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प्रश्न
(i) An alternating emf of 200 V, 50 Hz is applied to an L-R ciruit, having a resistance R of 10 Ω and an inductance L of 0.05H connected in series. Calculate :
(1) Impedance
(2) Current flowing in the circuit
(ii) Draw a labelled graph showing the variation of inductive reactance (`X_L`) verses frequency (`f`) .
थोडक्यात उत्तर
संख्यात्मक
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उत्तर
(i) Given :
e = 200V f = 50Hz
R = 10Ω L = 0.05 H
Z = ? I = ?
`omega = 2pif = 2 xx 3.14 xx 50 = 314 rad"/"s`
`Z = sqrt(R^2 + X_L^2)`
= `sqrt(R^2 + (omegaL)^2)`
= `sqrt(10^2 + (3.14 xx 0.05)^2)`
= `sqrt(100 + 246.5)`
= `sqrt(346.5) = 18.61 Omega`
`because Z = e/I`
`therefore I = e/Z = 200/18.61 = 10.75 A`
(ii) Graph of inductive reactance (`X_L`) vs frequency (f) :
⇒ `X_L = (2piL)·f`
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