Now, $\quad \frac{\text{ Output Signal Voltage }(\Delta V_0)}{\text{ Input Singal Voltage }(\Delta V_i)}=$ Total voltage amplification
$=A v_x \times A v_y \times A v_z$
$\Rightarrow$ $ \Delta V_0 =A v_x \times A v_y \times A v_z \times \Delta V_i $
$ =10 \times 20 \times 30 \times 10^{-3}=6 V $
(i) If DC supply voltage is $10 V$, then output is $6 V$, since theoretical gain is equal to practical gain, i.e., output can never be greater than $6 V$.
(ii) If $DC$ supply voltage is $5 V$, i.e., $V _{c c}=5 V$. Then, output peak will not exceed $5 V$.
Hence $V_0=5 V$.