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
Show Answer
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 $
