(b) We know that for step down transformer
$V_{p} > V_{s}$ but $\frac{V_{p}}{V_{s}} = \frac{i_{s}}{i_{p}} \Rightarrow i_{s} > i_{p}$
Current in the secondary coil is greater than the primary
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.