mywbut.com 32 The superposition theorem At the end of this chapter you should be able to: ž solve d.c. and a.c. networks using the superposition theorem 32.1 Introduction The superposition theorem states: ‘In any network made up of linear impedances and containing more than one source of e.m.f. the resultant current flowing in any branch is the phasor sum of the currents that would flow in that branch if each source were considered separately, all other sources being replaced at that time by their respective internal impedances.’ 32.2 Using the The superposition theorem, which was introduced in Chapter 13 for d.c. superposition theorem circuits, may be applied to both d.c. and a.c. networks. A d.c. network is shown in Figure 32.1 and will serve to demonstrate the principle of application of the superposition theorem. To find the current flowing in each branch of the circuit, the following six-step procedure can be adopted: (i) Redraw the original network with one of the sources, say E2, removed and replaced by r2 only, as shown in Figure 32.2. (ii) Label the current in each branch and its direction as shown Figure 32.1 in Figure 32.2, and then determine its value. The choice of current direction for I1 depends on the source polarity which, by convention, is taken as flowing from the positive terminal as shown. R in parallel with r2 gives an equivalent resistance of 5 ð 2/5 C 2 D 10/7 D 1.429 as shown in the equivalent network of Figure 32.3. From Figure 28.3, Figure 32.2 E1 8 current I1 D D D 3.294 A r1 C 1.429 2.429 1 mywbut.com From Figure 32.2, r2 2 current I2 D I1 D 3.294 D 0.941 A R C r2 5 C 2 5 and current I D 3.294 D 2.353 A 3 5 C 2 (iii) Redraw the original network with source E1 removed and replaced by r1 only, as shown in Figure 32.4. Figure 32.3 (iv) Label the currents in each branch and their directions as shown in Figure 32.4, and determine their values. R and r1 in parallel gives an equivalent resistance of 5 ð 1/5 C 1 D 5/6 or 0.833 , as shown in the equivalent network of Figure 32.5. From Figure 32.5, E2 3 current I4 D D D 1.059 A r2 C 0.833 2.833 Figure 32.4 From Figure 32.4, 1 current I D 1.059 D 0.177 A 5 1 C 5 5 and current I D 1.059 D 0.8825 A 6 1 C 5 (v) Superimpose Figure 32.2 on Figure 32.4, as shown in Figure 32.6. Figure 32.5 Figure 32.6 (vi) Determine the algebraic sum of the currents flowing in each branch. (Note that in an a.c. circuit it is the phasor sum of the currents that is required.) 2 mywbut.com From Figure 32.6, the resultant current flowing through the 8 V source is given by I1 I6 eD 33. V294 source 0.8825 is givenD 2 by 41 A (discharging, i.e., flowing from the positive terminal of the source). The resultant current flowing in th I3 I4 D 2.353 1.059 D 1.29 A (charging, i.e., flowing into the positive terminal of the source). The resultant current flowing in the 5 resistance is given by I2 C I5 D 0.941 C 0.177 D 1.12 A The values of current are the same as those obtained on page 536 by using Kirchhoff’s laws. The following problems demonstrate further the use of the superposi- tion theorem in analysing a.c. as well as d.c. networks. The theorem is straightforward to apply, but is lengthy. Thevenin’s´ and Norton’s theo- rems (described in Chapter 33) produce results more quickly. Problem 1. A.c. sources of 1006 0° V and internal resistance 25 , and 506 90° V and internal resistance 10 , are connected in parallel across a 20 load. Determine using the superposition theorem, the current in the 20 load and the current in each voltage source. (This is the same