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}+{\left(2\pi \nu L\right)}^{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\text{ A} $
BETA
