In this page, we take up the study of a simple ac circuit. As shown in Fig. 42, this circuit consists of a voltage source, feeding power to a resistor and an inductor connected in series.
The aim is to analyze the circuit in Fig. 42. Given the source voltage, the values of resistance and reactance, we determine the following:
It can be seen that the analysis can be complicated, even though the circuit is simple.
The voltage source in the circuit in Fig. 42 is a sinusoidal source, defined by equation (227). It can be expressed as a phasor as shown by equation (228). The phasor contains information about the amplitude and the phase of the time domain signal at t = 0. The impedance seen by the source is presented by equation (229). The impedance seen by the source is a complex value, its real part being the resistance and the imaginary part the reactance of the inductor. When two impedances are in series, they can be considered as complex numbers and added accordingly. The impedance of the resistor is a real quantity expressed by the resistance. The impedance of the inductor is purely imaginary, equal to j times the reactance of the inductor. The reactance of a inductor is a positive value. Hence the impedance seen by the source is expressed by equation (229).
Using the phasor expression for the source voltage and the impedance Z of the circuit, the phasor expression for current is shown by equation (230).
Phasor expression for current is obtained as the ratio of phasor value of voltage over impedance, expressed as a complex value. Equation (231) expresses the impedance in the polar form.
Equation (232) shows the expression for current. Express both the voltage and the impedance as complex values in polar form and the current phasor is expressed by equation (232). Current lags the source voltage by an angle, equal to a . Equation (233) presents the steady-state solution for current through the RL circuit. The chapter on Time response provides an explanation for what steady-state solution is. This response can also be called as the forced response. The chapter on Time response provides an explanation for what the forced response is. Equation (233) presents the time-domain expression for current, whereas equation (234) presents the current phasor in its trigonometric form.
We can express the impedance of the circuit, as seen by the source in polar form. Equation (231) expresses the impedance in polar form. The impedance diagram is shown in Fig. 43. The magnitude is the hypotenuse and the angle is designated as a, where a is a positive value. Equation (231) specifies the magnitude of impedance and its angle.
We can obtain the phasor for current in polar form using the expressions for voltage and impedance in trigonometric form. Equation (i) expresses both voltage and the impedance as complex values in the polar form. We can express them in the trigonometric form, as shown by equation (ii). Multiply both the numerator and the denominator, by the conjugate of the denominator. Simplify both the numerator and the denominator, and get equation (iii). The trigonometric identities that are used for getting equation (iv) from equation (iii) are shown next. Equation (v) expresses the current in polar form. It is seen that we obtain the same result, as the one we got earlier. But this technique is slightly more difficult.
The circuit and the phasor diagram are shown in Fig. 44. The phasor sum of the resistor voltage and the inductor voltage equals the phasor value of the source voltage, as seen from the circuit. The phasor diagram shows the relative phase difference of the resistor voltage and the inductor voltage with respect to the source voltage. As shown by the phasor diagram, the resistor voltage lags the source voltage. The resistor voltage and the current in the circuit are in phase. We have seen earlier that the current in the circuit lags the source voltage by an angle, equal to a. Since the resistor and the inductor are in series, the current through them is the same. We know that in an inductor, the current lags the voltage by 90o. Since the resistor voltage and its current are in phase, the resistor voltage leads the inductor voltage by 90o. Hence in this case, the parallelogram used for phasor diagram turns out to be a rectangle. For this rectangle, the resistor voltage and the inductor voltage are the two adjacent sides, and the source voltage is its diagonal. The source voltage has a phase of fo with respect to the reference phasor, which is represented by the horizontal axis, to the right of origin. Since the phase difference the resistor voltage and the inductor voltage is 90o, the summing point lies on the circumference of semi-circle, drawn with the source voltage as the diameter. This aspect is shown in Fig. 44 by the diagram in the middle.
Equation (235) states that the source voltage is equal to the phasor sum of resistor and inductor voltages. Equation (236) states that the resistor voltage and its current are in phase. We can state that the phasor relationship between inductor voltage and the resistor voltage are as shown below.
The inductor voltage is the product of its current and its impedance. Since multiplying by j amounts to rotating the phasor by 90o in the anti-clockwise direction, the inductor voltage leads its current by 90o. We get equation (238) from equations (233) and (238).
The aim of this section is to find out expressions for the active power, the reactive power and the apparent power associated with the load, consisting of the resistor and the inductor. For the circuit in Fig. 45, we have obtained equation (239) already.
First let us obtain an expression for the power consumed by the the load. By definition, the word, power, refers to the average power. The instantaneous power associated with the load is expressed by equation (240).
