Circuit Analysis Using Complex Numbers

Section 41.7 Circuit Analysis Using Complex Numbers

The equations resulting from the application of Faraday’s loop rule to an AC circuit results in the summation of sine and cosine functions. These calculations using the trigonometric functions are usually messy and an alternative method is desirable. It turns out that the algebra of complex numbers is better at these calculations than the sine and cosine functions.

The connection between complex analysis and sines and cosines comes from Euler’s formula:

\begin{equation} e^{i\theta} = \cos\,\theta + i\,\sin\,\theta,\tag{41.41} \end{equation}

where \(i = \sqrt{-1}\text{.}\) Here, \(\cos\,\theta\) is the real part of \(e^{i\theta}\) and \(\sin\,\theta\) is the imaginary part of \(e^{i\theta}\text{.}\) Thus, in AC circuit calculations, we replace the cosine of the source voltage by a complex exponential. Then, we perform regular algebra. After completing the calculations, the real part of the final result corresponds to the answer we seek.

To distinguish complex voltage representing a real voltage, we will use the same symbol, but place a tilde (~) sign above the symbol, and similarly for currents. Thus, we will replace

\begin{equation*} V = V_0\,\cos(\omega t + \phi) \end{equation*}

\begin{equation*} \tilde V = V_0\, e^{ i ( \omega t + \phi ) }, \end{equation*}

whose real part is the original real voltage. As a result, the complex voltage \(\tilde{V}\) across an element and complex current \(\tilde{I}\) through it, are related by a complex version of Ohm’s law, we can call Complex Ohm’s Law.

\begin{equation} \tilde{V} = \tilde{Z} \tilde{I},\tag{41.42} \end{equation}

where the complex “resistance” denoted by \(\tilde{Z}\) is called the impedance or the complex impedance of the element. In the following we will work out expressions for complex impedances of resistor, inductor and capacitor. We will find that complex impedance of inductor and capacitor are imaginary with absolute values equal to their reactances.

The inverse of the impedance is called admittance and is denoted by \(\tilde{Y}\text{.}\)

\begin{equation*} \tilde{Y} =\frac{1}{\tilde{Z}}. \end{equation*}

We will see that the complex impedance allows one to treat resistors, capacitors and inductors in an AC circuit on an equal footing.

Subsection 41.7.1 Rectangular and Polar Forms of Complex Numbers

You may already know that a complex number has two parts and can be written using \(i=\sqrt{-1}\text{.}\) For instance, \(3+i4\) is a complex part whose real part is \(3\) and imaginary part \(4\text{.}\) This is called complex number written in rectangular or Cartesian form. You can imagine this as point \((3,4)\) in a complex plane. You can think of complex plane as Cartesina \(xy\)-plane.

Just as you can represent a point in \(xy\)-pane by \((x,y)\) coordinates or corresponding polar coordinates \((r,\theta)\text{,}\) you can do the same with a complex number. Thus, number \(3+i4\) will have polar representation \(re^{i\theta}\) with

\begin{equation*} r = \sqrt{3^2 + 4^2} = 5,\ \ \theta=\tan^{-1}\left( 4/3\right) = 51^\circ. \end{equation*}

We call \(r\) absolute value or amplitude of the complex number \(x+iy\) and \(\theta\) phase of the complex number. Suppose, we denote complex number \(x+iy\) by letter \(z\text{.}\) Then, we will often write its \(r\) as \(r = |z|.\) For instance, absolute value of complex impedance is equal to the amplitude of the impedance we have defined above.

\begin{equation*} |Z| = |\tilde{Z}|. \end{equation*}

Also, note that absolute value of a complex number with \(r=1\) will be 1. That is, if a complex number is purely exponential, i.e, if \(z=e^{i\theta}\text{,}\) then

\begin{equation*} |z| = \left| z=e^{i\theta} \right| = 1. \end{equation*}

Yoou can immediately verify this by using Euler’s formula given above and \(\cos^2\theta + \sin^2\theta = 1\text{.}\)

