- 2 The Controlled Rectifier Of Fig 3 Has An Ac Source Of 240 V At 60 Hz R 51 And An Inductor Large Enough To Cause 1 (171.81 KiB) Viewed 109 times
2. The controlled rectifier of Fig. 3 has an AC source of 240 V at 60 Hz, R = 51, and an inductor large enough to cause
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2. The controlled rectifier of Fig. 3 has an AC source of 240 V at 60 Hz, R = 51, and an inductor large enough to cause
2. The controlled rectifier of Fig. 3 has an AC source of 240 V at 60 Hz, R = 51, and an inductor large enough to cause continuous current. . a. Calculate the delay angle for an average DC current of 10A. b. Determine the value of inductance that will limit the peak-to-peak load current variations to 2 A. Hints: Decreasing amplitude of the voltage terms and increasing magnitude of the impedance both contribute to diminishing AC currents as n increases, the peak-to-peak current variation can be estimated from the first AC term (n = 2) Fourier contents of the two-pulsed output voltage: 2Vm [cos ((n + 1)a) cos ((n − 1)a) an = n + 1 n = 2k = 2,4,6, ... 2Vm (sin((n + 1)a) sin((n − 1)a) K EN n + 1 n-1 . TT n - 1 bn = TT + T1 T3 A R well Vsto Vout T $12 Tz - Fig. 3