The top line shows p(t) as the product of v(t) and i(t). Using the well-know trigonometric identities, we get the second, the third, the fourth and the fifth line of equation (240). We can deduce from the the last line that the instantaneous power contains two components. The first is a constant or a fixed value, reflecting the average power consumed by the load. The second is an oscillating component, oscillating at twice the source frequency. We can derive an expression for the average power as follows.
The average power, or simply the power, is the energy consumed per second and hence we get equation (241). The energy consumed per cycle can be determined and multiplying that energy by the frequency of the source yields the average power. We know that the cycle period is the reciprocal of the frequency. The energy consumed per cycle can be obtained by integrating the expression for the instantaneous power over a cycle period. We can use expression in the fourth line of equation (240) for the instantaneous power. The integral of the oscillating component over a cycle is zero and hence we get equation (242). The peak values can be replaced by the corresponding root mean square values, as shown by equation (242). In equation (242), cos(a) is known as the power factor. We call a as the power factor angle. If is a positive, the load has leading power factor. If is a negative, the load has lagging power factor. A load with net inductive reactance has lagging power factor, and a load with net capacitive reactance has leading power factor.
From the last line of equation (240), we can make the following observations.
Equation (243) expresses the average power of the load. The expressions for the minimum instantaneous power and the maximum instantaneous power are expressed by equation (244). They are obtained using the lowest and the highest value of the oscillating component, defined by the last line of equation (240). Equation (245) defines what the power factor is. If the power factor angle is 0o, we get equation (246). When the load is purely resistive, the power factor angle is 0o. In this case the minimum instantaneous power is zero, and the maximum instantaneous power is double the average power. The peak value of the instantaneous power equals the average power, and the instantaneous power is either zero or positive. It is never ever negative, given that the load is resistive.
If the power factor angle is plus or minus 90o, we get equation (247). When the load is purely reactive, the power factor angle is plus or minus 90o. In this case the average power is zero. The minimum instantaneous power is equal to the negative of the product of the root mean square values of voltage and current, and the maximum instantaneous power is equal to the product of the root mean square values of voltage and current. Whatever be the power factor, the magnitude of the oscillating component is equal to the product of the root mean square values of voltage and current.
The apparent power associated with the load is called Sckt. It is defined to be equal to the product of the root mean square values of voltage and current, as shown by equation (248). The unit for apparent power is VAR, representing reactive volt-amperes. The apparent power can also be expressed as a function of active power P and reactive power Q, as shown by equation (249). Since the value of Sckt and P are known, an expression for reactive power Q can be obtained as shown by equation (250). The only problem here is that we cannot assign the correct polarity to reactive power. Hence this problem is approached differently.
The complex power associated with the load can be defined as shown by equation (251). The real part of complex power is the active power P, and the imaginary part is the reactive power Q. It is shown as equal to the product of two phasors, which are defined as follows.
Both the phasors, Veff and Ieff, are defined by their root mean square values and the phase, as shown by equations (252), (253) and (254). When Ieff is known, its conjugate can be defined, as shown by equation (255).
The complex power associated with the load can be defined as shown by equation (251). Substitute the phasors by the corresponding expressions, shown in equations (252) and (255). Then we obtain equation (256). It can be simplified and the result is expressed by equation (257). We can express the complex power in trigonometric form, as shown by equation (258). The real part is the active power P, and the imaginary part is the reactive power Q. We also find that the polarity of reactive power is positive. It means that the inductor acts as a consumer of reactive power and hence it is positive. It can also be seen why the complex power is defined as the product of the voltage phasor and the conjugate of the current phasor, as shown by equation (251). This definition leads to the correct evaluation of both the active power and the reactive power.
The power diagram in Fig. 45 shows the complex power. Expression for the complex power is presented by equation (258). Complex power is the phasor sum of active power and reactive power, as shown by equation (259). From equations (258) and (259), we get expressions for active power and the reactive power of the circuit, which are expressed by equations (260), and (261). The complex power of the source is expressed by equation (262) and equation (263) states the source power in trigonometric form. Because current flows out of the source, there is a negative sign, which is in keeping with the passive sign convention. The active power of the source is negative, since it supplies the active power to the resistor. The reactive power of the source is also negative, indicating that it supplies reactive power supplied by the inductor. The phasor sum of the complex power of the source, and the complex power of the load is zero.
This page derives analytical expressions for most of the parameters. Along with these expressions, applets are provided to enhance the understanding with a numerical example.
An applet is presented below, with default values. You can edit and change the values of parameters. Click on the Next Step button, before proceeding to the next stage. Note that the inductance is specified in milliHenry, and the unit used is not microHenry.
