Module 4. Single-phase AC Circuits. Version 2 EE IIT, Kharagpur 1

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1 Module 4 Single-phase A ircuits ersion EE IIT, Kharagpur

2 esson 4 Solution of urrent in -- Series ircuits ersion EE IIT, Kharagpur

3 In the last lesson, two points were described:. How to represent a sinusoidal (ac) quantity, i.e. voltage/current by a phasor?. How to perform elementary mathematical operations, like addition/ subtraction and multiplication/division, of two or more phasors, represented as complex quantity? Some examples are also described there. In this lesson, the solution of the steady state currents in simple circuits, consisting of resistance, inductance and/or capacitance connected in series, fed from single phase ac supply, is presented. Initially, only one of the elements / /, is connected, and the current, both in magnitude and phase, is computed. Then, the computation of total reactance and impedance, and the current, in the circuit consisting of two components, & / only in series, is discussed. The process of drawing complete phasor diagram with current(s) and voltage drops in the different components is described. astly, the computation of total power and also power consumed in the components, along with the concept of power factor, is explained. Keywords: Series circuits, reactance, impedance, phase angle, power, power factor. After going through this lesson, the students will be able to answer the following questions;. How to compute the total reactance and impedance of the -- series circuit, fed from single phase ac supply of known frequency?. How to compute the current and also voltage drops in the components, both in magnitude and phase, of the circuit? 3. How to draw the complete phasor diagram, showing the current and voltage drops? 4. How to compute the total power and also power consumed in the components, along with power factor? Solution of Steady State urrent in ircuits Fed from Singlephase A Supply Elementary ircuits. Purely resistive circuit ( only) The instantaneous value of the current though the circuit (Fig. 4.a) is given by, i v where, m sin ω t I m sin ω t I m and m are the maximum values of current and voltage respectively. ersion EE IIT, Kharagpur 3

4 The rms value of current is given by I m m / I In phasor notation, ( + j ) + j I I I ( + j ) I + j The impedance or resistance of the circuit is obtained as, Z + j I I Please note that the voltage and the current are in phase ( φ ), which can be observed from phasor diagram (Fig. 4.b) with two (voltage and current) phasors, and also from the two waveforms (Fig. 4.c). In ac circuit, the term, Impedance is defined as voltage/current, as is the resistance in dc circuit, following Ohm s law. The impedance, Z is a complex quantity. It consists of real part as resistance, and imaginary part as reactance X, which is zero, as there is no inductance/capacitance. All the components are taken as constant, having linear -I characteristics. In the three cases being considered, including this one, the power ersion EE IIT, Kharagpur 4

5 consumed and also power factor in the circuits, are not taken up now, but will be described later in this lesson.. Purely inductive circuit ( only) For the circuit (Fig. 4.a), the current i, is obtained by the procedure described here. As di v m sin ω t sin ω t, dt di sin ( ω t) dt Integrating, i cos ω t sin ( ω t 9 ) I m sin ( ω t 9 ) I sin ( ω t 9 ) ω ω It may be mentioned here that the current i, is the steady state solution, neglecting the constant of integration. The rms value, I is I I 9 ω + j ; I I 9 j I ersion EE IIT, Kharagpur 5

6 The impedance of the circuit is Z φ jω + j X X 9 9 ω I I 9 j I where, the inductive reactance is X ω π f. Note that the current lags the voltage by φ + 9. This can be observed both from phasor diagram (Fig. 4.b), and waveforms (Fig. 4.c). As the circuit has no resistance, but only inductive reactance X ω (positive, as per convention), the impedance Z is only in the y-axis (imaginary). 3. Purely capacitive circuit ( only) The current i, in the circuit (Fig. 4.3a), is, dv i dt Substituting v sinω t m sinω t, i is d i ( sinω t) ω cosω t ω sin ( ω t + 9 ) dt I m sin ( ω t + 9 ) The rms value, I is I ω I 9 /( ω ) + j ; I I 9 + The impedance of the circuit is j I I sin ( ω t + 9 ) j Z φ j X X 9 9 I I 9 j I jω ω ω where, the capacitive reactance is X. ω π f Note that the current leads the voltage by φ 9 (this value is negative, i.e. φ 9 ), as per convention being followed here. This can be observed both from phasor diagram (Fig. 4.3b), and waveforms (Fig. 4.3c). As the circuit has no resistance, but only capacitive reactance, X c /( ω ) (negative, as per convention), the impedance Z is only in the y-axis (imaginary). ersion EE IIT, Kharagpur 6

