The equation of A-C- voltage is E-220sin - t- -6 and the A-C- current is I-10sin - t- -6 - The average power dissipated is -

The correct option is B 550 WWe know that , #160; Z=E_0/I_0 Given, #160; E_0=220V and #160; I_0=10A so #160; Z=220/10=22 ohm ϕ= [ ...

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Standard XII

Physics

Power in an AC Circuit

The equation ...

Question

The equation of A.C. voltage is $E = 220 sin (ω t + π / 6)$ and the A.C. current is $I = 10 sin (ω t - π / 6)$ . The average power dissipated is :

150 W

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550 W

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250 W

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50 W

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Solution

The correct option is B 550 W
We know that , $Z = \frac{E_{0}}{I_{0}}$
Given, $E_{0} = 220 V$ and $I_{0} = 10 A$
so $Z = \frac{220}{10} = 22$ ohm
$ϕ = [\frac{π}{6} - (- \frac{π}{6})] = \frac{π}{3}$
$p_{a} = \frac{E_{0}}{\sqrt{2}} \times \frac{I_{0}}{\sqrt{2}} \times cos ϕ$
$= \frac{220}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times cos \frac{π}{3} = 550 W$

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The equation of A.C. voltage is E=220sin(ωt+π/6) and the A.C. current is I=10sin(ωt−π/6). The average power dissipated is :

The correct option is B 550 WWe know that , Z=E0I0Given, E0=220V and I0=10Aso Z=22010=22 ohmϕ=[π6−(−π6)]=π3pa=E0√2×I0√2×cosϕ=220√2×10√2×cosπ3=550W

The equation of A.C. voltage is $E = 220 sin (ω t + π / 6)$ and the A.C. current is $I = 10 sin (ω t - π / 6)$ . The average power dissipated is :