The amplifiers $X, Y$ and $Z$ are connected in series. If the voltage gains of $X, Y$ and $Z$ are $10, 20$ and $30$ respectively,and the input signal is $1 \, mV$ peak value,then what is the output signal voltage (peak value) $(i)$ if $DC$ supply voltage is $10 \, V$? $(ii)$ if $DC$ supply voltage is $5 \, V$?

(A) The total voltage gain of amplifiers connected in series is the product of individual gains: $A_v = (A_v)_1 \times (A_v)_2 \times (A_v)_3 = 10 \times 20 \times 30 = 6000$.
The output voltage is given by $v_0 = A_v \times v_i$.
Given input signal $v_i = 1 \, mV = 10^{-3} \, V$.
Therefore,the theoretical output voltage is $v_0 = 6000 \times 10^{-3} \, V = 6 \, V$.
$(i)$ If the $DC$ supply voltage is $10 \, V$,the amplifier can support an output of $6 \, V$ because $6 \, V < 10 \, V$. Thus,the output is $6 \, V$.
$(ii)$ If the $DC$ supply voltage is $5 \, V$,the amplifier cannot produce an output higher than the supply voltage. Since the required output $6 \, V$ exceeds the supply $5 \, V$,the output is clipped at the supply voltage limit. Thus,the output is $5 \, V$.