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1- A single-phase 120 V, 60 Hz source is supplying two parallel loads one with an
impedance of 5245 ohm and the other one 6.260 ohm.
a) Find the real (P) and reactive (Q) power supplied by the source and the (resultant) input
power factor.
b) Ifa capacitor is connected in parallel to the loads and supplies 2000Var, Find P and Q
supplied by the source and resultant power factor
c) Find the value of the capacitor which should be added to original two parallel loads in
order for the power factor to be 0.95 lagging.
a) P= OKA cO845 + 120x129 _ cos 60
total 5 2
Prot 2 3236 W b 4 120x120 %n60
2 120N20 4% y
Cr otal “ :
= 4S Vol
ae ares a 62
PFe Paotal
_~ an
bd) aaw=2 - ame =
total 337
PF we CO (tan aoe Daman
P otal
/
C) ~ : RQ reud
can (OS (0.95))= > a
tot
= Anew ~ 1063-6 YQ, 305! {
2. at 4 Fl ohm Cx wWryTl
3- A 1OKVA, 8000/240 V, 60 HZ, single-phase transformer has a per unit resistance of
0.012 pu and a per unit reactance of 0.04 pu. The excitation branch elements are Xm=30
pu and R.=S0pu
a) find the transformer’s equivalent circuit referred to the low voltage side
b) If the primary of this transformer is connected to a constant 8000 V ac source and if a
6230’ Q load is connected to the secondary, what will be the voltage, current and power
at the load?
_(240N _ 20-76 ohm
base 0 10,000
= 0.069)2 490.2304
“4 ¥ side
HBOX Ze pp cgl720B = A%y oml-V Sia
288 ohm “Ew site
(0.01% 450-04) rey ase
50 x pases lays
0.06012 40-2304
YV
4A single-phase 300-kVA, 220/4400 V, 60 Hz transformer yielded the following
information when tested:
High voltage winding open: Voltage=220 V, Current=40 A Power=1000W
Low voltage terminal shorted: Voltage=195 V, Current=68.18A Power=4000W
a) Find the equivalent circuit of the transformer as viewed from the high voltage side.
b) Calculate the efficiency of the transformer when it delivers its rated load at rated terminal
voltage and 0.8 power factor lagging.
From Open cicest test calvlak &
Re aN Kw referred To LV Sid OF primary
c
z
Re= ve 22 BOG OHM
Poe. (a@o9
Xen = a _ 5536
—
\ ( 4O yh ~
22.0 R.*
referred to WW side
a> Re = 19360 ohm
Xm = RANG. de ohm
From short crlcurt test , Req end Key.
ce ferred to Hv $tde of secondary
Req + 000 = 0-86 Cha.
+
eg = |/( 2%, > - Ree a2 F272 chm
0-86 52.3727
AW
we
£ \9 340 224.4 ws
= Boo -}
trate { = ——— <- 9 o &
44-99
= 6R1R <-36.87
a”
Same cypenl a5 Fes f-
. B00 x0 Se 98 /
3 00x OF 4 4090941090 ‘
CIN ™ ee Ne a ae — en Oe a
or OFner Wag W
Poy = = Keg, eu *~ C Ftedytqe) = 4-909
v- Vor (keg ¢ I* <4) Lrited
“p=
a Z. 0 >
a (@)
560 L143 fen EL elOT4
me 2 ES fe
J 2 —.% 98 /
300 xO. b+ 4000 1107
