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<h2 class="hd hd-2 unit-title">Introduction to AC Circuits</h2>
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<p><center><b> Alternating Current, AC current</b></center></p><p>When a loop of wire rotates inside a magnetic field, Faraday's law predicts that the changing magnetic flux induces an emf that oscillates sinusoidally in time with a frequency determined by the angular speed of the coil. This is a source of <i>alternating current</i> (AC). More precisely, the emf in the coil is a source of alternating <i>voltage</i>, which will create alternating current in whatever circuit is connected to the coil. </p><p> In this lesson you will learn how to calculate the current flowing in a circuit when resistors, capacitors, and inductors are connected to an AC source. The symbol used to represent an AC source is shown below. </p><p><center><img src="/assets/courseware/v1/f34440e24db662259a04b2507416077b/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson31_01a.svg" width="250"/></center></p><br/><hr/><p><center><b> Sinusoidal source, amplitude and phase.</b></center></p><p>The AC voltage source is a sinusoidal function of time with a frequency \(\omega\) and amplitude \(V_0\), determined by the characteristics of the electric generator (the rotating coil in the B-field). A plot of the voltage as a function of time, \(V(t)\), is shown in the figure below. The period \(T = \dfrac{2\pi}{\omega}\) and the amplitude \(V_0\) are shown. </p><p><center><img src="/assets/courseware/v1/ecf41044f8bd905473888fd76aef5049/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson31_01b.svg" width="600"/></center></p><p> One way to write a mathematical representation of the voltage is \(V(t) = V_0\sin(\omega t - \theta)\), where \(\theta\) is an arbitrary phase. For example, if we decide to set up the zero of time at the instant when the source is at its maximum, that can be achieved by setting \(\theta = -\pi/2\) so that \(V(t) = V_0\cos(\omega t )\) as is shown in the left figure below. If, as another example, we decide to set the zero of time at the instant when the voltage is zero, we can pick \(\theta = 0\) leading to \(V(t) = V_0\sin(\omega t )\), shown in the right figure below. Note that the time we chose to call \(t=0\) is arbitrary, so no option for \(\theta\) is any more valid than any other.</p><p><center><img src="/assets/courseware/v1/fec1a515a986b39d1b9b2cfa345f1d61/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson31_01c.svg" width="700"/></center></p><p> For simplicity, in this lesson and the following ones, we will restrict the mathematical representation of an AC source to either \(V(t) = V_0\cos(\omega t )\) or \(V(t) = V_0\sin(\omega t )\). </p><br/><hr/><p><center><b> Driven AC circuit, Sinusoidal Current, amplitude and phase SHIFT.</b></center></p><br/><p><center><img src="/assets/courseware/v1/80595b2be13b94b4ce4d19f70204fc42/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson31_01d.svg" width="700"/></center></p><p> In the left figure, the AC source is connected to a circuit where the box can contain a resistor, a capacitor, an inductor, or a combination of these elements in series or in parallel. The goal is to determine the current \(I(t)\) flowing through the circuit. If the source is oscillating with an angular frequency \(\omega\), we expect the current to oscillate with the same frequency, therefore the current is also a sinusoidal function of time given by:</p><p>\[I(t) = I_0 \sin(\omega t - \phi)\]</p><p>where \(I_0\) is the amplitude and \(\phi\) the <b>phase shift</b> of the current with respect to the voltage. The seemingly odd choice of a minus sign will be explained below</p><p> The angular frequency is called the <b>driving frequency</b>, the voltage source is <b>driving</b> the circuit with frequency \(\omega\).</p><p>In the right figure shown above, we set the zero of time at the instant for which the voltage is zero, so \(V(t) = V_0\sin(\omega t )\), shown as a black line. If we consider the voltage to be the reference signal, we observe that the plot of the current (dashed red line) is shifted to the <i>right</i>, i.e. the peaks in the dashed red curve (such as the first maximum) are shifted to the right relative to the peaks in the black curve. The current is not in phase with the voltage. Since the current peaks <i>after</i> the voltage, we say that it "lags" the voltage (the opposite shift is called "leading" the voltage).</p><p>The shift between the two curves is given by the phase shift \(\phi\). In the expression \( \sin(\omega t - \phi)\), a positive value of \(\phi\) will correspond to the peaks occurring at a <i>larger</i> value of \(t\). The convention is that we think of a shift to later time as a "positive" shift that should correspond to a positive value of \(\phi\), and this is why the expression \(\sin(\omega t - \phi)\) has a minus sign. </p><p> In this lesson, you will learn how to calculate \(\phi\) and \(I_0\), the phase shift of the current with respect to the voltage and the amplitude of the current. You will find that the phase shift and the amplitude depend on the type of circuit element that is connected to the source as well as on the driving angular frequency \(\omega\).