Netwrk Therems - lternating urrent examples - J. R. Lucas n the previus chapter, we have been dealing mainly with direct current resistive circuits in rder t the principles f the varius therems clear. s was mentined, these theries are equally valid fr a.c. Example Ω 2 E 00 0 V 9.89 µf j60 Ω E 2 70-30 V Fr the circuit shwn in the figure, if the frequency f the supply is 50 Hz, determine using Ohm s Law and Kirchff s Laws the current in the 60 Ω capacitr. mpedance f capacitr and inductr at 50 Hz are j X L j 2π 50 63.66 0 3 j and - j X /( j 2π 50 9.89 0 6 ) j 60 Ω Using Kirchff s current law 2 Using Kirchff s vltage law 00 0 j60 ( 2 ) 5 (-j8) +j8 2... () and 70-30 j 2 j 60 ( 2 ) 7-30 j 6 + j 4 2... (2) multiplying equatin () by 7 and equatin (2) by 4 and subtracting gives 35-28 -30 (7 j 56 + j 64) + 0 0.75 + j 4 (7 + j 8) i.e. 7.65 52.48.660 3.67 0.63 48.8 substituting in (), j8 2 5 (-j8).660 3.67 i.e. 8 90 2 5 8.062-82.87.660 3.67 5 3.383-79. 2.492 + j3.46 3.38 79.27 i.e. 2.673-0.73 8 90 Thus the required current is.660 3.67.673-0.73.657 + j 0.06.644 + j 0.3 0.03 +j 0.47 0.42 88.2 The prblem culd prbably have been wrked ut with lesser steps, but have dne it in this manner s that yu can get mre familiarised with the slutin f prblems using cmplex numbers. Netwrk Therems- a c examples rfessr J R Lucas Nvember 0
Example 2 Let us slve the same prblem as earlier, but using Superpsitin therem. Ω 2 E 00 0 V 9.89 µf j60 Ω E 2 70-30 V This circuit can be brken int its tw cnstituent cmpnents as shwn. Ω 2 Ω 2 E 00 0 V 9.89 µf j60 Ω + 9.89 µf j60 Ω E 2 70-30 V Using series parallel additin f impedances, we can btain the supply currents as fllws. Equivalent Z s + (-j60)//, Z s2 + //(-j60) j60 ( j60) +, + j40 j60 + j 22.857, + 9.692 j 2.462 9.692 + j7.538 30.372 48.8 Ω, 26.370 4.69 Ω 00 0 70 30 surce current, 2 30.372 48.8 26.370 4.69 3.293-48.8, 2.655-7.69, Using the current divisin rule (nte directins f currents and signs), 3.293 48.8, 2.655 7.69 j40 j60 0.470 3.9 53.0 7.69, 0.329.8 6.25 82.87 Using superpsitin therem, the ttal current in 0.470 3.9 + 0.329.8-0.30 + j 0.354 + 0.323 + j0.064 0.03 + j 0.49 0.42 88.2 which is the same answer btained in the earlier example. Netwrk Therems- a c examples rfessr J R Lucas 2 Nvember 0
Example 3 Let us again cnsider the same example t illustrate Thevenin s Therem. Ω 2 E 00 0 V 9.89 µf j60 Ω E 2 70-30 V nsider the capacitr discnnected at and. urrent flwing in the circuit under this cnditin 00 0 70 30 + 00 60.62 + j35 52.69 4.63.863 3.37 + 28.28 45 Thevenin s vltage surce 00 0.863-3.37 62.80 + j 2.9 ls, Thevenin s impedance acrss // Thevenin s equivalent circuit is E th V 0 + j 0 Ω 9.89 µf j60 Ω + 4.42 45 0 + j 0 Frm this circuit, it fllws that 0 + j0 j60 50.33 86.9 0.48-88.2 which is again the same result. Example 4 Let us again cnsider the same example t illustrate Nrtn s Therem. Ω 2 E 00 0 V 9.89 µf j60 Ω E 2 70-30 V nsider the capacitr shrt-circuited at and. The Nrtn s current surce is given as 3.25 j 3.03 4.444-43.00 00 0 70 30 + 5.75 j3.03 Netwrk Therems- a c examples rfessr J R Lucas 3 Nvember 0
