Question

An ideal transformer steps down 220 V to 22 V in order to operate a device with an impedance of 220 W. The

current in the primary is

• A. 0.01 A
• B. 0.1 A
• C. 0.5 A
• D. 0 A

Answer 1

Answer: (A)

0.01 A

Answer 2

V_p = 220 V, V_s = 22 V, R_s = 220 W secondary current$i_{s} = \frac{V_{s}}{R_{s}} = \frac{22}{220} = \frac{1}{10}amp$

So by using the relation $\frac{V_{p}}{V_{s}} = \frac{i_{s}}{i_{p}},i_{p} = 0.01A$