# Applied Electrotechnology for Engineers by C.H. Laycock

By C.H. Laycock

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Express both the voltage and current in complex form . Impedance Z = = Current I = V (R 2 + XL 2) V (9 + 16) = 5 n V 250 Z 5 50 A Phase angle ¢ arctan XL R 4 arctan 3 53° 7' lagging The current lags the voltage since the circuit is an inductive one. 9a. Because resistors or inductors in series may be added arithmetically the circuit . 9b. The complex impedance of Z is thus (a + c) + j (b + d). The total impedance of two series impedances is thus the complex sum of the individual impedances. 10a) 1 1 1 Z Z, h -;- = ...

Phasor addition of h and Ie gives (h - Ie) which when combined with IR gives the total current I as shown (assuming Xc> XL). By Pythagoras or II I = I V I x j [~2 + (;L - ;e) Hence IZ I IVI - II I =1J~,+GL 2J ;JJ! - 1 n and ¢ = arctan IL - Ie IR = arctan (I/XL) - (I/Xd l/R the phase angle lagging in this case since lt. > Ie (Xc> Xd. Again if the circuit is only a two-component one, say RL , we can use the above equations putting Xc equal to infinity. Similarly for RC parallel circuits we can put XL to infinity.

Voltage between any pair of lines. 26a it is clear that for a balanced load (equal loads in each phase), points Nand N' are at the same potential hence a neutral wire connecting the generator and load star po ints would carry no current. In practice this connection is usually included because of the difficulty of ensuring exact equality in the three loads. s. 3 Delta Connection .. 26b the sum of the three phase voltages VRN , VY N and VBN is clearly zero. 27a. Before connection to the load there will be no circulating currents in the generator coils because, being connected end-to-end, we have effectively summed the three voltages.