Module 4 Single-phase circuits ersion EE T, Kharagpur
esson 6 Solution of urrent in Parallel and Seriesparallel ircuits ersion EE T, Kharagpur
n the last lesson, the following points were described:. How to compute the total impedance/admittance in series/parallel circuits?. How to solve for the current(s) in series/parallel circuits, fed from single phase ac supply, and then draw complete phasor diagram?. How to find the power consumed in the circuit and also the different components, and the power factor (lag/lead)? n this lesson, the computation of impedance/admittance in parallel and series-parallel circuits, fed from single phase ac supply, is presented. Then, the currents, both in magnitude and phase, are calculated. The process of drawing complete phasor diagram is described. The computation of total power and also power consumed in the different components, along with power factor, is explained. Some examples, of both parallel and series-parallel circuits, are presented in detail. Keywords: Parallel and series-parallel circuits, impedance, admittance, power, power factor. fter going through this lesson, the students will be able to answer the following questions;. How to compute the impedance/admittance, of the parallel and series-parallel circuits, fed from single phase ac supply?. How to compute the different currents and also voltage drops in the components, both in magnitude and phase, of the circuit?. How to draw the complete phasor diagram, showing the currents and voltage drops? 4. How to compute the total power and also power consumed in the different components, along with power factor? This lesson starts with two examples of parallel circuits fed from single phase ac supply. The first example is presented in detail. The students are advised to study the two cases of parallel circuits given in the previous lesson. Example 6. The circuit, having two impedances of (8 + j5) and (6 j8) in parallel, is connected to a single phase ac supply (Fig. 6.a), and the current drawn is 0. Find each branch current, both in magnitude and phase, and also the supply voltage. ersion EE T, Kharagpur
(8 + j5) B 0 (6 j8) Fig. 6. (a) ircuit diagram Solution (8 + j5) 7 6. 9 φ (6 j8) 0 5. 0 ( O) 0 (0 + j 0) The admittances, using impedances in rectangular form, are, 8 j5 8 j5 Y φ (7.68 j 5.9) 0 8 + j5 8 + 5 89 6 + j 8 6 + j 8 Y (60.0 + j 80.0) 0 φ 6 j 8 6 + 8 00 lternatively, using impedances in polar form, the admittances are, Y φ 0.0588 6.9 7.0 6.9 (7.68 Y j 5.9) 0 0. 5. (60.0 + j 80.0) 0 φ 0.0 5. The total admittance is, Y Y + Y [(7.68 j 5.9) + (60.0 + j 80.0)] 0 (87.68 + j 8.) 0 9.07 0 7.77 The total impedance is, φ 0.86 7.77 (0.4 j.5) Y 9.07 0 7.77 The supply voltage is φ ( B ) φ (0 0.86) 7.77 08.6 7. 77 ( 0.4 j.5) The branch currents are, φ 08.6 θ ( OD) (7.77 + 6.9 ) 6.9 79. 7 7.0 ersion EE T, Kharagpur
