A sinusoidal voltage of peak value 283 V and angular frequency 320 rad/s is applied to a series LCR circuit. Given that $R = 5 Ω, L = 25 m H$ and $C = 1000 μ F$ .The total impedance, and phase difference between the voltage across the source and the current approximately

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$10 Ω a n d \tan^{- 1} (\frac{5}{3})$

$7 Ω a n d 45^{0}$

$10 Ω a n d \tan^{- 1} (\frac{8}{3})$

$7 Ω a n d \tan^{- 1} (\frac{5}{3})$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given,
$V_{0} = 283 v o l t, ω = 320, R = 5 Ω, L = 25 m H, C = 1000 μ F$
$X_{L} = ω L = 320 \times 25 \times 10^{- 3} = 8 Ω$
$X_{C} = \frac{1}{ω C} = \frac{1}{320 \times 1000 \times 10^{- 6}} = 3.1 Ω$
Total impedance of the circuit :
$Z = \sqrt{R^{2} + {(X_{L} - X_{C})}^{2}} = \sqrt{25 + {(4.9)}^{2}} = 7 Ω$
Phase difference between the voltage and current
$\tan ϕ = \frac{X_{L} - X_{C}}{R}$
$\tan ϕ = \frac{4.9}{5} \approx 1 \Rightarrow ϕ = 45^{0}$