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
A. 27.5 V
B. 86.5 V
C. 30 V
D. 42.5 V
Answer: Option D
If the arrow of crystal diode symbol is positive w.r.t. bar, then diode is ___________ biased.
A. forward
B. reverse
C. either forward or reverse
D. none of the above
In the breakdown region, a zener didoe behaves like a ___________ source.
A. constant voltage
B. constant current
C. constant resistance
D. none of the above
When the crystal current diode current is large, the bias is ___________
A. forward
B. inverse
C. poor
D. reverse
If the temperature of a crystal diode increases, then leakage current ___________
A. remains the same
B. decreases
C. increases
D. becomes zero
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
4. none of the above
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