(.4 j 6.86) θ ( OE) 0 θ ( O OD O E) (0.0 + j 0.0) (.4 j6.86) (8.857 + j6.86) 0.86 5.6 lternatively, the current is, φ 08.6 θ( OE) ( 7.77 + 5. ) 0.86 5.6 φ 0.0 ( 8.857 + j 6.85) The phasor diagram with the total (input) current as reference is shown in Fig. 6.b. 0.86 E 5. O θ 79.7 θ 5.5 6.90 Ф 7.8 0 B 6.4 D Fig. 6. (b) Phasor diagram 08.6 lternative Method + φ (8 + j5) + (6 j8) (4 + 7) 5.65 6. 565 j φ 7.0 0.0 φ (6.9 5. 6.565 ) + φ 5.65 0.86 7.77 (0.4 j.5) The supply voltage is φ ( ) (0 0.86) 7.77 08.6 7.77 B (0.4 j.5) The branch currents are, 0.0 0.0 θ( OD) ( 5. 6.565 ) 6.9 79.7 + 5.65 (.4 j6.86) ersion EE T, Kharagpur
θ ( OE) ( O OD O E) (0.0 + j0.0) (.4 j6.86) (8.858 + j6.86) 0.86 5.6 lternatively, the current is, 0.0 7.0 OE θ( ) (6.9 6.565 ) 0.86 5..6 + 5.65 (8.858 + j6.86) Example 6. The power consumed in the inductive load (Fig. 6.a) is.5 kw at 0.7 lagging power factor (pf). The input voltage is 0, 50 Hz. Find the value of the capacitor, such that the resultant power factor of the input current is 0.866 lagging. + - O D Solution 0 Fig. 6. (a) ircuit diagram P.5 KW.5 0 500 W 0 f 50 Hz The power factor in the inductive branch is cos φ 0.7 ( lag) The phase angle is φ cos (0.7) 44.77 45 P cos φ 0 ( cosφ ) 500 P 500 5. cosφ 0 0.7 cos φ 5. 0.7 0.87 ; sin φ 5. sin 45 0. 87 The current is, φ 5. 45 (0.87 j0.87) The power consumed in the circuit remains same, as the capacitor does not consume any power, but the reactive power in the circuit changes. The active component of the total current remains same as computed earlier. cos φ cosφ 0. 87 The power factor of the current is cos φ 0.866 ( lag) The phase angle is φ cos (0.866) 0 The magnitude of the current is 0.87 / 0.866. 55 The current is φ.55 0 (0.87 j 6.76) ersion EE T, Kharagpur
The current in the capacitor is 90 φ φ (0.87 j 6.76) (0.87 j0.87) j 4.504 4.504 90 This current is the difference of two reactive currents, sinφ sinφ 6.76 0.87 4. 504 0 The reactance of the capacitor, is X 5. 066 π f 4.504 The capacitor, is 6. 0 6 6. μf π f X π 50 5.066 The phasor diagram with the input voltage as reference is shown in Fig. 6.b. 4.5 45 φ 0 Example 6. 5. B Fig. 6. (b) Phasor diagram n inductive load (R in series with ) is connected in parallel with a capacitance of.5 μ F (Fig. 6.a). The input voltage to the circuit is 00 at.8 Hz. The phase angle between the two branch currents, ( ) and ( ) is 0, and the current in the first branch is 0. 5. Find the total current, and also the values of R &. 00 + - 0.5 R D.5 μf Fig. 6. (a) ircuit diagram B ersion EE T, Kharagpur
Solution f.8 Hz ω π f π.8 00 rad / s 00 0. 5 F 6.5μ.5 0 F 6 X /( ω ) /(00.5 0 ) 400 The current in the branch no. is 90 /( j X ) 00 / 400 90 (00 / 400) 90 0.5 90 (0.0 + j 0.5) The current in the branch no. is φ 0. 5 φ The phase angle between and is 90 + φ 0 So, φ 0 90 0 0 0.5 0 (0.4 j 0.5) The impedance of the branch no. is, R + j X ) / 0 (00 / 0.5) 00 ( (7. + j00.0) R 7. X ω 00. 0 So, X / ω 00 / 00 0.5 H 500 0 500 mh The total current is, 0 + 90 (0.4 j 0.5) + j 0.5 (0.4 + j 0.0) 0 0.4 The total impedance is, ( R + j 0) / (00 / 0.4).0 (.0 + j 0.0) The current, is in phase with the input voltage,. The total admittance is Y Y φ + Y 90 (/ ) + (/ 90 ) The total impedance is ( 90 ) /( + 90 ) ny of the above values can be easily calculated, and then checked with those obtained earlier. The phasor diagram is drawn in Fig. 6.b. Solution of urrent in Series-parallel ircuit Series-parallel circuit The circuit, with a branch having impedance, in series with two parallel branches having impedances, and, shown in Fig.,, is connected to a single phase ac supply. The impedance of the branch, B is B B Y φ ; Y φ ersion EE T, Kharagpur