</p><p> In summary: In an AC circuit, the voltage oscillates with an angular frequency \(\omega\) and if we use it as a reference signal we can describe it as \(V(t) = V_0\cos(\omega t )\) or \(V(t) = V_0\sin(\omega t )\) depending on our choice of zero for the time. We remark that the resulting current is not necessarily in phase with the driving voltage, therefore we will assume that is expressed as \(I(t) = I_0\cos(\omega t -\phi)\) or \(I(t) = I_0\sin(\omega t -\phi)\), with the choice of trig function being the same for voltage and current. Our goal is to find \(I_0\) and \(\phi\).</p>
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<h2 class="hd hd-2 unit-title">L31v1: Introduction to AC Circuits</h2>
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<center><b>Introduction to Driven RLC circuits.</b></center><br/><p> A driven RLC circuit is an AC source connected to a resistor, an inductor, and a capacitor in series. In the following videos, we will use an applet to help you visualize the presentation of the new concepts. We advise you to use the applet while watching the videos. You can find the it at
<a href="https://mathlets.org/mathlets/series-rlc-circuit/">MIT
mathlets, RLC circuits.</a> An example of the interactive window of the applet is shown below.</p><p><center><figure><img src="/assets/courseware/v1/4505a7bfaf978f51cbff6b25c9898378/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson31_02.png" width="600"/><figcaption>Mathlet: RLC circuit.</figcaption></figure></center></p>
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<h3 class="hd hd-2">L31v1: Introduction to AC Circuits</h3>
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<h2 class="hd hd-2 unit-title">L31v2: Definition of Impedance</h2>
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Complexifying the voltage and current
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<p>
An AC source is connected to a circuit element that can be a resistor, an inductor, a capacitor, or a combination of those. Assume the voltage source to be given by [mathjaxinline]V(t) = V_0\cos (\omega t)[/mathjaxinline], where [mathjaxinline]V_0[/mathjaxinline] is the amplitude in volts, and [mathjaxinline]\omega[/mathjaxinline] is the angular frequency in radians/sec. The resulting current in the circuit is [mathjaxinline]I(t) = I_0\cos (\omega t - \phi )[/mathjaxinline], where [mathjaxinline]I_0[/mathjaxinline] is the current amplitude in amperes, and [mathjaxinline]\phi[/mathjaxinline] is the phase shift of the current with respect to the voltage in radians. </p>
<p>
Consider now the complex function of time [mathjaxinline]V_ c(t) = V_0 e^{i\omega t}[/mathjaxinline] and [mathjaxinline]I_ c(t) = I_0 e^{i(\omega t -\phi )}[/mathjaxinline]. These functions, with the subscript "c", are the <b class="bfseries">complexified</b> versions of the physical real functions [mathjaxinline]V(t)[/mathjaxinline] and [mathjaxinline]I(t)[/mathjaxinline] defined above. </p>
<p><b class="bfseries">(Part a)</b> Calculate the real part of [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline]. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], I_0 for [mathjaxinline]I_0[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline], and phi for [mathjaxinline]\phi[/mathjaxinline], as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\text {Re}(V_ c)=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\text {Re}(I_ c)=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Calculate the imaginary part of [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline]. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], I_0 for [mathjaxinline]I_0[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline], and phi for [mathjaxinline]\phi[/mathjaxinline], as needed. </p>
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<p style="display:inline">[mathjaxinline]\text {Im}(V_ c)=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\text {Im}(I_ c)=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> We now introduce a complex number [mathjaxinline]z[/mathjaxinline] called <b class="bfseries">impedance</b> such that [mathjaxinline]V_ c = z\; I_ c[/mathjaxinline]. If the Euler expression of the impedance is [mathjaxinline]z = |z| \; e^{i\delta }[/mathjaxinline], find [mathjaxinline]|z|[/mathjaxinline], the modulus of [mathjaxinline]z[/mathjaxinline], and [mathjaxinline]\delta[/mathjaxinline], the argument of [mathjaxinline]z[/mathjaxinline] in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], I_0 for [mathjaxinline]I_0[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline], and phi for [mathjaxinline]\phi[/mathjaxinline], as needed. </p>
