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

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

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)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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
$\tan ϕ = \frac{X_{L} - X_{C}}{R}$
$\tan ϕ = \frac{4.9}{5} \approx 1 \Rightarrow ϕ = 45^{0}$