The admittance of the parallel branch, B is YB φb Y φ + Y φ + The impedance of the parallel branch, B is + + B B YB φ B ( ) ( φ + φ ) The total impedance of the circuit is B B + B B + B B The supply current is φ The current in the impedance is 4 φ ( + ) Thus, the currents, along with the voltage drops, in all branches are calculated. The phasor diagram cannot be drawn for this case now. This is best illustrated with the following examples, where the complete phasor diagram will also be drawn in each case. Example 6.4 0.5 0.4 00 B 86.6 0 50 0.5 D 0.5 Fig. 6. (b) Phasor diagram Find the input voltage at 50 Hz to be applied to the circuit shown in Fig. 6.4a, such that the current in the capacitor is 8? R 5 5.5 mh 8 D R 7 8. mh B R 8 8 μf Fig. 6.4 (a) ircuit diagram ersion EE T, Kharagpur
Solution f 50 Hz ω π f π 50 4.6 rad / s 0. 055 H 0. 08 H F 6 F 8 μ 8 0 X ω 4.6 0.055 8 X ω 4.6 0.08 6 X /( ω ) /(4.6 8 0 ) 0 0 8 (8 + j 0) R + j X (5 + j 8) 9.44 58 φ R j X (8 j0).806 5. 4 R + j X (7 + ).89 59. 74 j φ φ (64 j 80) φ θ (.6 j0.5) θ θ +.54 66.77 B θ B 54.9 7.0 B θ B (8.0.806) 5.4 0.45 5.4 0.45 (5.4 + 58 ) 0.86 09.4 9.44 θ (.6 + j0.5) + (8.0 + (5.764 j8.96) j 0.0) (4.4 (.54.89) ( 66.77 + 59.74 ) φ + θ (7.764 j 90.96) 9. 4.44 The phasor diagram with the branch current, B B (64.0 j 80.0) + (5.764 j0.5) j8.96) as reference, is shown in Fig. 6.4b. 8 7.0 58 5.4 66.77 DB 54.9 0.86.5 D 0.45 B 9. Fig. 6.4 (b) Phasor diagram ersion EE T, Kharagpur
Example 6.5 resistor of 50 in parallel with an inductor of 0 mh, is connected in series with a capacitor, (Fig. 6.5a). voltage of 0, 50 Hz is applied to the circuit. Find, (a) the value of to give unity power factor, (b) the total current, and (c) the current in the inductor R 50 D B 0 mh 0 50 Hz Fig. 6.5 (a) ircuit diagram Solution f 50 Hz ω π f π 50 4.6 rad / s R 50 0 0 mh 0 0 0. 0 H X ω 4.6 0.0 94. 4 Y D The admittance, is, YD φ D + R j X l 50 + j 94.4 (0.0 j0.6) 0 0.064 7.95 The impedance, D is, D D /( YD D ) /(0.064 7.95 ) 44.7 7.95 (9.0 + j 0.7) The impedance of the branch (DB) is DB j X j[ /( ω ). s the total current is at unity power factor (upf), the total impedance, resistive only. B RB + j 0 D D + DB 90 9.0 + j (0.7 X ) Equating the imaginary part, X c /( ω ) 0. 7 The value of the capacitance is, B is ersion EE T, Kharagpur
5.8 0 6 5.8 μf ω X 4.6 0.7 So, B RB + j 0 (9.0 + j 0.0) 9.0 The total current is, 0 / B (0.0 / 9.0) (5.64 + j 0.0) 5.64 D The voltage, is, D D (0.0 + j6.7) D D (5.64 44.7) 7.95 49.05 7.95 The current in the inductor, is, θ / X 90 (49.05/ 94.4) (7.95 90 ) D D.64 6.05 (.4 j.5) The phasor diagram is shown in Fig. 6.5b. 6 8 R 4.98 5.64 49.0 D 6.7 8 B.64 6.7 0 (i) (ii) Fig. 6.5 (b) Phasor diagram Example 6.6 n the circuit (Fig. 6.6a) the wattmeter reads 960 W and the ammeter reads 6. alculate the values of S,,,, and X. W R 0 D R 6 E + - jx j 8 B Fig. 6.6 (a) ircuit diagram ersion EE T, Kharagpur