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<p style="display:inline">[mathjaxinline]|z|=[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L31v3: Impedance for a Resistor</h2>
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Impedance for a Resistor
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<p>
An AC source is connected to a resistor of resistance [mathjaxinline]R[/mathjaxinline]. The voltage source is given by [mathjaxinline]V(t) = V_0\cos (\omega t)[/mathjaxinline] and the current in the circuit is [mathjaxinline]I(t) = I_0\cos (\omega t - \phi )[/mathjaxinline]. The real physical voltage [mathjaxinline]V(t)[/mathjaxinline] and the real physical current [mathjaxinline]I(t)[/mathjaxinline] are related 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 V(t) = R\; I(t)[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
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<span>(<span>1</span>)</span>
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<p>
and the complexified voltage and current, [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline] satisfy: </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 V_ c(t) = R\; I_ c(t)[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
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<span>(<span>2</span>)</span>
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<p><b class="bfseries">Note:</b> You showed that equation (2) is true in lesson 30. </p>
<p><b class="bfseries">(Part a)</b> What is the Euler expression of [mathjaxinline]V_ c[/mathjaxinline], the complexified voltage. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]V_ c=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> What is the Euler expression of [mathjaxinline]I_ c[/mathjaxinline], the complexified current. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], I_0 for [mathjaxinline]I_0[/mathjaxinline], phi for [mathjaxinline]\phi[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]I_ c=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Use the results obtained in parts (a) and (b) in equation (2) to calculate [mathjaxinline]I_0[/mathjaxinline] and [mathjaxinline]\phi[/mathjaxinline]. Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], R, and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
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<p><b class="bfseries">(Part d)</b> If the impedance is a complex number [mathjaxinline]z = |z|\; e^{i\delta }[/mathjaxinline] such that </p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle V_ c(t) = z\; I_ c(t)\, ,[/mathjaxinline]
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use the results obtained in parts (a) to (c) in the equation above to calculate the modulus [mathjaxinline]|z|[/mathjaxinline] and the argument [mathjaxinline]\delta[/mathjaxinline] of the impedance of a resistor. Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]|z|=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e)</b> If the AC voltage source is expressed in terms of a [mathjaxinline]\sin ( )[/mathjaxinline] instead of a [mathjaxinline]\cos ( )[/mathjaxinline], that is, if [mathjaxinline]V(t) = V_0\sin (\omega t)[/mathjaxinline] and the current is [mathjaxinline]I(t) = I_0\sin (\omega t - \phi )[/mathjaxinline], will your answers in parts (c) and (d) change? </p>
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<p><b class="bfseries">Note:</b> In the video shown below, the subscript "c" for the complex functions [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline] is not used. </p>
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<h3 class="hd hd-2">L31v3: Impedance for a Resistor</h3>
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<h2 class="hd hd-2 unit-title">L31v4: Impedance for an Inductor</h2>
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Impedance for an Inductor
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<p>
An AC source is connected to an inductor of inductance [mathjaxinline]L[/mathjaxinline]. The voltage source is given by [mathjaxinline]V(t) = V_0\cos (\omega t)[/mathjaxinline] and the current in the circuit is [mathjaxinline]I(t) = I_0\cos (\omega t - \phi )[/mathjaxinline]. The real physical voltage [mathjaxinline]V(t)[/mathjaxinline] and the real physical current [mathjaxinline]I(t)[/mathjaxinline] are related by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle V(t) = L\dfrac {dI(t)}{dt}[/mathjaxinline]
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<span>(<span>1</span>)</span>
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<p>
and the complexified voltage and current, [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline] satisfy: </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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[mathjaxinline]\displaystyle V_ c(t) = L\dfrac {dI_ c(t)}{dt}[/mathjaxinline]
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<span>(<span>2</span>)</span>
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<p><b class="bfseries">Note:</b> You showed that equation (2) is true in Lesson 30. </p>
