A sinusoidal voltage V ( t )=210 sin 3000 t volt is applied to a series LCR circuit in which L =10 mH , C =25 μ F and R =100 Ω. The phase difference (Φ) between the applied voltage and resultant current will be :

Question

A sinusoidal voltage V ( t )=210 sin 3000 t volt is applied to a series LCR circuit in which L =10 mH , C =25 μ F and R =100 Ω. The phase difference (Φ) between the applied voltage and resultant current will be : (A)  tan -1(0.17) (B)  tan -1(9.46) (C)  tan -1(0.30) (D)  tan -1(13.33)

Tardigrade Admin · Accepted Answer

X L =10-2 × 3000=30 Ω X C =(1/3000 × 25 × 10-6)=(40/3) Ω X = X L - X C =30-(40/3)=(50/3) tan δ=( X / R )=(50/3 × 100)=(1/6) δ= tan -1((1/6))= tan -1(0.17)

Q. A sinusoidal voltage $V (t) = 210 sin 3000 t$ volt is applied to a series $L CR$ circuit in which $L = 10$ $m H, C = 25 μ F$ and $R = 100 Ω$ . The phase difference $(Φ)$ between the applied voltage and resultant current will be :

$tan^{- 1} (0.17)$

$tan^{- 1} (9.46)$

$tan^{- 1} (0.30)$

$tan^{- 1} (13.33)$

$X_{L} = 1 0^{- 2} \times 3000 = 30 Ω$
$X_{C} = \frac{1}{3000 \times 25 \times 1 0 ^{- 6}} = \frac{40}{3} Ω$
$X = X_{L} - X_{C}$
$= 30 - \frac{40}{3} = \frac{50}{3}$
$tan δ = \frac{X}{R} = \frac{50}{3 \times 100} = \frac{1}{6}$
$δ = tan^{- 1} (\frac{1}{6}) = tan^{- 1} (0.17)$

Q. A sinusoidal voltage V(t)=210sin3000t volt is applied to a series LCR circuit in which L=10 mH,C=25μF and R=100Ω. The phase difference (Φ) between the applied voltage and resultant current will be :

tan−1(0.17)

tan−1(9.46)

tan−1(0.30)

tan−1(13.33)