A 200 mH inductor is connected in series to a resistor of 10 ohm. An AC supply of 220V,50 Hz is connected. Calculate i) the rms value of current ii) the peak value of current iii) the power factor of the circuit and write the equation for instantaneous value of current. Calculate time lag b/w eff value of emf and current

It is not time lag, it is phase lag.

Given,Inductance of inductor, L=200 mH=200×10-3 HResistance of resisor, R=10 ΩFrequency of AC source, v=50 HzVoltage of AC supply, Ev=220 VImpedence of RL circuitZ=R2+ωL2Z=102+2×3.14×50×200×10-32Z=100+62.82Z=100+3943.84Z=63.59 Ωi R.M.S value of current Irms is given byIrms=VZIrms=22063.59=3.46 Aii Peak value of current I0 is given byI0=Irms2I0=3.46×1.41=4.88 Aiii Power factor of the circuit is given bycosϕ=RZ=1063.59cosϕ=0.157iv Equation of instantaneous value of currentI=I0sinωtI=4.88sin314tv tanθ=XLRtanθ=62.810=6.28θ=tan-16.28θ=80.95°

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