A series circuit consisting of a variable resistor in series with a capacitance of 80uF is connected across a 120V, 50hz supply. To what value should the resistor R be adjusted so that the power absorbed by the series circuit will be 100W
Solution;
Since;
P=VIP=VIP=VI
Given;
P=100WP=100WP=100W
V=120VV=120VV=120V
I=PV=100120=0.833AI=\frac PV=\frac{100}{120}=0.833AI=VP=120100=0.833A
Also;
Xc=12πfCX_c=\frac{1}{2πfC}Xc=2πfC1
C=80×10−6FC=80×10^{-6}FC=80×10−6F
f=50Hzf=50Hzf=50Hz
Xc=12π×50×80×10−6=39.789ΩX_c=\frac{1}{2π×50×80×10^{-6}}=39.789\OmegaXc=2π×50×80×10−61=39.789Ω
And;
Z=VI=R2+Xc2Z=\frac{V}{I}=\sqrt{R^2+X_c^2}Z=IV=R2+Xc2
Z=1200.833=R2+39.7892Z=\frac{120}{0.833}=\sqrt{R^2+39.789^2}Z=0.833120=R2+39.7892
144.058=R2+39.7892144.058=\sqrt{R^2+39.789^2}144.058=R2+39.7892
20752.6=R2+1583.16520752.6=R^2+1583.16520752.6=R2+1583.165
R2=19169.44R^2=19169.44R2=19169.44
R=138.45ΩR=138.45\OmegaR=138.45Ω
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