Electro Magnetic Induction And Alternating Currents Question 174

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

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Answer:

Correct Answer: A

Solution:

  • Given: L=20mH; $ C=50\mu F $ ;

    $ R=40\Omega $

    $ V=10\sin 340t $

    $ \therefore V_{runs}=\frac{10}{\sqrt{2}} $

    $ 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 $

    Impedance, $ Z=\sqrt{R^{2}+{{(X_{C}-X_{L})}^{2}}} $

    $ =\sqrt{40^{2}+{{(58.8-6.8)}^{2}}}=\sqrt{4304}\Omega $

    Power loss in A.C. circuit, $ 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}\simeq 0.51W $



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