The input signal given to C.E. amplifier having a voltage gain of 126 is $V_i=2 \cos \left(12…
- $252 \cos \left(12 \mathrm{t}+\frac{4 \pi}{3}\right)$
- $252 \cos \left(12 t+\frac{\pi}{3}\right)$
- $\quad 63 \cos \left(12 t+\frac{2 \pi}{3}\right)$
- $2 \cos \left(12 t+\frac{5 \pi}{3}\right)$
Solution
In CE amplifier, $\mathrm{V}_0$ and $\mathrm{V}_{\mathrm{i}}$ have a phase difference of $\pi$ between them. $\begin{aligned} V_0 & =252 \cos \left(12 t+\frac{\pi}{3}+\pi\right) \\ & =252 \cos \left(12 t+\frac{4 \pi}{3}\right) \end{aligned}$ .
Asked in: MHT CET 2024 (16 May Shift 1)