Alternating Current Ques 9
9. 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 :
[2016]
(a) $0.51$ $ W$
(b) $0.67$ $ W$
(c) $0.76 $ $W$
(d) $0.89 $ $W$
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Answer:
Correct Answer: 9.(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} $
BETA
