Question

The current through an inductor of impedance $10 \, \Omega$ lags behind the voltage by a phase of $60^\circ $ when just the inductor is connected to the AC source. Now the inductor is connected to a $5\Omega$ resistance in series, then the net impedance of the circuit is

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

Answer 1

Answer: (C)

$10=\sqrt{r^{2} + X_{L}^{2}}$ and $\frac{X_{L}}{r}=tan 60^\circ $
$10=\sqrt{r^{2} + \left(r \sqrt{3}\right)^{2}}$ or $r=5\Omega, \, X_{L}=5\sqrt{3}\Omega$
$Z=\sqrt{\left(5 + 5\right)^{2} + \left(5 \sqrt{3}\right)^{2}}=\sqrt{175}=13.2 \, \Omega$