7 Series ircuits. Inductive circuit ( and in series) The voltage balance equation for the - series circuit (Fig. 4.4a) is, di v i + dt where, v sinω t m sinω t sinθ, θ being ω t. The current, i (in steady state) can be found as i I sin ( ω t φ) I m sin ( ω t φ) I sin ( θ φ) The current, i(t) in steady state is sinusoidal in nature (neglecting transients of the form shown in the earlier module on dc transients). This can also be observed, if one sees the expression of the current, i Im sin ( ω t) for purely resistive case (with only), and i I m sin ( ω t 9 ) for purely inductive case (with only). Alternatively, if the expression for i is substituted in the voltage equation, the equation as given here is obtained. sinω t I sin ( ω t φ) + ω I cos( ω t φ) If, first, the trigonometric forms in the HS side is expanded in terms of sin ω t and cos ω t, and then equating the terms of sin ω t and cos ω t from two (HS & HS) sides, the two equations as given here are obtained. ( cosφ + ω sinφ) I, and ersion EE IIT, Kharagpur 7

8 ( sinφ + ω cosφ) From these equations, the magnitude and phase angle of the current, I are derived. From the second one, tanφ ( ω / ) So, phase angle, φ tan ( ω / ) Two relations, cosφ ( / Z ), and sinφ ( ω / Z ), are derived, with the term (impedance), Z + ( ω ) If these two expressions are substituted in the first one, it can be shown that the magnitude of the current is I / Z, with both and Z in magnitude only. The steps required to find the rms value of the current I, using complex form of impedance, are given here. ersion EE IIT, Kharagpur 8

9 The impedance (Fig. 4.5) of the inductive (-) circuit is, Z φ + j X + jω where, X Z + X + ( ω ) and ω φ tan tan + j + j I φ Z φ + j X + jω I Z + X + ( ω ) Note that the current lags the voltage by the angle φ, value as given above. In this case, the voltage phasor has been taken as reference phase, with the current phasor lagging the voltage phasor by the angle, φ. But normally, in the case of the series circuit, the current phasor is taken as reference phase, with the voltage phasor leading the current phasor by φ. This can be observed both from phasor diagram (Fig. 4.4b), and waveforms (Fig. 4.4c). The inductive reactance X is positive. In the phasor diagram, as one move from voltage phasor to current phasor, one has to go in the clockwise direction, which means that phase angle, φ is taken as positive, though both phasors are assumed to move in anticlockwise direction as shown in the previous lesson. The complete phasor diagram is shown in Fig. 4.4b, with the voltage drops across the two components and input (supply) voltage (OA), and also current (OB ). The voltage phasor is taken as reference. It may be observed that O ( I ) + A [ I ( j X )] OA ( I Z), using the Kirchoff s second law relating to the voltage in a closed loop. The phasor diagram can also be drawn with the current phasor as reference, as will be shown in the next lesson. Power consumed and Power factor From the waveform of instantaneous power ( W the above circuit, the average power is, v i ) also shown in Fig. 4.4c for ersion EE IIT, Kharagpur 9

10 π π W v i dθ sinθ I sin ( θ φ) dθ I π π π [ cos φ cos (θ φ) ] π I π φθ θ φ π I cos sin ( ) I I cos φ ( π ) [ sin (π φ) sinφ] I cosφ π + Note that power is only consumed in resistance, only, but not in the inductance,. So, W I. Power factor average power apparent I cos φ cos φ power I π Z + ( ω ) The power factor in this circuit is less than (one), as φ 9, φ being positive as given above. For the resistive () circuit, the power factor is (one), as φ, and the average power is I. For the circuits with only inductance, or capacitance, as described earlier, the power factor is (zero), as φ ± 9. For inductance, the phase angle, or the angle of the impedance, φ + 9 (lagging), and for capacitance, φ 9 (leading). It may be noted that in both cases, the average power is zero (), which means that no power is consumed in the elements, and. The complex power, olt-amperes (A) and reactive power will be discussed after the next section.. apacitive circuit ( and in series) This part is discussed in brief. The voltage balance equation for the - series circuit (Fig. 4.6a) is, v i + i dt The current is i I sin ( ω t + φ) sinω t The reasons for the above choice of the current, i, and the steps needed for the derivation of the above expression, have been described in detail, in the case of the earlier example of inductive (-) circuit. The same set of steps has to be followed to derive the current, i in this case. Alternatively, the steps required to find the rms value of the current I, using complex form of impedance, are given here. The impedance of the capacitive (-) circuit is, Z φ j X j ω dθ ersion EE IIT, Kharagpur

11 where, + + X Z ω and X ω ω φ tan tan tan ) (/ j j X j j Z I ω φ φ + + ( ) / X Z I ω + + ersion EE IIT, Kharagpur

