The given figure shows a step-down chopper switched at 1 kHz with a duty ratio D = 0.5. The peak-peak ripple in the load current is close to

0.25 A

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10 A

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0.5 A

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0.125 A

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Solution

The correct option is D 0.125 A
Here, $T = \frac{1}{1 \times^{3}} = 10^{- 3}$ sec
$T_{a} = \frac{L}{R} = \frac{200 m H}{5} = 40 s e c$

$R i p p l e = \frac{V_{s}}{R} [\frac{(1 - e^{α T / T_{a}}) (1 - e^{- (1 - α) T / T_{a}})}{1 - e^{- T / T_{a}}}$

$Δ I_{m a x} = \frac{V_{s}}{4 f L} = \frac{100}{4 \times 10^{3} \times 200 \times 10^{- 3}} = 0.125 A$

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The given figure shows a step-down chopper switched at 1 kHz with a duty ratio D = 0.5. The peak-peak ripple in the load current is close to

The correct option is D 0.125 AHere, T=11×3=10−3sec Ta=LR=200mH5=40sec Ripple=VsR[(1−eαT/Ta)(1−e−(1−α)T/Ta)1−e−T/Ta ΔImax=Vs4fL=1004×103×200×10−3=0.125A