MGVCL Exam Paper (30-07-2021 Shift 2)
A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element.
Frequency of natural oscillation is given by, fn = {((dPe/dδ)at(δo))/M)} dPe/dδ = ((V1*V2)/X*(cosδ)) = (11/05)*cosδ = (11/0.5)*0.5 M = (H*s)/(πf) = 4/(50π)
Here, ω = 500 rad/sec = 2πf. At Resonance condition, ωL = 1/(ωC) L 0.2 H Also at series resonace 1. fr = 1/[2π√(LC)] 2. PF = 1 3. Maximum current 4. Source voltage = voltage across resistor.