<p><b class="bfseries">(Part a)</b> What is the Euler expression of [mathjaxinline]V_ c[/mathjaxinline], the complexified voltage. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]V_ c=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> What is the Euler expression of [mathjaxinline]\dfrac {dI_ c}{dt}[/mathjaxinline], the time derivative of the complexified current. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], I_0 for [mathjaxinline]I_0[/mathjaxinline], phi for [mathjaxinline]\phi[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. Note that the correct Euler format has a positive <i class="it">real</i> expression multiplied by [mathjaxinline]e[/mathjaxinline] to an imaginary power. </p>
<p>
<p style="display:inline">[mathjaxinline]\dfrac {dI_ c}{dt}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Use the results obtained in parts (a) and (b) in equation (2) to calculate [mathjaxinline]I_0[/mathjaxinline] and [mathjaxinline]\phi[/mathjaxinline]. Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </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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<p><b class="bfseries">(Part d)</b> If the impedance is a complex number [mathjaxinline]z = |z|\; e^{i\delta }[/mathjaxinline] such that </p>
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[mathjaxinline]\displaystyle V_ c(t) = z\; I_ c(t)[/mathjaxinline]
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Use the results obtained in parts (a) to (c) in the equation above to calculate the modulus [mathjaxinline]|z|[/mathjaxinline] and the argument [mathjaxinline]\delta[/mathjaxinline]. Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]|z|[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e)</b> If the AC voltage source is expressed in terms of a [mathjaxinline]\sin ( )[/mathjaxinline] instead of a [mathjaxinline]\cos ( )[/mathjaxinline], that is, if [mathjaxinline]V(t) = V_0\sin (\omega t)[/mathjaxinline] and the current is [mathjaxinline]I(t) = I_0\sin (\omega t - \phi )[/mathjaxinline], will your answers in parts (c) and (d) change? </p>
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<p><b class="bfseries">Note:</b> In the video shown below, the subscript "c" for the complex functions [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline] is not used. </p>
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<h2 class="hd hd-2 unit-title">L31v5: Impedance for a Capacitor</h2>
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Impedance for a Capacitor
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<p>
An AC source is connected to a capacitor of capacitance [mathjaxinline]C[/mathjaxinline]. The voltage source is given by [mathjaxinline]V(t) = V_0\cos (\omega t)[/mathjaxinline] and the current in the circuit is [mathjaxinline]I(t) = I_0\cos (\omega t - \phi )[/mathjaxinline]. At the instant shown in the figure, the capacitor is charging and therefore the current [mathjaxinline]I[/mathjaxinline] and the charge in the positive plate, [mathjaxinline]q[/mathjaxinline], are related by: </p>
<p>
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<text> [mathjaxinline]I=\dfrac {dq}{dt}[/mathjaxinline]</text>
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<text> [mathjaxinline]I=-\dfrac {dq}{dt}[/mathjaxinline]</text>
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<p>
The voltage across the capacitor is [mathjaxinline]V(t)=\dfrac {q}{C}[/mathjaxinline], and the time derivative of this relationship 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 \dfrac {dV(t)}{dt} = \dfrac {1}{C}I(t)[/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>
where [mathjaxinline]V(t)[/mathjaxinline] and [mathjaxinline]I(t)[/mathjaxinline] are the real and physical voltage and current. The complexified voltage and current, [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline] satisfy: </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 \dfrac {dV_ c}{dt} = \dfrac {1}{C}I_ c[/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><b class="bfseries">Note:</b> You showed that equation (2) is true in lesson 30. </p>
<p><b class="bfseries">(Part a)</b> What is the Euler expression of [mathjaxinline]\dfrac {dV_ c}{dt}[/mathjaxinline], the time derivative of the complexified voltage. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. Note that the correct Euler format has a <i class="it">real</i> expression multiplied by [mathjaxinline]e[/mathjaxinline] to an imaginary power. </p>
<p>
