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

Answer 1

Answer: (C)

$X_C=\frac{1}{\omega C}=\frac{1}{340\times 50\times {10}^{-6}}=58.8\Omega$

$X_L=\omega L=340\times 20\times {10}^{-3}=6.8\Omega$

$Z=\sqrt{R^2+(X_C-X_L{)}^2}$

$=\sqrt{{40}^2+(58.8-6.8{)}^2}$

$=\sqrt{4304}\Omega$

Power, $P=i_{rms}^{2}R={\left(\frac{V_{rms}}{Z}\right)}^{2}R$

$={\left(\frac{10/\sqrt{2}}{\sqrt{4304}}\right)}^2\times 40$

$=\frac{50\times 40}{4304}\approx 0.51W$