For a series R L C circuit, R=XL=2 XC The impedance of the circuit and phase difference between V and i will be

Question

For a series R L C circuit, R=XL=2 XC The impedance of the circuit and phase difference between V and i will be (A) (√5 R/2), tan -1(2) (B) (√5 R/2), tan -1((1/2)) (C) √5 XC, tan -1(2) (D) √5 R, tan -1((1/2))

Tardigrade Admin · Accepted Answer

XL=R, XC=R / 2 ∴ tan φ=(XL-XC/R)=(R-(R/2)/R)=(1/2) ⇒ φ= tan -1(1 / 2) Also, Z=√R2+(XL-XC)2 =√R2+(R2/4)=(√5/2) R

Q. For a series $R L C$ circuit, $R = X_{L} = 2 X_{C}$ The impedance of the circuit and phase difference between $V$ and $i$ will be

$\frac{5 R}{2}, tan^{- 1} (2)$

$\frac{5 R}{2}, tan^{- 1} (\frac{1}{2})$

$5 X_{C}, tan^{- 1} (2)$

$5 R, tan^{- 1} (\frac{1}{2})$

$X_{L} = R, X_{C} = R /2$
$∴ tan ϕ = \frac{X _{L} - X _{C}}{R} = \frac{R - \frac{R}{2}}{R} = \frac{1}{2}$
$\Rightarrow ϕ = tan^{- 1} (1/2)$
Also, $Z = R^{2} + (X_{L} - X_{C})^{2}$
$= R^{2} + \frac{R ^{2}}{4} = \frac{5}{2} R$

Q. For a series RLC circuit, R=XL​=2XC​ The impedance of the circuit and phase difference between V and i will be

25​R​,tan−1(2)

25​R​,tan−1(21​)

5​XC​,tan−1(2)

5​R,tan−1(21​)