Explanation: In the negative half, only D2 is active. Hence use KVL, Vo = 6 + 0.7 Volts. 7. Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle
A. Vo = Vr
B. Vo = 0
C. Vo = Vdc+Vr
D. Vo = Vdc
Answer:A
Explanation: In the negative half, only D2 is active. Hence use KVL, Vo = 6 + 0.7 Volts. 7. Consider the diode to be an ideal one and Vo = Vr + Vdc during positive half cycle. Thus, during the negative half cycle
A. Vo = Vr B. Vo = 0 C. Vo = Vdc+Vr D. Vo = Vdc