<p style="display:inline">[mathjaxinline]\dfrac {dV_ c}{dt}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> What is the Euler expression of [mathjaxinline]I_ c[/mathjaxinline], the complexified current. Express your answer in terms of [mathjaxinline]t[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], I_0 for [mathjaxinline]I_0[/mathjaxinline], phi for [mathjaxinline]\phi[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]I_ c=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Use the results obtained in parts (a) and (b) in equation (2) to calculate [mathjaxinline]I_0[/mathjaxinline] and [mathjaxinline]\phi[/mathjaxinline]. Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], C, and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </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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<p><b class="bfseries">(Part d)</b> If the impedance is a complex number [mathjaxinline]z = |z|\; e^{i\delta }[/mathjaxinline] such that </p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle V_ c(t) = z\; I_ c(t)[/mathjaxinline]
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<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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<p>
Use the results obtained in parts (a) to (c) in the equation above to calculate the modulus [mathjaxinline]|z|[/mathjaxinline] and the argument [mathjaxinline]\delta[/mathjaxinline]. Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]|z|[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e)</b> If the AC voltage source is expressed in terms of a [mathjaxinline]\sin ( )[/mathjaxinline] instead of a [mathjaxinline]\cos ( )[/mathjaxinline], that is, if [mathjaxinline]V(t) = V_0\sin (\omega t)[/mathjaxinline] and the current is [mathjaxinline]I(t) = I_0\sin (\omega t - \phi )[/mathjaxinline], will your answers in parts (c) and (d) change? </p>
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<p><b class="bfseries">Note:</b> In the video shown below, the subscript "c" for the complex functions [mathjaxinline]V_ c[/mathjaxinline] and [mathjaxinline]I_ c[/mathjaxinline] is not used. </p>
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<h2 class="hd hd-2 unit-title">L31Q1: An Unknown Element</h2>
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An Unknown Element
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An AC source is connected to a single unknown circuit element as shown above. The element can be a resistor, a capacitor, or an inductor. The driving frequency [mathjaxinline]\omega =100[/mathjaxinline] rad/s. The voltage and the current are both plotted in the figure below. The scales are indicated on the figure, where the lines are separated by 100 mA for the current and 10 V for the voltage. Only a portion of the signals are shown over an unknown time interval. Note that the dashed signal is behind (i.e. lags) the solid one. </p>
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Which of the following statements could be true? </p>
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<text>The element is a resistor with [mathjaxinline]R =200 \Omega[/mathjaxinline], the solid line is [mathjaxinline]V(t)[/mathjaxinline] and the dashed line is [mathjaxinline]I(t)[/mathjaxinline]</text>
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<text>The element is a resistor with [mathjaxinline]R = 50 \Omega[/mathjaxinline], the solid line is [mathjaxinline]I(t)[/mathjaxinline] and the dashed line is [mathjaxinline]V(t)[/mathjaxinline]</text>
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<text>The element is an inductor with [mathjaxinline]L = 0.5[/mathjaxinline] H, the solid line is [mathjaxinline]V(t)[/mathjaxinline] and the dashed line is [mathjaxinline]I(t)[/mathjaxinline]</text>
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<text>The element is an inductor with [mathjaxinline]L = 2[/mathjaxinline] H, the solid line is [mathjaxinline]I(t)[/mathjaxinline] and the dashed line is [mathjaxinline]V(t)[/mathjaxinline]</text>
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<text>The element is an capacitor with [mathjaxinline]C=0.2[/mathjaxinline] mF, the solid line is [mathjaxinline]V(t)[/mathjaxinline] and the dashed line is [mathjaxinline]I(t)[/mathjaxinline]</text>
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<text>The element is an capacitor with [mathjaxinline]C=50 \mu[/mathjaxinline]F, the solid line is [mathjaxinline]I(t)[/mathjaxinline] and the dashed line is [mathjaxinline]V(t)[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L31Q2: Impedance for a Series RLC Circuit</h2>
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Impedance for a Series RLC Circuit
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An AC source is connected to a resistor of resistance [mathjaxinline]R[/mathjaxinline], an inductor of inductance [mathjaxinline]L[/mathjaxinline] , and a capacitor of capacitance [mathjaxinline]C[/mathjaxinline] as shown. The voltage source 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]. </p>
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Calculate the modulus [mathjaxinline]|z_{eq}|[/mathjaxinline], and the phase [mathjaxinline]\delta[/mathjaxinline] of the equivalent impedance [mathjaxinline]z_{eq}[/mathjaxinline]. Express your answer in terms of [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], [mathjaxinline]i[/mathjaxinline], arctan ( ) for [mathjaxinline]\tan ^{-1}()[/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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<p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L31v8: LR Circuit with Complex Numbers</h2>
