# Network Theorems - Alternating Current examples - J. R. Lucas

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1 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 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 j and - j X /( j 2π ) j 60 Ω Using Kirchff s current law 2 Using Kirchff s vltage law 00 0 j60 ( 2 ) 5 (-j8) +j () and j 2 j 60 ( 2 ) 7-30 j 6 + j (2) multiplying equatin () by 7 and equatin (2) by 4 and subtracting gives (7 j 56 + j 64) j 4 (7 + j 8) i.e substituting in (), j8 2 5 (-j8) i.e j i.e Thus the required current is j j j 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

2 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 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 µf j60 Ω E 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 , j j Ω, Ω surce current, , , Using the current divisin rule (nte directins f currents and signs), , j40 j , Using superpsitin therem, the ttal current in j j j which is the same answer btained in the earlier example. Netwrk Therems- a c examples rfessr J R Lucas 2 Nvember 0

3 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 V nsider the capacitr discnnected at and. urrent flwing in the circuit under this cnditin j Thevenin s vltage surce 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 j 0 Frm this circuit, it fllws that 0 + j0 j 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 V nsider the capacitr shrt-circuited at and. The Nrtn s current surce is given as 3.25 j j3.03 Netwrk Therems- a c examples rfessr J R Lucas 3 Nvember 0

4 Nrtn s admittance j0. 05 S 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 j j0 j 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 V N V SN Y. V Y j60 j j j j j j V /(-j60) 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

5 Y Y Y j60 j60 j60 j60 j60 j , Z j, Z 60 j 40 Ω j60 60 j40 j Example 7 Determine the star equivalent f the delta cnnected netwrk shwn., Z 40 j 60 Ω j Ω S 40 j60ω 60 j40 Ω Z Z Z Z (7.5 + )( 40 j60) j j j280 Z (same as riginal value in Ex 6) (7.5 + )(60 j40) j j j Ω (same as riginal value in Ex 6) (60 j40)( 40 j60) Z j j j j60 Ω (same as riginal value in Ex 6) 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

6 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 V Slutin µf crrespnds t π 50 -j59.5 Ω change f impedance Z j59.5 ( j60) j0.85 Ω. frm earlier calculatins using cmpensatin therem,. Z j 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 j // j j j µf j59.5 Ω Ω , giving as , r i.e. crrect current j j j mparing result using Thevenin s equivalent circuit derived in example j 0 Ω Frm this circuit, it fllws that V µf j59.5 Ω 0 + j0 j which is the same result. Netwrk Therems- a c examples rfessr J R Lucas 6 Nvember 0

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