Average power in an A-C circuit is 50 watt- Resistance of circuit is 32 - and source voltage is 50V- Power factor of this circuit is

Average power lt;P gt; = i_rms^2 R Since i_rms = v_rms cos ϕR ∴ lt;p gt; = v_rms^2 cos^2 ϕR^2 . R ⇒cos^2 ϕ = lt;p gt; Rv_rms^2 cosϕ = √(50 × 32)/50 ...

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Byju's Answer

Standard XI

Physics

Power Factor

Average power...

Question

Average power in an A.C circuit is 50 watt. Resistance of circuit is $32 Ω$ , and source voltage is 50V. Power factor of this circuit is

0.8

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0.6

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\frac{3}{5}

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\frac{3}{4}

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Solution

The correct option is A $0.8$
Average power $< P >= i_{r m s}^{2} R$
Since $i_{r m s} = \frac{v_{r m s} c o s ϕ}{R}$
$∴< p >= \frac{v_{r m s}^{2} c o s^{2} ϕ}{R^{2}} . R \Rightarrow {cos}^{2} ϕ = \frac{< p > R}{v_{r m s}^{2}}$
$cos ϕ = \frac{\sqrt{50 \times 32}}{50} = \frac{4}{5} = 0.8$

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Average power in an A.C circuit is 50 watt. Resistance of circuit is 32 Ω, and source voltage is 50V. Power factor of this circuit is

The correct option is A 0.8Average power <P>=i2rmsR Since irms=vrms cosϕR ∴<p>=v2rmscos2ϕR2.R⇒cos2ϕ=<p>Rv2rms cosϕ=√50×3250=45=0.8

Average power in an A.C circuit is 50 watt. Resistance of circuit is $32 Ω$ , and source voltage is 50V. Power factor of this circuit is

The correct option is A $0.8$
Average power $< P >= i_{r m s}^{2} R$
Since $i_{r m s} = \frac{v_{r m s} c o s ϕ}{R}$
$∴< p >= \frac{v_{r m s}^{2} c o s^{2} ϕ}{R^{2}} . R \Rightarrow {cos}^{2} ϕ = \frac{< p > R}{v_{r m s}^{2}}$
$cos ϕ = \frac{\sqrt{50 \times 32}}{50} = \frac{4}{5} = 0.8$