Alternating Current - Result Question 13
13. An inductor $20 mH$, a capacitor $100 \mu F$ and a resistor $50 \Omega$ are connected in series across a source of emf, $V=10 \sin 314 t$. The power loss in the circuit is
(a) $0.79 W$
(b) $0.43 W$
(c) $1.13 W$
(d) $2.74 W$
[2018]
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Answer:
Correct Answer: 13. (a)
Solution:
- (a) Power dissipated in an LCR series circuit connected to an a.c. source of emf E
$P=E _{\text{rms }} i _{rms} \cos \phi=\frac{E _{r m s}^{2} R}{Z^{2}}=\frac{E _{r m s}^{2} R}{R^{2}+(\omega L-\frac{1}{C \omega})^{2}}$
$=\frac{(\frac{10}{\sqrt{2}})^{2} \times 50}{(50)^{2}+(314 \times 20 \times 10^{-3}-\frac{1}{314 \times 100 \times 10^{-6}})^{2}}$
Solving we get, $P=0.79 W$