problem as problem 1 on page 536 and problem 6 on page 553 and a comparison of methods may be made.) The circuit diagram is shown in Figure 32.7. Following the above procedure: (i) The network is redrawn with the 506 90° V source removed as shown in Figure 32.8 (ii) Currents I1, I2 and I3 are labelled as shown in Figure 32.8. Figure 32.7 Figure 32.8 3 mywbut.com 1006 0° 1006 0° I D D D 3.1586 0° A 1 25 C 10 ð 20/10 C 20 25 C 6.667 10 I D 3.1586 0° D 1.0536 0° A 2 10 C 20 20 I D 3.1586 0° D 2.1056 0° A 3 10 C 20 (iii) The network is redrawn with the 1006 0° V source removed as shown in Figure 32.9 (iv) Currents I4, I5 and I6 are labelled as shown in Figure 32.9. 6 ° 6 ° I D 50 90 D 50 90 4 10 C 25 ð 20/25 C 20 10 C 11.111 D 2.3686 90° Aorj2.368 A 25 I D j2.368 D j1.316 A 5 20 C 25 20 I D j2.368 D j1.052 A 6 20 C 25 (v) Figure 32.10 shows Figure 32.9 superimposed on Figure 32.8, giving the currents shown. Figure 32.9 Figure 32.10 (vi) Current in the 20 load, I2 C I5 D 1.053 C j1.316 Aor 1.696 51.33° A 6 ° Current in the 100 0 V source, I1 I6 D 3.158 j1.052 Aor 3.336 −18.42° A 6 ° Current in the 50 90 V source, I4 I3 D j2.368 2.105 or 3.176 131.64° A 4 mywbut.com Problem 2. Use the superposition theorem to determine the current in the 4 resistor of the network shown in Figure 32.11. (i) Removing the 20 V source gives the network shown in Figure 32.12. Figure 32.11 Figure 32.12 (ii) Currents I1 and I2 are shown labelled in Figure 32.12. It is unnec- essary to determine the currents in all the branches since only the current in the 4 resistance is required. From Figure 32.12, 6 in parallel with 2 gives 6 ð 2/6 C 2 D 1.5 , as shown in Figure 32.13. 2.5 in series with 1.5 gives 4 ,4 in parallel with 4 gives 2 , and 2 in series with 5 gives 7 . 12 Thus current I D D 1.714 A and 1 7 Figure 32.13 4 current I D 1.714 D 0.857 A 2 4 C 4 (iii) Removing the 12 V source from the original network gives the network shown in Figure 32.14. (iv) Currents I3, I4 and I5 are shown labelled in Figure 32.14. From Figure 32.14, 5 in parallel with 4 gives 5 ð 4/5 C 4 D 20/9 D 2.222 , as shown in Figure 32.15, 2.222 in series with 2.5 gives 4.722 , 4.722 in parallel Figure 32.14 with 6 gives 4.722 ð 6/4.722 C 6 D 2.642 , 2.642 in series with 2 gives 4.642 . 20 Hence I D D 4.308 A 3 4642 6 I D 4.308 D 2.411 A, from Figure 32.15 4 6 C 4.722 5 I5 D 2.411 D 1.339 A, from Figure 32.14 Figure 32.15 4 C 5 5 mywbut.com (v) Superimposing Figure 32.14 on Figure 32.12 shows that the current flowing in the 4 resistor is given by I5 I2 (vi) I5 I2 D 1.339 0.857 D 0.48 A, flowing from B toward A (see Figure 32.11) Problem 3. Use the superposition theorem to obtain the current flowing in the 4 C j3 impedance of Figure 32.16. (i) The network is redrawn with V2 removed, as shown in Figure 32.17. Figure 32.16 Figure 32.17 (ii) Current I1 and I2 are shown in Figure 32.17. From Figure 32.17, 4 C j3 in parallel with j10 gives an equivalent impe- dance of C j j j 6 . ° 4 3 10 D 30 40 D 50 53 13 4 C j3 j10 4 j7 8.0626 60.26° D 6.2026 7.13° or 6.154 C j0.770 Total impedance of Figure 32.17 is 6.154 C j0.770 C 4 D 10.154 C j0.770 or 10.1836 4.34° 306 45° Hence current I D D 2.9466 40.66° A 1 10.1836 4.34° j10 and current I D 2.9466 40.66° 2 4 j7 6 ° . 6 . ° D 10 90 2 946 40 66 8.0626 60.26° D 3.6546 10.92° Aor3.588 C j0.692A 6 mywbut.com (iii) The original network is redrawn with V1 removed, as shown in Figure 32.18. (iv) Currents I3 and I4 are shown in Figure 32.18. From Figure 32.18, 4 in parallel with 4 C j3 gives an equivalent impedance of C j C j 6 .