When the secondary of a 120/440 V transformer is open, primary current is 0.2
amps at a PF of .3. The transformer is a 5 kVA transformer. Find: (a) IP, (b) IE,
(c) IH, and (d) Im.
(a) Full load current
(b) IP at no load is equal to IE
IE = 0.2 amp
IE cos q
IE x PF
IE sin q
(0.2) sin 72.5°
The symbol for a transformer gives no indication of the phase of the voltage across the
secondary. The phase of that voltage depends on the direction of the windings around the core.
In order to solve this problem, polarity dots are used to show the phase of primary and secondary
signals. The voltages are either in phase (Figure 7a) or 180° out of phase with respect to primary
voltage (Figure 7b).