An applet that accepts and verifies your answers is shown below. You can enter your answers into the respective boxes and click the VERIFY button. If your answer, an appropriate message is shown in the text box on the bottom row. You can correct your answer and verify again. When you have the correct answers, click on the Next Step button, before proceeding to the next stage.
You can calculate the real and the imaginary parts of the resistor voltage, the inductor voltage and the source voltage and enter them into the respective boxes of the applet presented below. To verify your answers, you should have done correctly the calculations for the previous applet and you should have enabled the applet shown below by clicking on the NEXT STEP button.
You can calculate the active power, the reactive power, the power factor angle, the power factor, the minimum instantaneous power and the maximum instantaneous power for the values of components specified in the first applet, and enter them into the respective boxes of the applet shown below. To see the plots of voltage, current, and the instantaneous power, click on the NEXT STEP button.
As a matter of interest, we can find out the locus of impedance, as the value of inductance changes from zero to infinity. When the inductance is infinite, it acts as an open circuit for ac signals and hence the impedance becomes infinite. When the inductance is zero, it acts as a short circuit, and, in this case, the impedance of the circuit is equal to the resistance. Since the real part of impedance equals the resistance, which remains constant, the real part of impedance for any value of inductance remains the same. Hence the trajectory of the impedance is along the vertical line, as shown in Fig. 45. The imaginary part has positive value, since the reactance of an inductor is positive, as shown by equation (264).
Next, we find out the locus diagram of the resistor voltage, as the inductance changes form zero to infinity. For the circuit in Fig. 46, the source voltage is the phasor sum of the resistor voltage and the inductor voltage. Given that the phase of source voltage is f radians, it can be represented as shown by the phasor diagram in Fig. 41. The phasor diagram shows the relative phase difference of the resistor voltage, and the capacitor voltage with respect to the source voltage. As shown by the phasor diagram, the resistor voltage lags the source voltage. Since the resistor and the inductor are in series, the current through them is the same. We know that in an inductor, the current lags the voltage by 90o. Since the resistor voltage and its current are in phase, the resistor voltage lags the inductor voltage by 90o. Hence, in this case, the parallelogram used for phasor diagram turns out to be a rectangle. For this rectangle, the resistor voltage and the inductor voltage are the two adjacent sides, and the source voltage is its diagonal. The source voltage has a phase of f degrees with respect to the reference phasor, which is represented by the horizontal axis, to the right of origin.
Equation (265) states that source voltage is the phasor sum the resistor voltage and the inductor voltage. The resistor current is in phase with its voltage, as shown by equation (266). The inductor voltage is obtained as shown by equation (267). We know that multiplying a phasor by j corresponds to phase-shifting that phasor by 90o in the anti-clockwise direction. In other words, the inductor voltage leads its current by 90o. The resistor voltage is in phase with its current, which is the same as the inductor current, since they are in series. Hence the phase difference between the resistor voltage, and the inductor voltage is 90o. This fact can be expressed in another way. Draw a horizontal line, to the right of origin. This horizontal line represents the reference phasor. Draw a line at angle of f with respect to the horizontal axis, its length proportionate to the phasor magnitude of source voltage. Let the source voltage correspond to the diameter of a circle, and let the centre be at mid-point of the line corresponding to the source voltage. Draw a semi-circle, as shown in Fig. 46. Draw lines corresponding to the phasor values of resistor voltage, and the inductor voltage. The tip of resistor voltage has to lie on the circumference of the semi-circle, since the angle between the resistor voltage, and the inductor voltage is 90o degrees. You should be able to recall what you have learned in geometry. If you join a point on the circumference of a circle to either ends of a diameter, the angle subtended at the point on the circumference is 90o.
Now let us find out what the trajectory of the resistor voltage is, as the inductance changes from zero to infinity. When the inductance is zero, it is a short circuit, and the voltage across the resistor equals the source voltage. When the inductance tends to infinity, its reactance tends to infinity. In this case, the inductor voltage equals the source voltage, and the resistor voltage is zero. Hence the locus of the resistor voltage lies on the circumference of a semi-circle, starting at the end of the diameter when the inductance is zero, and ending up equal to zero when the inductance is nearly infinite. The semi-circle is located as shown in Fig. 46, since the resistor voltage lags the source voltage. You may recall that the positive direction for angle is the anti-clockwise direction. You may wonder whether you should learn about the locus diagram, because most of the textbooks do not touch upon this aspect. It is not essential, but learning about the locus diagram may help you to understand the frequency response of circuits, to be explained later. Locus diagram presents an aspect of the behaviour of the RL circuit, and it may lead to better understanding.
This page has presented in detail the analysis of a series connected RL circuit. The next page is on simple ac circuits, such as R and C in parallel, R and L in parallel, L and C in series, and L and C in parallel.