u=Umsinωt
84=(120√2)sin{(100π)t}
sin{(100π)t}≈0.5
(100π)t1=π/6 或 (100π)t2=5π/6
Δt=t2-t1=(5π/6-π/6)/(100π)≈0.0067s
“大于”激发电压是“霓虹灯”点亮.
“小于”是熄灯电压“霓虹灯”熄灭.