(3) 12.5 mA
The secondary, supplying a voltage of 22V, operates a device of impedance 200 Ohms.
Hence current in the secondary 22/220 A = 0.1 A
∴ Power consumed in the secondary circuit = 22 x 0.1, W = 2.2 W
Efficiency = 80%
Power available in the primary circuit = 2.2 x 100/80 W = 2.75 W
∴ If I is the current in the primary coil, we have