A 200 mH inductor is connected in series to a resistor of 10 ohm An AC supply of 220V,50 - Physics - Alternating Current

Question

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

Govind · Accepted Answer

hi,
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+ωL2⇒Z=102+2×3
14×50×200×10-32⇒Z=100+62
82⇒Z=100+3943 84⇒Z=63 59 Ωi R M
S value of current Irms is given byIrms=VZ⇒Irms=22063
59=3
46 Aii Peak value of current I0 is given byI0=Irms2⇒I0=3
46×1 41=4
88 Aiii Power factor of the circuit is given bycosϕ=RZ=1063
59⇒cosϕ=0
157iv Equation of instantaneous value of current⇒I=I0sinωt⇒I=4
88sin314tv tanθ=XLR⇒tanθ=62 810=6
28⇒θ=tan-16 28⇒θ=80 95°