Question

A leaky capacitor $10\Omega /{60}^{\circ }$ is connected in series with a 10 $\Omega$ resistance. Find the overall impedance.

• A. $10\Omega$
• B. $10\sqrt{2}\Omega$
• C. $15\Omega$
• D. $10\sqrt{3}\Omega$

Answer 1

The correct option is D $10\sqrt{3}\Omega$

$\sqrt{r^2+X_c^2}=10\text{and}\frac{X_c}{r}=\tan 60$
$\therefore \sqrt{r^2+(\sqrt{3}r{)}^2}=10$ or $r=5\Omega ,X_c=5\sqrt{3}\Omega$

$|Z|=\sqrt{(10+5{)}^2+5\sqrt{3}}=10\sqrt{3}\Omega$;
$\Phi ={\tan }^{-1}\frac{X_c}{R}=ta{n}^{-1}\left(\frac{5\sqrt{3}}{15}\right)$
${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$ or $\Phi ={30}^{\circ }$