Question

An inductor 10 W/60⁰ is connected to a 5W resistance in series. Find net impedance –

• A. 15 W
• B. 12 W
• C. 13.2 W
• D. 18 W

Answer 1

Answer: (C)

13.2 W

Answer 2

10 =$\sqrt{r^{2} + X_{L}^{2}}$and $\frac{X_{L}}{r}$= tan 60

10 =$\sqrt{r^{2} + (r\sqrt{3})^{2}}$ or r = 5 W, X_L = 5$\sqrt{3}$W

Z =$\sqrt{(5 + 5)^{2} + (5 + \sqrt{3})^{2}}$=$\sqrt{175}$= 13.2 W

tan f =$\frac{X_{L}}{R + r}$=$\frac{5\sqrt{3}}{10}$

or f = tan^–1$\left( \frac{\sqrt{3}}{2} \right)$.