In the R – C circuit shown below,

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The power factor of the circuit is $\frac{3}{\sqrt{10}}$

The phase difference between the current in the source and source voltage is $\frac{π}{4}$

The power factor of the circuit is $\frac{1}{\sqrt{2}}$

The average power dissipated is 300 watt

answer is A, C.

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Detailed Solution

$\begin{array}{l} I_{1} = \frac{E_{0}}{100} \sin ωt = 2 \sin ωt; I_{2} = \frac{E_{0}}{100 \sqrt{2}} \sin (ωt + \frac{π}{4}) = \sqrt{2} \sin (ωt + \frac{π}{4}) \\ I_{net} = I_{0} \sin (ωt + ϕ) \\ I_{0} = \sqrt{(2 + 1)^{2} + (1)^{2}} = \sqrt{9 + 1} = \sqrt{10} \\ \tan ϕ = - \frac{1}{3} \Rightarrow ϕ = \tan^{- 1} (\frac{- 1}{3}) = - \tan^{- 1} (\frac{1}{3}) \end{array}$

$\begin{array}{l} \cos ϕ^{'} = \frac{3}{\sqrt{10}} E = 200 \sin ωt \Rightarrow I = \sqrt{10} \sin [ωt - \tan^{- 1} (\frac{1}{3})] \\ P_{avg} = \frac{1}{2} \times 200 \times \sqrt{10} \cos ϕ = \frac{1}{2} \times 200 \times \sqrt{10} \times \frac{3}{\sqrt{10}} = 300 watt \end{array}$