Respuesta :
Answer:
[tex]V_{p (load)} = 28,3 V - 0,7 V = 27,6 V[/tex]
[tex]V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V[/tex]
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
[tex]I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA[/tex]
Explanation:
The peak voltage after the 6 to 1 step down is [tex]V_{p} = \frac{120}{6} \sqrt{2} = 28,3V[/tex]. Then, the peak voltage of the rectified output is [tex]V_{p} - [tex]V_{avg} = \frac{2V_{p (load)} }{\pi} = \frac{55,2 V}{\pi } = 17,6 V[/tex]V_{d}[/tex] and according to the statement, the diodes can be modeled to be [tex]V_{d} = 0,7 V[/tex]. Then, the peak voltage in the load is [tex]V_{p (load)} = 28,3 V - 0,7 V = 27,6 V[/tex].
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
The average output voltage is calculated as:
[tex]V_{p (load)} = 27,6 V\\V_{avg} = 17,57 V[/tex]
The average current in the load is calculated as:
[tex]I_{avg} = \frac{V_{avg}}{R_{load}} = \frac{17,57 V}{1000 Ω} = 17,57 mA[/tex]
