A series LCR circuit is connected across an AC source E=10 sin100πt−π6. Current in the circuit is I=2sin100πt+π12. What is the average power dissipated ?

ϕ=π12−−π6=π4Avg power =Vrmsirmscosϕ=102×22×12=52W

A series LCR circuit is connected across an AC source E=10 sin100πt−π6. Current in the circuit is I=2sin100πt+π12. What is the average power dissipated ?

ϕ=π12−−π6=π4Avg power =Vrmsirmscosϕ=102×22×12=52W

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A series LCR circuit is connected across an AC source $E = 10 \sin (100 π t - \frac{π}{6})$ . Current in the circuit is $I = 2 \sin (100 π t + \frac{π}{12})$ . What is the average power dissipated ?

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$5 W$

$\frac{5}{\sqrt{2}} W$

$\sqrt{2} W$

$5 \sqrt{2} W$

answer is C.

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Detailed Solution

$ϕ = \frac{π}{12} - (\frac{- π}{6}) = \frac{π}{4}$

Avg power $= V_{r m s} i_{r m s} \cos ϕ$

$= \frac{10}{\sqrt{2}} \times \frac{2}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 5 \sqrt{2} W$

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A series LCR circuit is connected across an AC source E=10 sin100πt−π6. Current in the circuit is I=2sin100πt+π12. What is the average power dissipated ?

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A series LCR circuit is connected across an AC source $E = 10 \sin (100 π t - \frac{π}{6})$ . Current in the circuit is $I = 2 \sin (100 π t + \frac{π}{12})$ . What is the average power dissipated ?