When the input voltage $(20 \sin \omega t)$ is equal to or less than $5 \mathrm{~V}$, diode will be reverse biased. It will offer high resistance in comparison to resistance (R) in series. Now, diode appears in open circuit. So input signal will not pass through diode. The input signal is then passed to the output terminals. Then the result with sin wave input is to dip off all positive going portion above $5 \mathrm{~V}$.
If input voltage $(20 \sin \omega \mathrm{t})$ is more than $+5 \mathrm{~V}$, diode will be conducting as if forward biased offering low resistance in comparison to $\mathrm{R}$. But there will be no voltage in output beyond $5 \mathrm{~V}$ as the voltage beyond $+5 \mathrm{~V}$ will appears across R. So the current passes through diode and battery and output remian $5 \mathrm{~V}$.
When input voltage is negative. there will be opposition to $5 \mathrm{~V}$ battery in $\mathrm{p}-\mathrm{n}$ junction input voltage becomes more than $-5 \mathrm{~V}$, the diode will be revese biased. It will offer high resistance in comparison to resistance $\mathrm{R}$ in series. Now junction diode appears in open circuit. The input wave form is then passed on to output terminals.
The output waveform is shown.