# An LCR circuit has L = 10 mH. R = 3 ohm and C = 1μF connected in series to a source of 15 cosωt volt. What is average power dissipated per cycle…

Q: An LCR circuit has L = 10 mH. R = 3 ohm and C = 1μF connected in series to a source of E = 15 cosωt volt. What is average power dissipated per cycle at a frequency that is 10% lower than the resonant frequency ?

Sol: Here, L = 10-2 H, R = 3 Ω, C = 10-6 F

Resonant frequency, $\large \omega_0 = \frac{1}{\sqrt{L C}}$

$\large \omega_0 = \frac{1}{\sqrt{10^{-2} \times 10^{-6}}}$

Actual frequency, ω = (90%)ω0

XL = ω L = 9 × 103 × 10-2 = 90 Ω

XC = 1/ωC = 1/(9 × 103 × 10-6) = 1000/9 ohm

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

$\large Z = \sqrt{3^2 + (\frac{1000}{9} – 90)^2 }$

Z = 21.3 ohm

Power dissipated / cycle = Ev Iv cosφ

$\large = E_v (\frac{E_v}{Z}) \frac{R}{Z} = (\frac{E_v}{Z})^2 R$

$\large = (\frac{E_0}{\sqrt{2}Z})^2 R = (\frac{15}{\sqrt{2}\times 21.3})^2 \times 3$

= 0.744 W