Electro Magnetic Induction And Alternating Currents Question 67
Question: An inductor of inductance L and resistor of resistance R are joined in series and connected by a source of frequency $ \omega $ . Power dissipated in the circuit is [AIEEE 2002; RPET 2003]
Options:
A) $ \frac{(R^{2}+{{\omega }^{2}}L^{2})}{V} $
B) $ \frac{V^{2}R}{(R^{2}+{{\omega }^{2}}L^{2})} $
C) $ \frac{V}{(R^{2}+{{\omega }^{2}}L^{2})} $
D) $ \frac{\sqrt{R^{2}+{{\omega }^{2}}L^{2}}}{V^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ P=Vi\cos \varphi =V\ ( \frac{V}{Z} )\ ( \frac{R}{Z} )=\frac{V^{2}R}{Z^{2}} $ $ =\frac{V^{2}R}{(R^{2}+{{\omega }^{2}}L^{2})} $
BETA