Subsection 41.7.2 Complex Impedance of a Resistor

Since current \(I(t)\) through a resistor and \(V_R(t)\text{,}\) the voltage drop across a resistor are in-phase, when we take their ratio to get impedance, the exponetials will cancel out, leaving just the real resistance \(R\text{.}\)

\begin{equation*} \tilde{Z}_R = \dfrac{\tilde{V}_R}{\tilde{I}} = R. \end{equation*}

Subsection 41.7.3 Complex Impedance of an Inductor

Since current \(V_L(t)\) across an inductor leads current \(I(t)\) through it by a phase of \(\pi/2\) radians, their ratio will give reactance times an exponential with a positive phase constant.

\begin{equation*} \tilde{Z}_L = \dfrac{\tilde{V}_L}{\tilde{I}} = \omega L e^{i\pi/2} = i \omega L, \end{equation*}

where I used \(e^{i \pi/2} = \cos\, \pi/2 + i\, \sin\, \pi/2 = 0 + i = i\text{.}\) Note that absolute value of this impedance is the expected reactance of the inductor.

\begin{equation*} |Z_L| = \omega L. \end{equation*}

The phase of complex \(Z_L\) is \(\frac{\pi}{2}\text{.}\)

Subsection 41.7.4 Complex Impedance of a Capacitor

Since current \(V_L(t)\) across a capacitor lags current \(I(t)\) through it by a phase of \(\pi/2\) radians, their ratio will give reactance times an exponential with a negative phase constant.

\begin{equation*} \tilde{Z}_C = \dfrac{\tilde{V}_C}{\tilde{I}} = \dfrac{1}{\omega C} e^{-i\pi/2} = i \omega L, \end{equation*}

where I used \(e^{i \pi/2} = \cos\, \pi/2 - i\, \sin\, \pi/2 = 0 - i = - i\text{.}\) Note that absolute value of this impedance is the expected reactance of the capacitor.

\begin{equation*} |Z_C| = \dfrac{1}{\omega C}. \end{equation*}

The phase of complex \(Z_C\) is \(-\frac{\pi}{2}\text{.}\)

Subsection 41.7.5 AC Circuit Problems

The first step in the analysis of an AC circuit is to replace circuit elements by their impedances. Each resistor \(R\) is replaced by its \(Z_R = R\text{,}\) each inductance \(L\) is replaced by its impedance \(Z_L = i\omega L\text{,}\) and each capacitance is replaced by its impedance \(Z_C =1/ i\omega C\text{.}\) Then, we subject the circuit by Kirchoff’s current law for complexe currents and Faraday’s loop rule for each loop in the circuit. We then, solve for complex currents and deduce the physical currents from the real parts of the complex currents. These steps are best illustrated by examples.

Example 41.42. Series RLC Circuit in Complex Notation.

Consider the series RLC-circuit. Find current in the circuit.

Figure 41.43. A series RLC-circuit driven by an alternating EMF.

Answer.

\(I_0 = \dfrac{V_0}{ \sqrt{ R^2 + \left( \omega L- \dfrac{1}{\omega C}\right)^2 } }\text{,}\) \(\phi_I = - \tan^{-1} \left( \dfrac{\omega L- \dfrac{1}{\omega C}}{R} \right)\text{.}\)

Solution.

First step in our solution is to replace original circuit by a complex circuit where we replace \(R\) by \(Z_R\text{,}\) \(C\) by \(Z_C\text{,}\) and \(L\) by \(Z_L\text{,}\) and treat currents and voltages complex. This is shown as a complex circuit in the figure below.

Figure 41.44. A series RLC-circuit driven by an alternating EMF with complex quantities.

Now, we think of impedances as if they were simple resistors. In this circuit, the loop equation will give

\begin{equation*} \tilde V - \tilde Z_R\tilde I - \tilde Z_L\tilde I - \tilde Z_C\tilde I = 0, \end{equation*}

which can be solved for complex current.

\begin{equation} \tilde I = \dfrac{\tilde V}{ \tilde Z_R + \tilde Z_L + \tilde Z_C}.\tag{41.43} \end{equation}

From this we note that complex impedance of the circuit is

\begin{align*} \tilde Z \amp = \tilde Z_R + \tilde Z_L + \tilde Z_C \\ \amp = R + i \omega L+ \dfrac{-i}{\omega C} \\ \amp = R + i \left( \omega L- \dfrac{1}{\omega C}\right) \end{align*}