Solution n this circuit, the power is consumed in two resistance, R and R only, but not consumed in inductance, and capacitance. These two components affect only the reactive power. P 960 W 6 R 0 6 8 R X Total power is, P R + R (6) 0 + 6 60 + 6 960 W or, 6 960 60 600 W So, 600 / 6 0 The impedance of the inductive branch is, R + j X (6 + j8) 0 5. The magnitude of the voltage in the inductive branch is, DB 0 0 00 ssuming DB 00 as reference, the current, is, φ DB / (00/0) 5. 0 5. (6 j8) The current, 90 j DB / X 90 ( / X ) 90 The total current is φ + 90 (6 j 8) + j 6 + j( 8) So, (6) + ( 8) 6 or, 6 ( 8) (6) 6 So, + 8 The capacitive reactance is, X DB / 00 / 8. 5 The total current is 0 (6 + j 0) 6 or, it can be written as, φ + 90 (6 j 8) + j 8 (6 + j 0) 6 0 D is. 0 ( R + j 0) ( R ) (6 0) 60 (60 + The voltage D The voltage, S B S B is. 0 + (60 + 00) 60 (60 + D DB j 0) j 0) The current, is in phase with S B, and also. The total impedance is, + / (60 / 6) 6.67 B D (6.67 + j 0.0) DB The impedance, is, DB B DB D B DB / (6.67 + j 0.0) Both the above impedances can be easily obtained using the circuit parameters by the method given earlier, and then checked with the above values. The impedance, can DB (00 / 6) 6.67 DB ersion EE T, Kharagpur
be obtained by the steps given in Example 6.. The phasor diagram is shown in Fig. 6.6b. 8 60 6 D 6 5. 8 B B 60 ( O ) Fig. 6.6 (b) Phasor diagram Starting with the examples of parallel circuits, the solution of the current in the seriesparallel circuit, along with the examples, was taken up in this lesson. The problem of resonance in series and parallel circuits will be discussed in the next lesson. This will complete the module of single phase ac circuits ersion EE T, Kharagpur
Problems 6. Find the impedance, ab in the following circuits (Fig. 6.7a-b): (check with admittance diagrams in complex plane) a l l a l jl - j -j j b b (a) (b) Fig. 6.7 6. resistor (R) of 50 in parallel with a capacitor () of 40 μf, is connected in series with a pure inductor () of 0 mh to a 00, 50 Hz supply. alculate the total current and also the current in the capacitor. Draw the phasor diagram. 6. n a series-parallel circuit (Fig.6.8), the two parallel branches and B, are in series with the branch. The impedances in are, 5+j, B 6-j8, and 0+j8. The voltage across the branch, is (50+j0). Find the branch currents, and B, and the phase angle between them. Find also the input voltage. Draw the phasor diagram. 6.4 total current of is drawn by the circuit (Fig.6.9) fed from an ac voltage, of 50 Hz. Find the input voltage. Draw the phasor diagram. (5 + j6) (0 + j8) P Q R B (6 j8) 50 Fig. 6.8 ersion EE T, Kharagpur
+ - R 5 60 μf B R 0 50 mh Fig. 6.9 ersion EE T, Kharagpur
ist of Figures Fig. 6. (a) ircuit diagram (Ex. 6.) Fig. 6. (a) ircuit diagram (Ex. 6.) Fig. 6. (a) ircuit diagram (Ex. 6.) Fig. 6.4 (a) ircuit diagram (Ex. 6.4) Fig. 6.5 (a) ircuit diagram (Ex. 6.5) Fig. 6.6 (a) ircuit diagram (Ex. 6.6) (b) Phasor diagram (b) Phasor diagram (b) Phasor diagram (b) Phasor diagram (b) Phasor diagram (b) Phasor diagram ersion EE T, Kharagpur