Alternating Current - Result Question 15
15. An inductor $20 mH$, a capacitor $50 \mu F$ and a resistor $40 \Omega$ are connected in series across a source of emf $=10 \sin 340 t$. The power loss in A.C. circuit is :
(a) $0.51 W$
(b) $0.67 W$
(c) $0.76 W$
(d) $0.89 W$
[2016]
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Answer:
Correct Answer: 15. (a)
Solution:
- (a) Given: $L=20 mH ; C=50 \mu F ; R=40 \Omega$
$V=10 \sin 340 t$
$\therefore \quad V _{\text{runs }}=\frac{10}{\sqrt{2}}$
$X_C=\frac{1}{\omega C}=\frac{1}{340 \times 50 \times 10^{-6}}=58.8 \Omega$
$X_L=\omega L=340 \times 20 \times 10^{-3}=6.8 \Omega$
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,
$ \begin{aligned} & P=i _{\text{rms }}^{2} R=(\frac{V _{\text{rms }}}{Z})^{2} R \\ & =(\frac{10 / \sqrt{2}}{\sqrt{4304}})^{2} \times 40=\frac{50 \times 40}{4304} \simeq 0.51 W \end{aligned} $