This is rectangular form of complex impedance, which we can also write in polar form. Let

\begin{equation*} R + i \left( \omega L- \dfrac{1}{\omega C}\right) = |Z|\,e^{i\phi_Z}. \end{equation*}

Then,

\begin{gather*} |Z| = \sqrt{ R^2 + \left( \omega L- \dfrac{1}{\omega C}\right)^2 } \\ \phi_Z = \tan^{-1} \left( \dfrac{\omega L- \dfrac{1}{\omega C}}{R} \right) \end{gather*}

Using polar form of complex impedance in Eq. (41.43) we get polar form of complex current.

\begin{equation*} \tilde I = \dfrac{\tilde V}{ \tilde Z } = \dfrac{V_0}{|Z|}\, e^{i\left(\omega t-\phi_Z\right)}, \end{equation*}

whose real part gives the physical current in the circuit.

\begin{equation*} I(t) = \dfrac{V_0}{|Z|}\,\cos(\omega t - \phi_Z ). \end{equation*}

Wrting this as \(I(t) = I_0\,\cos(\omega t + \phi_I)\) we have

\begin{gather*} I_0 = \dfrac{V_0}{ \sqrt{ R^2 + \left( \omega L- \dfrac{1}{\omega C}\right)^2 } },\\ \phi_I = - \tan^{-1} \left( \dfrac{\omega L- \dfrac{1}{\omega C}}{R} \right). \end{gather*}

Example 41.45. Parallel RL Circuit in Complex Notation.

Consider the circuit given in the Fig. \ref{fig:example-ac-RL-1} where a resistor \(R\) and an inductor \(L\) are connected in parallel to an AC power generator with voltage \(V(t) = V_0\cos(\omega t)\text{.}\) Find (a) current in each branch of the circuit, (b) average power dissipated in the resistor, and (c) average power produced by the AC generator.

Figure 41.46. A parallel RL-circuit driven by an alternating EMF.

Answer.

(a) \(I_1(t) = \dfrac{V_0}{R}\,\cos(\omega t)\text{,}\) \(I_2(t) = \dfrac{V_0}{\omega L}\,\sin(\omega t)\text{,}\) \(I = I_1 + I_2\text{,}\) (b) and (c) \(\dfrac{1}{2}\,\dfrac{V_0^2}{R}\)

Solution 1. (a)

(a) First step in our solution is to replace original circuit by a complex circuit where we replace \(R\) by \(Z_R\text{,}\) and \(L\) by \(Z_L\text{,}\) and treat currents and voltages complex. This is shown as a complex circuit in the figure below.

Figure 41.47. A parallel RL-circuit driven by an alternating EMF with complex quantities.

From loop with source and \(\tilde Z_R\) we get

\begin{equation*} \tilde I_1 = \dfrac{\tilde V}{\tilde Z_R} = \dfrac{V_0}{R}\, e^{i\omega t}, \end{equation*}

whose real part gives the physical current through \(R\text{.}\)

\begin{equation*} I_1(t) = \dfrac{V_0}{R}\,\cos(\omega t). \end{equation*}

From loop with source and \(\tilde Z_L\) we get

\begin{equation*} \tilde I_2 = \dfrac{\tilde V}{\tilde Z_L} = \dfrac{V_0}{i\omega L}\, e^{i\omega t}, \end{equation*}

whose real part gives the physical current through \(L\text{.}\) To obtain the real part of this complex quantity we note that

\begin{equation*} \dfrac{1}{i} = -i = e^{-i\pi/2}, \end{equation*}

which can be combined with \(e^{i\omega t}\) to give

\begin{equation*} \tilde I_2 = \dfrac{V_0}{\omega L}\, e^{i(\omega t - \pi/2)}, \end{equation*}

whose real part is

\begin{equation*} I_2(t) = \dfrac{V_0}{\omega L}\,\cos(\omega t -\pi/2), \end{equation*}

which we can also write as

\begin{equation*} I_2(t) = \dfrac{V_0}{\omega L}\,\sin(\omega t), \end{equation*}

