This document provides steps to calculate the β value for a half-wave rectifier circuit with an R-L load. It is given that Vs=240V, f=50Hz, R=100Ω, and L=0.1H. The document calculates the θ and ωτ values, then sets up and solves a function to find the value of β that satisfies the equation. Through iterative calculations using this function, it determines that the β value is approximately 3.4.

1. How to calculate β
Half-wave with R-L Load
GivenVs= 240V, f = 50Hz
R = 100Ω, L = 0.1H
1. To find β:
𝜃 = tan−1 (
𝜔𝐿
𝑅
) = 0.304 𝑟𝑎𝑑
𝜔𝜏 =
𝜔𝐿
𝑅
=
100𝜋(0.1)
100
= 0.314
sin(β-0.304) + sin(0.304)e-β/0.314
= 0
Let f(β) = sin(β-0.304) + sin(0.304)e-β/0.314
f(0) = 0
f(1) = 0.6535
f(2) = 0.9927
f(3) = 0.431
f(4) = -0.5264
 β liesbetween3and 4.
f’(β) = cos (β - 0.304) – 0.9533 e-β/0.314
𝛽 𝑛+1 = 𝛽 𝑛 −
𝑓( 𝛽 𝑛)
𝑓′( 𝛽 𝑛)
= 𝛽 𝑛 −
sin( 𝛽 𝑛 − 0.304) + 0.2993𝑒−𝛽 𝑛/0.314
cos( 𝛽 𝑛 − 0.304) − 0.9533𝑒−𝛽 𝑛/0.314
Let β0 = 3.1
β1 = 3.46
β2 = 3.45
β3 = 3.45  β = 3.4 ischosen.