An ideal transformer has primary:secondary turns 1:100. The primary is driven by 100 V rms, 50 Hz through the shown reactance. Find rms input current I in amperes.
Reflect the secondary impedance to the primary using the squared turns ratio: $(1/100)^2(80000-j40000)=8-j4$ Ω.
Add the primary series inductive reactance j10 Ω to get Zin=8+j6 Ω, of magnitude 10 Ω. The rms primary current is therefore 100/10=**10 A**, within the official 9.5–10.5 interval.
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