Question

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

Answer 1

Solution;

Since;

$P=VI$

Given;

$P=100W$

$V=120V$

$I=\frac PV=\frac{100}{120}=0.833A$

Also;

$X_c=\frac{1}{2πfC}$

Given;

$C=80×10^{-6}F$

$f=50Hz$

$X_c=\frac{1}{2π×50×80×10^{-6}}=39.789\Omega$

And;

$Z=\frac{V}{I}=\sqrt{R^2+X_c^2}$

$Z=\frac{120}{0.833}=\sqrt{R^2+39.789^2}$

$144.058=\sqrt{R^2+39.789^2}$

$20752.6=R^2+1583.165$

$R^2=19169.44$

$R=138.45\Omega$