The current and voltage function in an AC circuit areI = 100 sin ⁡100t mAV=100sin⁡100t+π3 volt the power dissipated in the circuit is

P=VrmsIrmscosϕ and cosϕ=cosπ3=12=1002×1002×10−3×12=2.5W

The current and voltage function in an AC circuit areI = 100 sin ⁡100t mAV=100sin⁡100t+π3 volt the power dissipated in the circuit is

P=VrmsIrmscosϕ and cosϕ=cosπ3=12=1002×1002×10−3×12=2.5W

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The current and voltage function in an AC circuit are
I = 100 sin ⁡100t mA
$V = 100 \sin (100 t + \frac{π}{3}) volt$
the power dissipated in the circuit is

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10 W

2.5 W

5 W

5 kW

answer is B.

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Detailed Solution

$P = V_{rms} I_{rms} \cos ϕ a n d \cos ϕ = \cos \frac{π}{3} = \frac{1}{2}$

$= \frac{100}{\sqrt{2}} \times \frac{100}{\sqrt{2}} \times 10^{- 3} \times \frac{1}{2} = 2.5 W$

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The current and voltage function in an AC circuit areI = 100 sin ⁡100t mAV=100sin⁡100t+π3 volt the power dissipated in the circuit is

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The current and voltage function in an AC circuit are
I = 100 sin ⁡100t mA
$V = 100 \sin (100 t + \frac{π}{3}) volt$
the power dissipated in the circuit is