Electro Magnetic Induction And Alternating Currents Question 72

Question: A coil has L = 0.04 H and $ R=12,\Omega $ . 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=\sqrt{R^{2}+4{{\pi }^{2}}{{\nu }^{2}}L^{2}} $

$ =\sqrt{{{(12)}^{2}}+4\times {{(3.14)}^{2}}\times {{(50)}^{2}}\times (0.04)} $ = 17.37

A Now current $ i=\frac{V}{Z} $ $ =\frac{220}{17.37}=12.7\Omega $



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