A step down DC chopper as shown in figure has a resistive load of R- 15- 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

Rms current, I_0r=√(α)(V_s V_d)R =√(0.5)(200 2.5)15 I_0r= 9.31 A Output power, P_o=I_0r^2× R = 9.31^2×15= 1300.14 Input power P_f=1T∫_0^TV_si_sdt =1T∑_0^α T(V_c ...

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A step down DC chopper as shown in figure has a resistive load of $R = 15 Ω$ 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

98.74%

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49.37%

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95%

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69.83%

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Solution

The correct option is A 98.74%
Rms current, $I_{0 r} = \frac{\sqrt{α} (V_{s} - V_{d})}{R}$
= $\frac{\sqrt{0.5} (200 - 2.5)}{15}$
$I_{0 r} = 9.31 A$
Output power, $P_{o} = I_{0 r}^{2} \times R$
$= {9.31}^{2} \times 15 = 1300.14$
Input power $P_{f} = \frac{1}{T} \int_{0}^{T} V_{s} i_{s} d t$
$= \frac{1}{T} \sum_{0}^{α T} \frac{(V_{c} - V_{d}) α T}{T \times R}$
$= \frac{V_{s} (V_{s} - V_{d}) α T}{T} \times R$
$P_{i} = \frac{0.5 \times 200 \times (200 - 2.5)}{15}$
$P_{i} = 1316.67 W$
Efficiency, $% η = \frac{P_{0}}{P_{i}} \times 100$
$= \frac{1300.14}{1316.67} \times 100$
$= 98.74 %$

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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 step down DC chopper as shown in figure has a resistive load of $R = 15 Ω$ 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