A 25mF capacitor, 0.10H inductor and 25Ω resistor are connected in series with a.c. source whose emf is e=310sin314t(volt). Find the following 1. Frequency of the source 2. Reactance of the circuit 3. Impedance 4. Current in the circuit 5. Phase angel of the current by which it leads or lags the applied emf 6. Expression for the instantaneous current 7. Construct a phasor diagram for these voltages 8. What value of inductance will make the impedance of the circuit minimum?[

Dear Student,given  e=310sin314t(volt)C =25mF L =0.10HR = 25 ohm1)comparing e with  x = Asinωt frequency  = ω2π =3142π =50 Hz2)Reactance  =XL - XC =ωL - 1ωC =314 × 0.1 -1314×25×10-3 =31.4 - 0.127 =31.27 ohm3)Impedance=Z = R2 +(XL - XC)2 =625 +977.81 = 1602.81 =40.035 ohm4) Irms = Io2Io = VoZ =31040.035= 7.74 A Irms =7.74 2 = 5.5 A5)φ =  tan-1(XL - XCR) =tan-1(40.03525) 6)instantaneous  current in the circuit I = Io sin(ωt +φ) I = 7.74sin(ωt +tan-1(40.03525))8)for minimum impedance circuit must be in resonance i.e.314 =1LCL=1314 × 314 ×25 × 10-3 =0.41mH

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