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})} $



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