Question

A 200mH inductor is connected in series to a resistor of 10Ω. An AC supply of 220V, 50Hz is connected across it.
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.

Answer 1

Answer:

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=220V VImpedence of RL circuitZ=(R)2+(ωL)2

Z=(10)2+(2×3.14×50×200×10−3)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√⇒Z=100+(62.8)2−−−−−−−−−−−√⇒Z=100+3943.84−−−−−−−−−−−√⇒Z=63.59 Ω(i) R.M.S value of current (Irms) is given byIrms=VZ⇒Irms=22063.59=3.46 A(ii) Peak value of current (I0) is given byI0=Irms2√⇒I0=3.46×1.41=4.88 A(iii) Power factor of the circuit is given bycosϕ=RZ=1063.59⇒cosϕ=0.157(iv) Equation of instantaneous value of current⇒I=I0sinωt⇒I=4.88sin314t(v) tanθ=XLR⇒tanθ=62.810=6.28⇒θ=tan−1(6.28)⇒θ=80.95°