In a common emitter amplifier circuit using n-p-n transistor, the...

In a common emitter amplifier, the input signal is applied between the base and emitter, and the output is taken across the collector and emitter.

The output voltage is given by $V_{CE} = V_{CC} - I_C R_C$.

During the positive half cycle of the input signal, the forward bias across the base-emitter junction increases. This causes the base current $I_B$ to increase, which in turn increases the collector current $I_C$.

As $I_C$ increases, the voltage drop across the collector resistor $R_C$ increases. Consequently, the output voltage $V_{CE}$ decreases.

Thus, a positive input voltage results in a negative output voltage, indicating a phase reversal.

Therefore, the phase difference between the input and output voltages in a common emitter amplifier is 180^{}^{\circ}.

Answer: 180^{}^{\circ}