A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated?

A series L-C-R circuit is connected across an AC source E = `10sin[100

1.	A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated?
Answer» Phase difference between voltage and current. `phi = pi/12-((-pi)/6)=pi/4` `rArr` Power factor `=cosphi = 1/sqrt(2)` Average power dissipated, `(V_(m)I_(m))/2 cosphi = (10 xx 2)/2 xx 1/sqrt(2) = 5sqrt(2)`W

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A series L-C-R circuit is connected across an AC source E = `10sin[100pit - pi/6]`. Current from the supply is I = `2sin[100pit + pi/12]`, What is the average power dissipated?

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