mathjaxinline]\phi[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]V_{out} =[/mathjaxinline] </p>
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Capacitor Voltage Amplitude
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<p><b class="bfseries">(Part f)</b> What is the amplitude of the voltage across the capacitor? Express your answer using R, C, V_0 for [mathjaxinline]V_{in,0}[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]V_{out,0} =[/mathjaxinline] </p>
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Output/Input Voltage Ratio
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<p><b class="bfseries">(Part g)</b> What is the ratio of the amplitude of the voltage across the capacitor to the amplitude of the AC voltage source, [mathjaxinline]V_{out,0} / V_{in,0}[/mathjaxinline]? Express your answer using R, C and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]V_{out,0} / V_{in,0} =[/mathjaxinline] </p>
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Voltage Ratio at High Frequencies
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<p><b class="bfseries">(Part h)</b> What is the limit of your answer to Part (g) when [mathjaxinline]\omega CR \gg 1[/mathjaxinline]? Expand your answer to the first non-zero term, do not include further terms. </p>
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<p style="display:inline">[mathjaxinline]V_{out,0} / V_{in,0} =[/mathjaxinline] </p>
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Low Pass Filter?
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<p><b class="bfseries">(Part i)</b> Why is this type of filter called a low pass filter? </p>
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<h3 class="hd hd-2">W12PS2: Worked Example - Low Pass Filter</h3>
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<h2 class="hd hd-2 unit-title">Impedance Mismatching</h2>
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Impedance Mismatching
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<p>
You have a car stereo amplifier which you can think of as an AC voltage generator in series with a resistance internal to the amplifier. The EMF of the amplifier is [mathjaxinline]\mathcal{E}(t)=\mathcal{E}_0\cos (\omega t)[/mathjaxinline] where the amplitude [mathjaxinline]\mathcal{E}_0=18\, V[/mathjaxinline], and the internal resistance [mathjaxinline]R_{amplifier}=0.5\, \Omega[/mathjaxinline]. We say that the amplifier has an output impedance of half an ohm. </p>
<p>
With this amplifier you want to drive a [mathjaxinline]75\, W[/mathjaxinline] monitor speaker, which you can think of as an [mathjaxinline]R_{speaker}=16\, \Omega[/mathjaxinline] resistor that can take up to a maximum of [mathjaxinline]75\, W[/mathjaxinline] of power (of course, the more power you dump in it the more sound you get out). </p>
<p>
If you just connect your amplifier to the speaker, what is the average power that you will deliver to the speaker? (Hint: In order to time average a function, you need to integrate the function over one period, then divide by the period.) </p>
<p>
<p style="display:inline">[mathjaxinline]P_{avg}=[/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline]W[/mathjaxinline])</p>
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<h3 class="hd hd-3 problem-header" id="finalexam-problem17-problem-title" aria-describedby="block-v1:MITx+8.02.3x+1T2019+type@problem+block@finalexam-problem17-problem-progress" tabindex="-1">
Circuit Element(s) in a Driven Circuit
</h3>
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<p>
The circuit shown above contains an AC generator with [mathjaxinline]V(t) = V_0 \sin \omega t[/mathjaxinline], a resistor with resistance [mathjaxinline]R=\sqrt {3} \rm {\Omega }[/mathjaxinline], and a "black box", which contains <b class="bfseries">either an inductor, or a capacitor, or both an inductor and capacitor connected in series</b>. In general, the current in the circuit is given by [mathjaxinline]I(t) = I_0 \sin (\omega t - \phi )[/mathjaxinline], where </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle I_0 = \frac{V_0}{(R^2 + (\omega L - 1/\omega C)^2)^{1/2}}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
</table>
<p>
and </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000005">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \tan \phi = \frac{\omega L - 1/\omega C}{R}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>2</span>)</span>
</td>
</tr>
</table>
<p>
In the above equations: </p>
<p>
1. If there is no capacitor in the circuit, set [mathjaxinline]C = \infty[/mathjaxinline], then [mathjaxinline]1/C = 0[/mathjaxinline], (note that infinite capacitance means that the capacitor can be treated like a resistanceless wire); </p>
<p>
2. If there is no inductor in the circuit, set [mathjaxinline]L = 0[/mathjaxinline]. </p>
<p>
We first measure the current in the circuit at an angular frequency [mathjaxinline]\omega _1 = 2\, \rm {rad \cdot s^{-1}}[/mathjaxinline] and find that the current leads the voltage source by exactly [mathjaxinline]\pi /6[/mathjaxinline] radians. We then measure the current in the circuit at an angular frequency [mathjaxinline]\omega _2 = 6\, \rm {rad \cdot s^{-1}}[/mathjaxinline] and find that the current lags the driving emf by exactly [mathjaxinline]\pi /6[/mathjaxinline] radians. [Note: [mathjaxinline](\pi /6) \rm { rad} = 30^{\circ }[/mathjaxinline], and [mathjaxinline]\tan (\pi /6) = 1/ \sqrt {3}[/mathjaxinline]]. </p>
<p><b class="bfseries">(Part a)</b> What is inside the black box? </p>
<p>
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<text> Only L.</text>
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<text> Only C.</text>
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<text> Both L &amp; C.</text>
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<br/>
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<p><b class="bfseries">(Part b)</b> What is (are) the value(s) of the inductance or (and) capacitance? </p>
<p>
<p style="display:inline">[mathjaxinline]L =[/mathjaxinline]</p>
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<p style="display:inline">H</p>
<br/>
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<p>