12 Note that the current leads the voltage by the angle φ, value as given above. In this case, the voltage phasor has been taken as reference phase, with the current phasor leading the voltage phasor by the angle, φ. But normally, in the case of the series circuit, the current phasor is taken as reference phase, with the voltage phasor lagging the current phasor by φ. This can be observed both from phasor diagram (Fig. 4.6b), and waveforms (Fig. 4.6c). The capacitive reactance X is negative. In the phasor diagram, as one move from voltage phasor to current phasor, one has to go in the anticlockwise direction, which means that phase angle, φ is taken as negative. This is in contrast to the case as described earlier. The complete phasor diagram is shown in Fig. 4.6b, with the voltage drops across the two components and input (supply) voltage, and also current. The voltage phasor is taken as reference. The power factor in this circuit is less than (one), with φ being same as given above. The expression for the average power is P I cosφ, which can be obtained by the method shown above. The power is only consumed in the resistance,, but not in the capacitance,. One example is included after the next section. omplex Power, olt-amperes (A) and eactive Power The complex power is the product of the voltage and complex conjugate of the current, both in phasor form. For the inductive circuit, described earlier, the voltage ( ) is taken as reference and the current ( I φ I cos φ j I sin φ ) is lagging the voltage by an angle, φ. The complex power is S * I I φ ( I ) φ I cos φ + j I sin φ P + j Q The olt-amperes (S), a scalar quantity, is the product of the magnitudes the voltage and the current. So, S I P + Q. It is expressed in A. The active power (W) is * P e ( S) e ( I ) I cos φ, as derived earlier. * The reactive power (Ar) is given by Q Im ( S) Im ( I ) I sin φ. As the phase angle, φ is taken as positive in inductive circuits, the reactive power is positive. The real part, ( I cos φ ) is in phase with the voltage, whereas the imaginary part, I sin φ is in quadrature ( 9 ) with the voltage. But in capacitive circuits, the current ( I φ ) leads the voltage by an angle φ, which is taken as negative. So, it can be stated that the reactive power is negative here, which can easily be derived Example 4. A voltage of at 5 Hz is applied to a resistance, in series with a capacitance, (Fig. 4.7a). The current drawn is A, and the power loss in the resistance is W. alculate the resistance and the capacitance. Solution I A P W f 5 Hz ersion EE IIT, Kharagpur

13 P / I / 5 Ω Z + X / I / 6 Ω X c /(π f ) Z (6) (5) Ω μf π f X π The power factor is, cos φ / Z 5 / 6.47 ( lead ) The phase angle is φ cos (.47) The phasor diagram, with the current as reference, is shown in Fig. 4.7b. The examples, with lossy inductance coil (r in series with ), will be described in the next lesson. The series circuit with all elements,. &, along with parallel circuits, will be taken up in the next lesson. ersion EE IIT, Kharagpur 3

14 Problems 4. alculate the power factor in the following cases for the circuit with the elements, as given, fed from a single phase ac supply. (i) With resistance, only, but no and (a). (Φ ) (c). leading (Φ-9 ) (b). lagging (Φ+9 ) (d) None of the above (ii) with only pure/lossless inductance,, but no and (a). (Φ ) (c). leading (Φ-9 ) (b). lagging (Φ+9 ) (d) None of the above (iii) with only pure capacitance,, but no and. (a). (Φ ) (c). leading (Φ-9 ) (b). lagging (Φ+9 ) (d) None of the above 4. alculate the current and power factor (lagging / leading) in the following cases for the circuits having impedances as given, fed from an ac supply of. Also draw the phasor diagram in all cases. (i) (ii) Z (5+j) Ω Z (4-j4) Ω (iii) Z + j (X X ), where Ω, X Ω, and X Ω. 4.3 A, 5 Hz supply is connected to a resistance () of Ω in series with an iron cored choke coil (r in series with ). The readings of the voltmeters across the resistance and across the coil are and 5 respectively. Find the loss in the coil. Also find the total power factor. Draw the phasor diagram. 4.4 A circuit, with a resistance, and a lossless inductance in series, is connected across an ac supply () of known frequency (f). A capacitance, is now connected in series with -, with and f being constant. Justify the following statement with reasons. The current in the circuit normally increases with the introduction of. Under what condition, the current may also decrease. Explain the condition with reasons. ersion EE IIT, Kharagpur 4

15 ist of Figures Fig. 4. esistive () load, connected to single phase ac supply (a) ircuit diagram (b) Phasor diagram (c) Waveforms. oltage (v),. urrent (i) Fig. 4. oad Inductance () only (a) ircuit diagram (b) Phasor diagram (c) Waveforms. oltage (v),. urrent (i) Fig. 4.3 oad apacitance () only (a) ircuit diagram (b) Phasor diagram (c) Waveforms. oltage (v),. urrent (i) Fig. 4.4 oad Inductive ( and in series) (a) ircuit diagram (b) Phasor diagram (c) Waveforms. oltage (v),. urrent (i), 3. Instantaneous power (W v i) Fig. 4.5 The complex form of the impedance (- series circuit) Fig. 4.6 oad apacitive ( and in series) (a) ircuit diagram (b) Phasor diagram (c) Waveforms. oltage (v),. urrent (i), Fig. 4.7 (a) ircuit diagram (Ex. 4.) (b) Phasor diagram ersion EE IIT, Kharagpur 5

Module 4. Single-phase AC Circuits

Module 4. Single-phase AC Circuits Module 4 Single-phase AC Circuits Lesson 14 Solution of Current in R-L-C Series Circuits In the last lesson, two points were described: 1. How to represent a sinusoidal (ac) quantity, i.e. voltage/current