AC chopper with 1-phase on-off control, which feeds a pure resistive load of 5, in 220V-50Hz network, with 10 Hz- 50% effective period.
a) Draw the circuit diagram exactly, check the current and voltage directions
show on it.
b) Draw the basic waveforms of the circuit with its values.
c) Calculate the basic parameters according to the drawn waveforms.
From the given question,
Resistance of load "(R)=5\\Omega"
Source voltage "(V_s)=220V"
Frequency (f)=50Hz
Effective frequency =10Hz
Effective period =50%
So, chopper on time "T=\\frac{1}{10}=0.1s"
Now, 50% of this time on off time T=0.05
Hence, "V_{avg}=V\\times f\\times T = 220\\times 50\\times0.05 V"
=550V
"I_{avg}=\\frac{V_{avg}}{R}=\\frac{550}{5}=110A"
"V_{rms}= V\\times \\sqrt{T\\times f }"
"=220\\times \\sqrt{0.05\\times 50}"
"=220\\sqrt{2.5}"
"I_{rms}=\\frac{V_{rms}}{R}=\\frac{220\\sqrt{2.5}}{5}=44\\sqrt{2.5}"
"P= I_{rms}^2R =(44\\times \\sqrt{2.5})^2\\times 5"
=24200watt
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