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An inductor 20 mH, a capacitor 50 $\mu F$ and a resistor 40 $\Omega$ are connected in series across a source of emf $V = 10\sin340t$. The power loss in AC circuit is
A
0.67 W
B
0.76 W
C
0.89 W
D
0.51 W
Explanation
$P = I_{rms}^2R$ with Z from $X_L$ and $X_C$.
Detailed Solution
$X_L = \omega L = 340\times20\times10^{-3} = 6.8\ \Omega$
$X_C = \frac{1}{\omega C} = \frac{1}{340\times50\times10^{-6}} = 58.82\ \Omega$
$Z = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{1600 + 2704} = \sqrt{4304}\ \Omega$
$I_v = \frac{V_0}{\sqrt{2}}\times\frac{1}{Z} = \frac{10}{\sqrt{2}}\times\frac{1}{\sqrt{4304}}$
$P = I_v^2R = \frac{100}{2\times4304}\times40 = \frac{4000}{8608} = 0.51$ W
$X_C = \frac{1}{\omega C} = \frac{1}{340\times50\times10^{-6}} = 58.82\ \Omega$
$Z = \sqrt{40^2 + (58.82 - 6.8)^2} = \sqrt{1600 + 2704} = \sqrt{4304}\ \Omega$
$I_v = \frac{V_0}{\sqrt{2}}\times\frac{1}{Z} = \frac{10}{\sqrt{2}}\times\frac{1}{\sqrt{4304}}$
$P = I_v^2R = \frac{100}{2\times4304}\times40 = \frac{4000}{8608} = 0.51$ W