Conservation of current at nodes gives the overall current.

\begin{align} I(t) \amp = I_1(t) + I_2(t) \notag\\ \amp = \dfrac{V_0}{R}\,\cos(\omega t) + \dfrac{V_0}{\omega L}\,\sin(\omega t). \tag{41.44} \end{align}

We can write this answer as \(I_0\,\cos(\omega t + \phi_I)\) with the following answer.

\begin{gather*} I_0 = V_0\,\sqrt{ \left( \dfrac{1}{R} \right)^2 + \left( \dfrac{1}{\omega L} \right)^2 },\\ \phi_I = -\tan^{-1}\left( \dfrac{R}{\omega L}\right). \end{gather*}

To see these expressions, you can either expand \(I_0\,\cos(\omega t + \phi_I)\text{,}\) equate the result to the answer in Eq. (41.44), separately equate coefficients of \(\cos(\omega t)\) and \(\sin(\omega t)\text{,}\) and then solve for the phase constant and amplitude.

Solution 2. (b)

(b) The instantaneous power dissipated in the resistor will be

\begin{equation*} P_R(t) = I_1(t) V_R(t) = R I_1(t)^2 = R \left( \dfrac{V_0}{R}\right)^2\,\cos^2(\omega t). \end{equation*}

Average of \(\cos^2(\omega t)\) is \(\dfrac{1}{2}\text{.}\) Therefore, average power dissipates is

\begin{align*} P_{R,\text{av}} \amp = \dfrac{1}{2}\,\dfrac{V_0^2}{R}. \end{align*}

Solution 3. (c)

\begin{equation*} P(t) = I(t) V(t) = I_0 V_0 \cos(\omega t) \cos(\omega t + \phi_I) \end{equation*}

Average of \(\cos(\omega t) \cos(\omega t + \phi_I)\) is \(\dfrac{1}{2}\cos\,\phi_I\text{.}\) Therefore, average power dissipates is

\begin{align*} P_{\text{av}} \amp = \dfrac{1}{2}\,I_0 V_0\cos\,\phi_I, \\ \amp = \dfrac{1}{2}\, \left[V_0\sqrt{\left( \dfrac{1}{R} \right)^2 + \left( \dfrac{1}{\omega L} \right)^2 }\right]\, V_0\, \left[\dfrac{\omega L}{\sqrt{R^2 + \omega^2 L^2}}\right]. \\ \amp = \dfrac{1}{2}\,\dfrac{V_0^2}{R}. \end{align*}

This is the same as the energy dissipated in the resistor, the only place where energy could have gone.

Example 41.48. Circuit Elements in Series and in Parallel.

(a) Find current through the capacitor in the circuit. (b) Use the following values to find the numerical values of the magnitude and phase of the current through the capacior.

\begin{align*} \amp V_0=170\text{ V},\ \ \omega = 2\pi\times 60\text{ sec}^{-1}, \\ \amp R = 1000\ \Omega;\ \ L = 1\; \text{mH};\ \ C = 1\; \mu\text{F}. \end{align*}

Answer.

(a) See solution, (b) \(64\; \text{mA}\text{,}\) \(\pi/2\)

Solution 1. (a)

(a) We start by first replacing the passive elements in the circuit by their impedances.

Since we just want the current through \(C\text{,}\) which is the same as current through the source, we will work out equivalent impedance of everything connected across the source. We see that \(\tilde{Z}_C\) is in series with parallel combinations of \(\tilde{Z}_R\) and \(\tilde{Z}_L\text{,}\) the equivalent impedance is

\begin{align*} \tilde{Z} \amp = \tilde{Z}_C + \left( \dfrac{1}{\tilde{Z}_R} + \dfrac{1}{\tilde{Z}_L} \right)^{-1},\\ \amp = \dfrac{-i}{\omega C} + \left( \dfrac{1}{R} + \dfrac{1}{i\omega L} \right)^{-1},\\ \amp = \dfrac{-i}{\omega C} + \dfrac{i\omega R L}{ R + i\omega L } ,\\ \amp = \dfrac{-i}{\omega C} + \dfrac{i\omega R L}{ R + i\omega L }\times \dfrac{R-i\omega L}{R-i\omega L},\\ \amp = \dfrac{\omega^2 R L^2}{ R^2 + \omega^2 L^2 } + i \left(\dfrac{\omega R^2 L}{ R^2 + \omega^2 L^2 } - \dfrac{1}{\omega C}\right). \end{align*}

