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 $



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