Question

An inductor $10\Omega /{60}^{\circ }$ is connected to a $5\Omega$ resistance in series. Find net impedance.

• A. $15\Omega$
• B. $12\Omega$
• C. $13.2\Omega$
• D. $18\Omega$

Answer 1

The correct option is C $13.2\Omega$

$10=\sqrt{r^2+X_L^2}and\frac{X_L}{r}=tan60$
$10=\sqrt{r^2+(r\sqrt{3}{)}^2}orr=5\Omega ,X_L=5\sqrt{3}\Omega .$
$Z=\sqrt{(5+5{)}^2+(5\sqrt{3}{)}^2}=\sqrt{175}=13.2\Omega$
$tan\Phi =\frac{X_L}{R+r}=\frac{5\sqrt{3}}{10}$
or$\Phi =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$