This gives real and imaginary parts of the equivalent complex impedance.

\begin{align*} \amp \text{Re}(Z) = \dfrac{\omega^2 R L^2}{ R^2 + \omega^2 L^2 } \equiv a,\\ \amp \text{Im}(Z) = \dfrac{\omega R^2 L}{ R^2 + \omega^2 L^2 } - \dfrac{1}{\omega C} \equiv b. \end{align*}

Therefore, we have \(Z = |Z|e^{i\phi_Z}\) with

\begin{align*} \amp |Z| = \sqrt{a^2 + b^2},\\ \amp \phi_Z = \tan^{-1}\left( \dfrac{b}{a}\right). \end{align*}

Using the magnitude and phase of equivalent impedance, we can immediately write the magnitude and phase of current through the course, \(I = I_0\cos(\omega t + \phi_I)\text{,}\) with

\begin{align*} \amp I_0 = V_0/|Z|,\\ \amp \phi_I = -\phi_Z. \end{align*}

Solution 2. (b)

(b) The value of impedances of the elements are:

\begin{equation*} Z_R = 1000\; \Omega;\ \ Z_L = i 0. 377\; \Omega; \ \ Z_C =-i 2652\; \Omega. \end{equation*}

The equivalent impedance becomes

\begin{equation*} \tilde{Z} \approx 1.4\times 10^{-4} \Omega - i 2652 \Omega. \end{equation*}

The amplitudes and phase of the current would be,

\begin{equation*} I_0 \approx 64\; \text{mA}. \end{equation*}

\begin{equation*} \phi_I \approx \frac{\pi}{2}\; \text{rad}. \end{equation*}

Exercises 41.7.6 Exercises

1. Complex Circuit Analysis of a Pure Inductor Circuit.

In the circuit shown in Figure 41.49, find: (a) the current in the circuit, (b) the phase angle between the current in the circuit and the EMF, and (c) average power. Assume \(R = 0\) in the circuit.

Answer.

(a) \((225\, \text{mA}) \cos(\omega t - \pi/2)\text{.}\)

Solution 1. a

Here, the complex impedance is \(Z = i \omega L\text{.}\) We write the source voltage as \(V= V_0\: e^{i\omega t}\text{,}\) where \(V_0 = 170\) V and \(\omega = 120\pi\,\text{sec}^{-1}\text{.}\) The current is obtained from the real part of \(I = V/Z\text{.}\)

\begin{equation*} (1)\ \ I = Re\left[\dfrac{V_0\: e^{i\omega t}}{i \omega L} \right] = \dfrac{V_0}{\omega L}\:\cos\left(\omega t - \dfrac{\pi}{2}\right) = I_0 \:\cos\left(\omega t - \dfrac{\pi}{2}\right). \end{equation*}

Putting in the numbers we find \(I_0 = \dfrac{V_0}{\omega L} = 225\) mA.

Solution 2. b

The phase angle is \(\pi/2\) with the EMF leading the current as in (1).

Solution 3. c

Since there are no dissipative elements in the circuit average power will be zero.

2. Complex Circuit Analysis of a Pure Capacitor Circuit.

In the circuit shown in Figure 41.50, find the following. (a) the current in the circuit, (b) the phase angle between the current in the circuit and the EMF, and (c) average power. Assume \(R = 0\) in the circuit.

Solution 1. a

Here, the complex impedance is \(Z = \dfrac{1}{i \omega C}\text{.}\) We write the source voltage as \(V= V_0\: e^{i\omega t}\text{,}\) where \(V_0 = 170\, \text{V}\) and \(\omega = 120\pi\, \text{sec}^{-1}\text{.}\) The current is obtained from the real part of \(I = V/Z\text{.}\)

\begin{equation*} (1)\ \ I = Re\left[ V_0\: e^{i\omega t} i \omega C \right] = V_0 \omega C\:\cos\left(\omega t + \dfrac{\pi}{2}\right) = I_0 \:\cos\left(\omega t + \dfrac{\pi}{2}\right). \end{equation*}

Putting in the numbers we find \(I_0 = V_0\omega L = 128\) mA.

