Question

How much current is drawn by the primary coil of a transformer which steps down 220 V to 22 V to operate a device with an impedance of 220?

Answer 1

I_s = \(\frac{V_s}{Z} = \frac{22}{220}\) 0.1 A

since V_pI_p = V_sI_s

∴ I_p = \(\frac{V_sI_s}{V_p} = \frac{22 \times 0.1}{220}\)  = 0.01 A