PEREC Inductive load, quantitative analysis Differential equation 180 L=°+Ri=√2U1 snot dt (45) 140 Solution [sin(at-)-sin(a-o)eg9] Z a<ot≤a+b 20 Considering io=0 when ot=ac+8 02060100140180 We have /(°) a+b-)=sn(a-0)e(4-7) The rms value of output voltage output current, and thyristor current can then be calculatedPower Electron cs i 10 Differential equation Differential equation Solution Solution Considering Considering i o=0 when =0 when ωt = α+ θ We have We have 0 2 sin d d o o 1 o = + = ω = α ω t i Ri U t t i L (4 -5) α ω α θ ω ϕ α ϕ ϕ α ω ≤ ≤ + = − − − − t t e Z U i tg t o [sin( ) sin( ) ] 2 1 (4 -6) ϕ θ α θ ϕ α ϕ tg sin( ) sin( ) − + − = − e Inductive load, quantitative analysis Inductive load, quantitative analysis 0 2 0 60 100 140 180 20 100 60 θ /(°) 180 140 α /(°) ϕ = 90° 75° 60° 45° 30° 15° 0° (4 -7) The RMS value of output voltage, output current, and The RMS value of output voltage, output current, and thyristor thyristor current can then be calculated. current can then be calculated