Solution 2. b

The phase angle is \(\pi/2\) with the EMF lagging the current as in (1).

Solution 3. c

Since there are no dissipative elements in the circuit average power will be zero.

3. Complex Circuit Analysis of an RC Circuit.

A \(200\text{-}\Omega\) resistor and \(150\text{-}\mu\)F capacitor are connected in series across a \(120\text{-V}\) (rms ) AC power source oscillating at \(60\text{-Hz}\) frequency.

Find the amplitude and phase of current in the circuit, assuming phase of source voltage to be zero at \(t = 0\text{.}\)
Find the voltage drops across the resistor and capacitor.
Find the impedance of the circuit.
Find the power dissipated in the resistor.
Find the power dissipated in the capacitor.
Find the power produced by the source.

Hint.

Note the peak voltage \(V_0 = \sqrt{2} V_{\textrm{rms}} = 169\ \textrm{V}\text{.}\) Use complex method with \(\tilde Z = \tilde{Z}_R + \tilde{Z}_C\text{.}\)

Solution 1. a

The source voltage is given to be \(V(t) = V_0\:\cos(\omega t)\text{,}\) with \(V_0 = 120\sqrt{2} = 169.7\:\textrm{V}\) and \(\omega = 2\pi\times 60 = 120\pi\:\textrm{sec}^{-1}\text{.}\) In complex notation this is replaced by \(V_0\:e^{i\omega t}\text{.}\) Now, we divide this by complex impedance to find the complex current. The complex impedance here is

\begin{equation*} Z = R -\dfrac{i}{\omega C} = |Z|\: e^{i\phi}, \end{equation*}

where

\begin{align*} \amp |Z| = \left( R^2 + \dfrac{1}{\omega^2 C^2}\right)^{1/2} = 200.8\:\Omega,\\ \amp \phi = -\tan^{-1}\left( \dfrac{1}{\omega R C}\right)= -0.088\:\textrm{rad}. \end{align*}

Therefore, the complex current is

\begin{equation*} I = \dfrac{V}{Z} = \dfrac{V_0}{|Z|}\: e^{i(\omega t -\phi)} = (0.845\:\textrm{A})\:e^{i(\omega t +0.088)}. \end{equation*}

The real part of this complex current gives the current in the circuit.

\begin{equation*} I = (0.845\:\textrm{A})\:\cos(\omega t +0.088). \end{equation*}

Therefore, the amplitude of current is \(0.845\:\textrm{A}\) and the phase constant 0.088 rad or \(5.1^{\circ}\) ahead of the source voltage.

Solution 2. b

Now, we can use the current to find the voltage drop across the resistor.

\begin{equation*} V_R = (0.845\:\textrm{A}\times 200\:\Omega)\:\cos(\omega t + 0.088) = (169\:\textrm{V})\:\cos(\omega t + 0.088). \end{equation*}

Solution 3. c

\(Z\) already found above.

Solution 4. d

\begin{equation*} P_R = \dfrac{1}{2}I_0^2 R = \dfrac{1}{2}\times (0.845\:\textrm{A})^2\times 200\:\Omega = 71.4\:\textrm{W} \end{equation*}

Solution 5. e

No power is dissipated in the capacitor since current through the capacitor and the voltage across the capacitor are \(\pi/2\) rad out phase.

Solution 6. f

This will equal the power dissipated in the resistor: \(71.4\, \text{W}\text{.}\)

4. Complex Circuit Analysis of an RL Circuit.

A \(1.5\text{-k}\Omega\) resistor and \(30\text{-mH}\) inductor are connected in series across a \(120\text{-V}\) (rms ) AC power source oscillating at \(60\text{-Hz}\) frequency.

Find the current in the circuit.
Find the voltage drops across the resistor and inductor.
Find the impedance of the circuit.
Find the power dissipated in the resistor.
Find the power dissipated in the inductor.
Find the power produced by the source.

