Question #204198

The voltage applied to a circuit is v=100 sin(wt+30 º) and current flowing in the circuit is i= 15 sin(wt+60 º). induced emf. Determine: 1. Impedance 2.Phase angle between voltage and current 3.Active Power 4.Power Factor.

1
Expert's answer
2021-06-08T09:35:07-0400

Gives

V=100sin(wt+30°)V=100sin(wt+30°)

I=15sin(wt+60°)I=15sin(wt+60°)

Impedance (Z)

ZR2+(XLXc)2Z\sqrt{R^2+{(X_L-X_c})^2}

XL=Xc=0X_L=X_c=0

Then

Z=R=10015=6.66Ω\frac{100}{15}=6.66\Omega

Phase difference


ϕ=(wt+60wt+30)=30°∆\phi=(wt+60-wt+30)=30°

Active power

P=IV2cosϕP=\frac{IV}{2}cos\phi

Put value

P=100×152cos30°P=\frac{100\times15}{2}cos30°

P=650 W

Power factor

Cosϕ=RZCos\phi =\frac{R}{Z}

R=VIR=\frac{V}{I}

R=10015R=\frac{100}{15}

Cosϕ=6.666.66=1Cos\phi=\frac{6.66}{6.66}=1

ϕ=cos11=0°\phi=cos^{-1}1=0°

Power factorϕ=0°\phi=0°


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