# CBSE Class 12-science Answered

**There is a inductor connected with voltage 10V. AC volt 50Hz such that the current flowing in the inductor is 0.1A. The phase angle between current and voltage is 60°. Find impedance, resistance, and inductance of conductor.**

Asked by sharmujjwal00 | 17 Oct, 2019, 06:59: AM

Expert Answer

current passing through inductor, i

_{m}= V / (R^{2}+ L^{2}ω^{2})^{1/2}.............. (1)where V is applied voltage, R is resistance of inductor, L is inductance and ω = 2πf , where f is frequency of AC voltage.

If the current is 0.1 A and applied voltage is 10 V, then we have, 0.1 = 10/ (R

^{2}+ L^{2}ω^{2})^{1/2}..............(2)Impedence = (R

^{2}+ L^{2}ω^{2})^{1/2}= 100 ΩFrom eqn.(2), we get, (R

^{2}+ L^{2}ω^{2}) = 10000 ..............(3)Phase angle φ is given by, tan φ = Lω/R ............(4)

if phase angle is 60°, then we have tan 60° = ( Lω / R ) = √3 , i.e, Lω = √3 R ...............(4)

using eqn.(4), we rewrite eqn.(3) as , R

^{2}+ 3 R^{2}= 4 R^{2}= 10000 i.e. R = 50 ΩInductance is obtained from eqn. (4) as , L = √3 R / ω = ( √3 × 50 ) /(2π×50) = 0.276 H

Answered by Thiyagarajan K | 17 Oct, 2019, 11:02: PM

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