Answer:

When signal $20\sin \omega t$provides an input voltage less than 5 volt (because after 5V, the diode will receive a positive voltage at its P-junction), the diode will be in reverse bias, and the resistance of the diode will remain infinity, preventing the input signal from passing through the diode and battery path. This results in an increase in output across A and B from 0-5V. (graph OC).
When the input voltage $20\sin \omega t$exceeds 5V, the path of the 5V battery diode offers very low resistance, allowing current to flow through the diode and battery while keeping the output (across A and B) at 5V. (graph CD).
As the input voltage drops, the diode will be in reverse bias, and the output will drop from 5V to 0V. (graph ED). When the input voltage becomes negative (opposite of a 5V battery in a p-n junction, the input voltage becomes more than 5V now), the diode is in reversed bias, and the output across AB will be the same as the input AC, i.e., for negative cycles, the diode offers infinite resistance as compared to R in the series graph E,F,G.
The input and output waveforms are repeated in the graph, which shows the output waveform.