Answer.

(a) \((0.113\, \text{A})\, \cos(2\pi\times 60 t - 0.0075)\) when \(V(t) = (120\sqrt{2}\, \text{V}) \cos(2\pi\times 60 t)\text{.}\)

Solution 1. a

\begin{equation*} Z = R + i\omega L = |Z|\: e^{i\phi},\quad\quad (1) \end{equation*}

where

\begin{align*} \amp |Z| = \left( R^2 + \omega^2 L^2 \right)^{1/2} = 1500\:\Omega,\\ \amp \phi = \tan^{-1}\left( \dfrac{\omega L}{ R }\right)= 0.0075\:\textrm{rad} \end{align*}

Therefore, the complex current is

\begin{equation*} I = \dfrac{V}{Z} = \dfrac{V_0}{|Z|}\: e^{i(\omega t -\phi)} = (0.113\:\textrm{A})\:e^{i(\omega t -0.0075)}. \end{equation*}

The real part of this complex current gives the current in the circuit.

\begin{equation*} I = (0.113\:\textrm{A})\:\cos(\omega t -0.0075). \end{equation*}

Therefore, the amplitude of current is \(0.113\:\textrm{A}\) and the phase constant \(0.0.0075\, \text{rad}\) or \(0.4^{\circ}\) behind the phase of the source voltage.

Solution 2. b

Now, we can use the current to find the voltage drop across the resistor.

\begin{equation*} V_R = (0.113\:\textrm{A}\times 1500\:\Omega)\:\cos(\omega t -0.0075) = (169\:\textrm{V})\:\cos(\omega t -0.0075). \end{equation*}

The phase of the voltage across the inductor will be ahead of the phase of the current by \(\pi/2\text{.}\) The amplitude of the voltage across the inductor will be equal to the product of the amplitude of the current and the reactance of the inductor, which is \(X_L = \omega L = 11.3\:\Omega\text{.}\)

\begin{equation*} V_L = (0.113\:\textrm{A}\times 11.3\:\Omega)\:\cos(\omega t -0.0075+\frac{\pi}{2}) = (1.28\:\textrm{V})\:\cos(\omega t -0.0075+\frac{\pi}{2}). \end{equation*}

Solution 3. c

See (1) in part (a).

Solution 4. d

\begin{equation*} P_R = \dfrac{1}{2}I_0^2 R = \dfrac{1}{2}\times (0.113\:\textrm{A})^2\times 1500\:\Omega = 9.6\:\textrm{W}. \end{equation*}

Solution 5. e

No power is dissipated in the inductor since current through the inductor and the voltage across the inductor are \(\pi/2\) rad out phase.

Solution 6. f

This will equal the power dissipated in the resistor: \(9.6\, \text{W}\text{.}\)

5. Complex Circuit Analysis of an RL Circuit 1.

In the circuit shown in the figure, find the followig (a) the current in the circuit, (b) phase angle between the current in the circuit and the voltage drop across the resistor,

(c) phase angle between the current in the circuit and the voltage drop across the inductor, (d) phase angle between the current in the circuit and the voltage drop across the capacitor, (e) the power factor, (f) real impedance of the circuit, and (g) average power dissipated in the circuit.

Answer.

(a) \((3.2 \times 10^{-7}\,\text{A})\, \cos(120\pi t + \pi/2)\text{.}\)

Solution 1. a

To find the current in the circuit, let us first evaluate the complex impedance of the circuit.

\begin{align*} \amp Z_R = 300\:\Omega,\ \ Z_L = i\omega L = i\:0.754\:\Omega,\\ \amp Z_C = -i/\omega C = -i\:5.3 \times 10^8\:\Omega, \end{align*}

Summing them we get the imegance of the circuit since they are in series.

\begin{align*} Z \amp = Z_R + Z_L + Z_C = \left( 300 -i\:5.3 \times 10^8\right)\:\Omega \\ \amp = \left( 5.3 \times 10^8\:\Omega \right)\angle -90^{\circ}. \end{align*}

Therefore, the current written as \(I_0\cos(\omega t +\phi)\) has

\begin{equation*} I_0 = 318.6\:\textrm{nA},\ \ \phi = \dfrac{\pi}{2}. \end{equation*}

Solution 2. b

Since the impedance of the resistor is real, the voltage across it and the current through it would be in-phase with respect to each other.

