NUMERICAL EXAMPLK 23 



Impedance. Admittance. Phase Angle. 



7 ... 319 0-00314 - 90 



... 18-6 0-0537 + 57 30' 



... 24-35 0-0411 +70 49' 



Applying, as before, the rule for the composition of parallel 

 admittances, we have 



0-00314 cos 90 -f 0-0537 cos 57 30' + 0-0411 cos 70 49' = 0-0424 



and 



- 0-00314 sin 90 + 0*0537 sin 57 30' + 0-0411 sin 70 49' = 0*081 



Thus the joint admittance is \/0-0424 a + 0'081 a = 0-0914, the 

 joint impedance 10'94, and the joint phase angle tan" 1 nTrnrjr 



= -f 62 22'. 



Having obtained the joint impedance between L and M, and also 

 that between M and N, and the corresponding joint phase angles, 

 we next proceed to compound these two impedances, according to 

 the law for the composition of series impedances. We thus find 



Joint resistance between L and M = 62'3 cos 54 = 36'62 



M N = 10-94 cos 62 22' = 5'08 



Total resistance between L and N = 41*70 

 Similarly 



Joint reactance between L and M = 62'3 sin 54 = 50'41 

 M N = 10-94 sin 62 22' = 9'69 



Total reactance between L and N = 6010 



We now find for the total -impedance between L and N the 

 value \/41-7 a 4- 60'1 2 = 73-15 ohms, and for the total phase angle 



tan-' |i = + 55 15'. 



The remaining part of the problem presents but little difficulty. 

 The total current is 7077 e = 6'84 amperes. The p.d. across LM is 



obtained by multiplying the total current by the impedance between 

 L and M, and similarly the p.d. across MN" is found by multiplying 

 the total current by the impedance between M and N. 



P.d. across LM = 6'84 x 62'3 = 426 volts 

 P.d. across MN = 6'84 x 10 94 = 75 volts 



