A step down DC chopper as shown in figure has a resistive load of R=15Ω and input voltage of 200 V. Voltage drop of chopper is 2.5 V and chopper frequency is 1 kHz. If the duty cycle is 50%, then the chopper efficiency is
A
98.74%
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B
49.37%
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C
95%
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D
69.83%
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Solution
The correct option is A 98.74% Rms current, I0r=√α(Vs−Vd)R
=√0.5(200−2.5)15 I0r=9.31A
Output power, Po=I20r×R =9.312×15=1300.14
Input power Pf=1T∫T0Vsisdt =1T∑αT0(Vc−Vd)αTT×R =Vs(Vs−Vd)αTT×R Pi=0.5×200×(200−2.5)15 Pi=1316.67W
Efficiency, %η=P0Pi×100 =1300.141316.67×100 =98.74%