Electro Magnetic Induction And Alternating Currents Question 72

Question: A coil has L = 0.04 H and R=12,Ω . When it is connected to 220V, 50Hz supply the current flowing through the coil, in amperes is [Kerala PMT 2004]

Options:

A) 10.7

B) 11.7

C) 14.7

D) 12.7

Show Answer

Answer:

Correct Answer: D

Solution:

Impedance Z=R2+4π2ν2L2

=(12)2+4×(3.14)2×(50)2×(0.04) = 17.37

A Now current i=VZ =22017.37=12.7Ω



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