<p style="display:inline">[mathjaxinline]C =[/mathjaxinline]</p>
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<p style="display:inline">F</p>
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<span id="solution_finalexam-problem17_solution_2"/>
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<br/>
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<p><b class="bfseries">(Part c)</b> Will the circuit resonate? If so, at what angular frequency? </p>
<p>
<p style="display:inline">[mathjaxinline]\omega _0 =[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]\rm {rad \cdot s^{-1}}[/mathjaxinline]</p>
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<h3 class="hd hd-2">W12PS3: Worked Example - Unknown Circuit Element in a Box</h3>
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<h2 class="hd hd-2 unit-title">Power in a Driven RLC Circuit</h2>
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Maximum Power in a Driven RLC Circuit
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<p>
Consider the RLC circuit in the figure above. The driving voltage is given by [mathjaxinline]V(t) = V_0\sin (\omega t)[/mathjaxinline] and the current in the circuit is [mathjaxinline]I(t) = I_0\sin (\omega t - \phi )[/mathjaxinline]. The amplitude of the current as a function of [mathjaxinline]\omega[/mathjaxinline] is given by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle I_0=\dfrac {V_0}{\sqrt {R^2+(\omega L - 1/\omega C)^2}}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
</table>
<p>
and its phase shift with respect to the voltage source as a function of [mathjaxinline]\omega[/mathjaxinline] is: </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000005">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \tan \phi = \dfrac {\omega L - 1/(\omega C)}{R}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>2</span>)</span>
</td>
</tr>
</table>
<p>
The time average power delivered by the battery is given by: </p>
<table id="a0000000006" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000007">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle &lt;P&gt; = \dfrac {1}{2}I_0 V_0\cos (\phi )[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>3</span>)</span>
</td>
</tr>
</table>
<p><b class="bfseries">(Part a)</b> Using the trigonometric identity [mathjaxinline]1+ \tan ^2(\phi )=\dfrac {1}{\cos ^2(\phi )}[/mathjaxinline], and eq. (2), obtain an expression of [mathjaxinline]\cos (\phi )[/mathjaxinline] in terms of [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\cos (\phi )=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Use the expression of [mathjaxinline]\cos (\phi )[/mathjaxinline] obtained in part (a) and [mathjaxinline]I_0[/mathjaxinline] from eq. (1) in eq. (3) to write [mathjaxinline]&lt;P&gt;[/mathjaxinline] in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]&lt;P_{battery}&gt;=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Which of the following figures best represents a plot of the time average power [mathjaxinline]&lt;P&gt;[/mathjaxinline] vs. [mathjaxinline]\omega[/mathjaxinline], </p>
<center>
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<text>Figure 1</text>
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<text>Figure 2</text>
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<text>Figure 3</text>
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<b class="bfseries">Principle of a tuner</b>
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At resonance, when [mathjaxinline]\omega = \omega _0=1/\sqrt {LC}[/mathjaxinline], the time average power transfered by the AC power source to the circuit is at its maximum. In the left figure below, a plot of [mathjaxinline]&lt;P&gt;[/mathjaxinline] vs. [mathjaxinline]\omega[/mathjaxinline] is shown for two different values of [mathjaxinline]R[/mathjaxinline]. As indicated on the figure, larger and smaller values of the resistance, correspond to smaller and larger values of the maximum average power input, respectively, as expected from the [mathjaxinline]1/R[/mathjaxinline] in the power formula. Not as obvious is the fact that the width of the peak is larger for larger [mathjaxinline]R[/mathjaxinline]. This happens because, if [mathjaxinline]R[/mathjaxinline] is big, the term [mathjaxinline](\omega L - 1/\omega C)[/mathjaxinline] has to change by a larger factor (hence needing a bigger change in [mathjaxinline]\omega[/mathjaxinline]) to make the same proportional change in the denominator. </p>
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<img src="/assets/courseware/v1/57768bcad685b9ddb74f4cfa8470bc27/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w13_31b.svg" width="495"/>
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As shown in the right figure above, the width of the peak [mathjaxinline]\Delta \omega[/mathjaxinline] is measured by the difference of frequencies obtained when [mathjaxinline]&lt;P&gt; = \dfrac {&lt;P(\omega _0)&gt;}{2}[/mathjaxinline]. Doing some math you can show that [mathjaxinline]\Delta \omega = \dfrac {R}{L}[/mathjaxinline]. </p>
<p>
The principle of a radio tuner is based on the resonance of an AC circuit with the frequency we desire to detect. The antenna of the radio will receive thousands of sine waves. The job of a tuner is to select one sine wave from the thousands of radio signals that the antenna receives. This is achieved by setting [mathjaxinline]\omega _0=1/\sqrt {LC}[/mathjaxinline] equal to the desired frequency. </p>
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Tuning Resonance
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<p>A variable capacitor within an LC circuit is used to tune a radio to a desired frequency. You are initially tuned to Boston's local NPR station, which transmits at a frequency of 89.7 MHz. If you switch your radio to a music station playing at 107.9 MHz by tuning your capacitor from a value of \(C_{A}\) to \(C_{B}\), how is the new capacitance related to the initial capacitance? Express your answer as the ratio \(C_{B}/C_{A}\). (Note: the answer field will parse mathematical operations OR you can use the calculator provided at the bottom of the screen.)</p>
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