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An inductor and a bulb are connected in series to an AC source of 220V, 50Hz. 7C A current of 11A flows in the circuit and phase angle between voltage and current is (pi)/(4)radians. Calculate the impedance and inductance of the circuit |
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Answer» Solution :Here ,`V=220 V , f=50 HZ I=11 A , phi= pi //4.Z =? L=?` `I_(rms) = 0.7071 I_0 ` ` I_0=(I_(rms))/(0.707 ) = (11)/(0.707 ) = 15.55 A` `V_(rms)=0.707 V_0` `V_0 =(V_(rms))/( 0.707 )= (220 )/( 0.707 ) = 311.17 V ` indedence`Z=(V_0)/(I_C) = (311.17)/( 15.55) = 20.01Omega ` `cosphi = (R )/(Z)` ` R=Z COS phi= 20.01cos ((pi)/(4))` ` R= 20.01 xx (1)/(sqrt(2)) = (20.01)/(1.414)= 14.151 OMEGA ` ` Z= sqrt(R^2+ ( omega L)^2)` ` Z^2= R^2 + (2 pi fL)^2 "" (:.omega= 2 pi f )` `(Z^2 -R^2)/( 4 pi ^2f^2) = L^2impliesL^2=((20.01)^2 - (14.151)^2)/( 4 xx (3.14 )^2 xx (50)^2)` `L^2 = ( 400.40- 200.25 ) /( 98596 )` `= (200.15 )/( 98596 )= 0.00202H` `L= 0.0449H= 44.9mH .` |
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