A half-wave rectifier has an input voltage of 240 V r.m.s. If the step-down transformer has a turns ratio of 8:1, what is the peak load voltage? Ignore diode drop.
A. 27.5 V
B. 86.5 V
C. 30 V
D. 42.5 V
Answer: Option D
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Secondary Voltage=240/8 = 30V
Peak Load Voltage= √2 x 30 = 42.5V
The
.... filter circuit
results in the best voltage regulation
1. choke input
2. capacitor input
3. resistance input
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Emf/ turn are constant in a transferred.
Secondary vtg = 240/8= 30
In half wave rectifier pic= ✓2 x Vm
✓2*30=42.5.
V1/V2 =N1/N2
Vrms = 0.707Vo
Vrms= 0.707 Vo
Vo=240/8 = 30
I want clarify in this