Electro Magnetic Induction And Alternating Currents Question 254
Question: The coils of a step down transformer have 500 and 5000 turns. In the primary coil an ac of 4 ampere at 2200 volts is sent. The value of the current and potential difference in the secondary coil will be [MP PET 1996]
Options:
A) 20 A, 220 V
B) 0.4 A, 22000 V
C) 40 A, 220 V
D) 40 A, 22000 V
Show Answer
Answer:
Correct Answer: C
Solution:
$ \frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}=\frac{i_{s}}{i_{p}} $ . The transformer is step-down type, so primary coil will have more turns. Hence $ \frac{5000}{500}=\frac{2200}{V_{s}}=\frac{i_{s}}{4}\Rightarrow V_{s}=220\ V,\ i_{s}=40\ amp $
