NEET Solved Paper 2016 Question 40
Question: An inductor 20 mH, a capacitor $ 50\mu F $ and a resistor $ 40\ \Omega $ are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is :-
Options:
A) 0.51 W
B) 0.67 W
C) 0.76 W
D) 0.89 W
Show Answer
Answer:
Correct Answer: A
Solution:
$ X _{C}=\frac{1}{\omega C}=\frac{1}{340\times 50\times {10^{-6}}}=58.8\ \Omega $
$ X _{L}=\omega L=340\times 20\times {10^{-3}}=6.8\ \Omega $
$ Z=\sqrt{R^{2}+{{(X _{C}-X _{L})}^{2}}} $
$ =\sqrt{40^{2}+{{(58.8-6.8)}^{2}}}=\sqrt{4304\Omega } $
$ P=i_rms^{2}R={{( \frac{V _{rms}}{Z} )}^{2}}R $ $ ={{( \frac{10/\sqrt{2}}{\sqrt{4304}} )}^{2}}\times 40=\frac{50\times 40}{4304}=0.47\ W $
So best answer (nearest answer) will be 0.51 W
BETA