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A Resistor and Inductor in Series-Part 1
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An AC source is connected to an inductor of inductance [mathjaxinline]L[/mathjaxinline] and a resistor of resistance [mathjaxinline]R[/mathjaxinline] as shown. The voltage source is given by [mathjaxinline]V(t) = V_0\cos (\omega t)[/mathjaxinline] and the current in the circuit is [mathjaxinline]I(t) = I_0\cos (\omega t - \phi )[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> What is [mathjaxinline]|z|[/mathjaxinline], the modulus of the equivalent impedance? Express your answer in terms of [mathjaxinline]i[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]|z|=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_checkpoint_w13_18_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part b)</b> What is [mathjaxinline]\delta[/mathjaxinline], the phase of the equivalent impedance? Express your answer in terms of arctan for [mathjaxinline]\tan ^{-1}( )[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_checkpoint_w13_18_3_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part c)</b> Calculate [mathjaxinline]I_0[/mathjaxinline], the amplitude of the current. Express your answer in terms of [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]I_0=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_checkpoint_w13_18_4_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part d)</b> Calculate [mathjaxinline]\phi[/mathjaxinline], the phase shift of the current with respect to the AC voltage source. Express your answer in terms of arctan for [mathjaxinline]\tan ^{-1}( )[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]\phi =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="inputtype_checkpoint_w13_18_5_1" class="text-input-dynamath capa_inputtype inline textline">
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<h3 class="hd hd-2">L31v8: LR Circuit with Complex Numbers</h3>
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<h2 class="hd hd-2 unit-title">L31Q3: LR Circuit</h2>
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A Resistor and Inductor in Series-Part 2
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An AC source is connected to an inductor of inductance [mathjaxinline]L[/mathjaxinline] and a resistor of resistance [mathjaxinline]R[/mathjaxinline] as shown. The voltage source 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]. </p>
<p><b class="bfseries">(Part a)</b> In the figure below [mathjaxinline]V(t)/V_0[/mathjaxinline] and [mathjaxinline]I(t)/I_0[/mathjaxinline] are plotted together as functions of time. Only a portion of the signals is shown for an unknown time interval. </p>
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Which of the following statements about the signals are true? </p>
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<text>The voltage is the dashed line and the current is the solid line</text>
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<text>The phase shift is in the range [mathjaxinline]0&lt;\phi &lt;\pi /2[/mathjaxinline]</text>
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<text>The phase shift is in the range [mathjaxinline]\pi /2&lt;\phi &lt;\pi[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> The amplitude of the voltage is [mathjaxinline]V_0 = 100[/mathjaxinline] V. If the real part of the equivalent impedance is [mathjaxinline]\text {Re}(z_{eq}) = 50 \; \Omega[/mathjaxinline] and the imaginary is [mathjaxinline]\text {Im}(z_{eq}) = 35 \; \Omega[/mathjaxinline], what is the magnitude of the current in the circuit at the instants when the voltage is zero. </p>
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<p style="display:inline">[mathjaxinline]|I|=[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_checkpoint_w13_18b_3_1">[mathjaxinline]\, \mathrm{A}[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">L31v9: RC Circuit with Complex Numbers</h2>
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A Resistor and Capacitor in Series-Part 1
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An AC source is connected to a capacitor of capacitance [mathjaxinline]C[/mathjaxinline] and a resistor of resistance [mathjaxinline]R[/mathjaxinline] as shown. The voltage source 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]. </p>