Solution 3. c

The \(Z_L\) has a phase constant of \(\pi/2\text{.}\) Therefore, current will be behind the voltage across the inductor in phase by an amount of \(\pi/2\) rad.

Solution 4. d

The \(Z_L\) has a phase constant of \(-\pi/2\text{.}\) Therefore, current will be ahead of the voltage across the capacitor in phase by an amount of \(\pi/2\) rad.

Solution 5. e

Since the phase constant of the impedance is \(\phi \approx \pi/2\text{,}\) the power factor will be \(\cos\phi \approx 0\text{.}\)

Solution 6. f

The real impedance of the circuit refers to the amplitude of the complex impedance, \(|Z|\text{,}\) which is \(5.3 \times 10^8\:\Omega\text{.}\)

Solution 7. g

Average power dissipated is

\begin{equation*} \dfrac{1}{2}\:I_0^2 R = \dfrac{1}{2}\:(318.6\:\textrm{nA})^2 \times 300\:\Omega = 15\:\textrm{pJ}. \end{equation*}

6. Complex Circuit Analysis of an RL Circuit 2.

Determine current in each branch, power dissipated in the resistor, and power produced by the source for the circuit given in Figure 41.52.

Solution.

As the hint suggests, we replace the resistance, the capacitance, and the inductance by their complex impedances and treat the circuit as a ``purely resistive’’ circuit. The net impedance of the resistor and capacitor in parallel is

\begin{equation*} Z_{\parallel} = \dfrac{-iR/\omega C}{R -i/\omega C} = \dfrac{-iR }{\omega RC -i}= |Z_{\parallel}|e^{i\phi_{\parallel}}, \end{equation*}

with

\begin{equation*} |Z_{\parallel}| = \dfrac{R}{\sqrt{1+\omega^2R^2C^2}},\ \ \phi_{\parallel} = -\tan^{-1} (\omega R C). \end{equation*}

We need to add the impedance of the inductor to obtain the impedance across the source.

\begin{align*} Z \amp = Z_L + Z_{\parallel} = i\omega L + \dfrac{-iR }{\omega RC -i} \\ \amp = (63.6 + i 1224)\:\Omega = 1226\:\Omega\angle 1.51\:\textrm{rad}. \end{align*}

Therefore, the current \(I\) written as \(I_0\:\cos(\omega t + \phi)\) will have

\begin{equation*} I_0 = \dfrac{170\:\textrm{V}}{1226\:\Omega} = 0.139\:\textrm{A},\ \ \phi = -1.51\:\textrm{rad}. \end{equation*}

To find the currents \(I_1\) and \(I_2\) we can use the current divider formula. For the complex currents we get

\begin{equation*} I_1 = \dfrac{Z_C}{Z_R + Z_C} I,\ \ I_1 = \dfrac{Z_R}{Z_R + Z_C} I. \end{equation*}

The calculation for the amplitude and phase of \(I_1\) will go as follows.

\begin{equation*} I_1 = \dfrac{-i/\omega C}{R -i/\omega C}\times I_0\angle \phi\:\textrm{rad}. \end{equation*}

Now, we can rewrite the multiplier of the current on the right side as an amplitude and an angle

\begin{equation*} \dfrac{-i/\omega C}{R -i/\omega C} = A\:\angle \theta. \end{equation*}

Then, the amplitude and the phase of the current \(I_1\) will be

\begin{equation*} I_1 = A\times I_0\:\angle (\phi+\theta). \end{equation*}

Other current can be determined similarly. From the currents and the corresponding impedances we can deduce the EMF across each element. A student can use these to calculate the power produced by the source and the power dissipated in the resistor.

7. Complex Circuit Analysis of an RL Circuit 3.

Determine current in each branch, power dissipated in the resistors, and power produced by the source in Figure 41.53.

Hint.

This is a bridge circuit. Replace the circuit elements by their complex impedances.

Prev Top Next