Question

An inductor $20mH$, a capacitor $50\mu F$ and a resistor $40\Omega$ are connected in series across a source of emf $V=10\sin 340t$. The power loss in A.C. circuit is :

• A. 0.67 W
• B. 0.76 W
• C. 0.89 W
• D. 0.51 W

Answer 1

Answer: (D)

$X_C=\frac{1}{\omega C}=\frac{1}{340\times 50\times 10^{-6}}=58.8\Omega$
$X_L=\omega L=340\times 20\times 10^{-3}=6.8\Omega$
$Z=\sqrt{R^2+{\left(X_C-X_L\right)}^2}$
$=\sqrt{40^2+(58.8-6.8{)}^2}$
$=\sqrt{4304}\Omega$
Power, $P=i_{rms}^{2}R={\left(\frac{V_{rms}}{Z}\right)}^{2}R$
$={\left(\frac{10/\sqrt{2}}{\sqrt{4304}}\right)}^2\times 40$
$=\frac{50\times 40}{4304}\approx 0.51W$