Nrtn s admittance + 0.05 j0. 05 S 4.444-43.0 r same as Thevenin s impedance 0 + j0 Ω The Nrtn s equivalent circuit is as shwn in the figure. The current thrugh the capacitr can be determined using the current divisin rule. 4.444 43.0 0 + j0 4.444 43.0 4.42 45 0 + j0 j60 50.33 86.9 0.48-88.2 j60 Example 5 Using Millmann s therem find the current in the capacitr. S Ω 2 E 00 0 V 9.89 µf j60 Ω E 2 70-30 V N V SN Y. V Y 00 0 + 0 + j60 j60 70 30 5 + 0.75 j3.03 3.25 j3.03 4.444 43.00 0.05 + j0.00625 j0.05 0.05 j0.04375 0.06643 4.9 66.89 -.8 V 66.89 -.8 /(-j60) 0.48 88.9 Example 6 Determine the delta equivalent f the star cnnected netwrk shwn. j60 Ω S Ω Y Y Y Netwrk Therems- a c examples rfessr J R Lucas 4 Nvember 0
Y Y Y j60 j60 j60 j60 j60 j 2.5 + 7.5 + 60 +, Z 7.5 + j, Z 60 j 40 Ω j60 60 j40 j60 + 60 40 Example 7 Determine the star equivalent f the delta cnnected netwrk shwn., Z 40 j 60 Ω j60 7.5+ Ω S 40 j60ω 60 j40 Ω Z Z Z Z (7.5 + )( 40 j60) 7.5 + 40 j60 + 60 j40 26.575 48.8 22.603 3.9 37.5 j280 Z 5650 82.38.00 0.0 (same as riginal value in Ex 6). 282.5 82.37 (7.5 + )(60 j40) 7.5 + 40 j60 + 60 j40 26.575 48.8 22.603 4.9 37.5 j280 5650 7.62.00 89.99 Ω (same as riginal value in Ex 6). 282.5 82.37 (60 j40)( 40 j60) 22.603 4.9 22.603 3.9 Z 7.5 + 40 j60 + 60 j40 37.5 j280 450 72.38 60.00 90.0 j60 Ω (same as riginal value in Ex 6). 282.5 82.37 n rder t shw that the wrking is crrect, have selected the reverse prblem fr this example and used the results f the previus example t find the riginal quantities. Yu can see that the answers differ nly due t the cumulative calculatin errrs. Netwrk Therems- a c examples rfessr J R Lucas 5 Nvember 0
Example 8 Determine using cmpensatin therem, the current, if the available capacitr is µf, instead f the 9.89 µf already assumed in the earlier prblems. Ω 2 E 00 0 V 9.89 µf j60 Ω E 2 70-30 V Slutin µf crrespnds t 0 6 2 π 50 -j59.5 Ω change f impedance Z j59.5 ( j60) j0.85 Ω. frm earlier calculatins 0.48-88.2 using cmpensatin therem,. Z 0.48-88.2 j0.85 0.355.8 V changes in current in the netwrk can be btained frm Nte that the directin f is marked in the same directin as the riginal, s that the surce wuld in fact send a current in the ppsite directin. i.e. 0.355.8 j59.5 + // 0.355.8 j59.5 + + 0.355.8 j59.5 + 0 + j0 0.355.8 µf j59.5 Ω Ω 2 0.355.8 49.48 86.6 0.00237 88.0, giving as 0.00237 88.0, r 0.00237 268.0 i.e. crrect current 0.48-88.2 + 0.00237 268.0 0.03 j 0.477 0.00008 j0.00237 0.03 j 0.4 0.4-88.2 mparing result using Thevenin s equivalent circuit derived in example 3 0 + j 0 Ω Frm this circuit, it fllws that V µf j59.5 Ω 0 + j0 j59.5 49.48 86.6 0.4-88.2 which is the same result. Netwrk Therems- a c examples rfessr J R Lucas 6 Nvember 0