Question

A $9/100\pi$ inductor and a $12\omega$ resistance are connected in series to a $225V,50Hz$ ac source. Calculate the current (in $A$) in the circuit.

Answer 1

$z=\sqrt{{X_L}^2+R^2}$

$=\sqrt{(\frac{9}{100\pi }\times 2\pi \times 50{)}^2+{12}^2}$

$=\sqrt{9^2+{12}^2}$

$=\sqrt{81+144}=\sqrt{225}=15\Omega$

$I=\frac{V}{Z}$

$=\frac{225}{15}$

$=15A$