<p><b class="bfseries">(Part a)</b> What is [mathjaxinline]|z|[/mathjaxinline], the modulus of the equivalent impedance? Express your answer in terms of [mathjaxinline]i[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]|z|=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_checkpoint_w13_17_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part b)</b> What is [mathjaxinline]\delta[/mathjaxinline], the argument of the equivalent impedance? Express your answer in terms of arctan for [mathjaxinline]\tan ^{-1}( )[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_checkpoint_w13_17_3_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part c)</b> Calculate [mathjaxinline]I_0[/mathjaxinline], the amplitude of the current. Express your answer in terms of [mathjaxinline]R[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], V_0 for [mathjaxinline]V_0[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]I_0=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_checkpoint_w13_17_4_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part d)</b> Calculate [mathjaxinline]\phi[/mathjaxinline], the phase shift of the current with respect to the AC voltage source. Express your answer in terms of arctan for [mathjaxinline]\tan ^{-1}( )[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], and omega for [mathjaxinline]\omega[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="inputtype_checkpoint_w13_17_5_1" class="text-input-dynamath capa_inputtype inline textline">
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<h3 class="hd hd-2">L31v9: RC Circuit with Complex Numbers</h3>
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<h2 class="hd hd-2 unit-title">L31Q4: RC Circuit</h2>
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A Resistor and Capacitor in Series-Part 2
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An AC source is connected to a capacitor of capacitance [mathjaxinline]C[/mathjaxinline] and a resistor of resistance [mathjaxinline]R[/mathjaxinline] as shown. The voltage source 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]. </p>
<p><b class="bfseries">(Part a)</b> Assuming that [mathjaxinline]\omega = \dfrac {1}{RC}[/mathjaxinline], calculate the value of [mathjaxinline]I[/mathjaxinline] at the instant when the voltage is zero for the third time (counting [mathjaxinline]t=0[/mathjaxinline] as the first time). Express your answer in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline], and [mathjaxinline]C[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]I=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Which of the following statements are true? </p>
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<text>The current is ahead of the voltage for all [mathjaxinline]\omega[/mathjaxinline]</text>
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<text>The current is behind the voltage for all [mathjaxinline]\omega[/mathjaxinline]</text>
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<text>The current is ahead of the voltage only for [mathjaxinline]\omega = \dfrac {1}{RC}[/mathjaxinline]</text>
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<text>The current is behind the voltage only for [mathjaxinline]\omega = \dfrac {1}{RC}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L31Q5: Network Reduction</h2>
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Network Reduction
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What is the equivalent impedance for the circuit elements shown in the figure? </p>
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<text> [mathjaxinline]Z_{eq} = Z_1 + Z_2 + Z_3[/mathjaxinline]</text>
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<text> [mathjaxinline]Z_{eq} = Z_1 + \frac{1}{Z_2 + Z_3}[/mathjaxinline]</text>
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<text> [mathjaxinline]Z_{eq} = Z_1 + \frac{Z_2 Z_3}{Z_2 + Z_3}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L31v11: Simplify a Circuit with Impedance and Admittance</h2>
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<h2 class="hd hd-2 unit-title">L31Q6: Series/Parallel RLC Circuit</h2>
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Series/Parallel RLC Circuit
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The circuit shown in the left figure above is being driven at angular frequency [mathjaxinline]\omega[/mathjaxinline]. The combination of the capacitor in parallel with the inductor can be replaced by an equivalent impedance [mathjaxinline]z_{eq}[/mathjaxinline]. Assume that [mathjaxinline]\omega &gt;\dfrac {1}{\sqrt {LC}}[/mathjaxinline]. Calculate the modulus [mathjaxinline]|z_{eq}|[/mathjaxinline], and the phase [mathjaxinline]\delta[/mathjaxinline] of [mathjaxinline]z_{eq}[/mathjaxinline]. Express your answer in terms of [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], arctan ( ) for [mathjaxinline]\tan ^{-1}()[/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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<p style="display:inline">[mathjaxinline]\delta =[/mathjaxinline] </p>
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Note that once you have done this step, you will now be ready to add the impedance of this LC part with the resistor the same way we have done before when adding impedances. </p>
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