Explanation: Load is S.C in the positive half cycle hence, voltage is zero. In negative half the peak value is (230 x 1.414) – 2(drop across the diode) = 323 Volts. 4. For the circuit shown in the figure below, V9 is the AC voltage source with peak value Vm. The waveform of the load voltage at the resistor has a
A. peak value of (Vm+V1) in the negative half cycle
B. peak value of (Vm-V1) in the positive half cycle
C. peak value of (Vm+V1) in the positive half cycle
D. peak value of (Vm-V1) in the negative half cycle
Answer:A
Explanation: Load is S.C in the positive half cycle hence, voltage is zero. In negative half the peak value is (230 x 1.414) – 2(drop across the diode) = 323 Volts. 4. For the circuit shown in the figure below, V9 is the AC voltage source with peak value Vm. The waveform of the load voltage at the resistor has a
A. peak value of (Vm+V1) in the negative half cycle B. peak value of (Vm-V1) in the positive half cycle C. peak value of (Vm+V1) in the positive half cycle D. peak value of (Vm-V1) in the negative half cycle