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
Explanation: Diode is reversed biased in the negative half cycle. In the positive half cycle, apply KVL to get the value of peak voltage at the load. Peak value = Vm + V1, as V1 is aiding the positive anode of the diode. 5. The circuit shown below has the following parameters: