Question 324: AC fundamentals

Question

For the series $L C R$ circuit, $R=\frac{X_L}{2}=2 \mathrm{X}_{\mathrm{c}}$. The impedance of the circuit and the phase difference between V and I will be

Accepted Answer

$\operatorname{Impedance}(\mathrm{Z})=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}ight)^2}\quad	ext{.... (i)}$
Given $R=\frac{X_L}{2} \Rightarrow X_L=2 R\quad	ext{.... (ii)}$
Also, $R=2 X_C \Rightarrow X_C=\frac{R}{2}\quad	ext{.... (iii)}$.
Substituting (ii) and (iii) in (i),
$$Z=\sqrt{R^2+\left(2 R-\frac{R}{2}ight)^2}=\sqrt{\frac{13}{4} R^2}=\frac{\sqrt{13}}{2} R$$
Phase difference $\phi=	an ^{-1}\left(\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}ight)=	an ^{-1}\left(\frac{3}{2}ight)$

Answer

$\frac{\sqrt{5}}{2} \mathrm{R}, 	an ^{-1}\left(\frac{1}{2}ight)$

Answer

$\frac{\sqrt{13}}{2} R, 	an ^{-1}\left(\frac{3}{2}ight)$

Answer

$\sqrt{5} R, 	an ^{-1}(1)$

Answer

$\sqrt{13} \mathrm{R}, 	an ^{-1}(2)$

AC fundamentals

For the series L C R circuit, R=\frac{X_L}{2}=2 \mathrm{X}_{\mathrm{c}}. The impedance of the circuit and the phase difference between V and I will be

Choose the correct answer: