The input voltage applied to a CE amplifier i | vedclass

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(D) Comparing the input voltage $v_i = 0.2 \sin(1000t)$ with the standard form $v = V_m \sin(\omega t)$,we get the peak input voltage $V_m = 0.2 	ext

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$2 \sin(1000t + \pi)$

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(D) Comparing the input voltage $v_i = 0.2 \sin(1000t)$ with the standard form $v = V_m \sin(\omega t)$,we get the peak input voltage $V_m = 0.2 	ext{ V}$.Given the voltage gain $A_v = 10$,the peak output voltage is $V_{out,m} = A_v 	imes V_m = 10 	imes 0.2 = 2 	ext{ V}$.In a Common Emitter $(CE)$ amplifier,there is a phase shift of $\pi$ radians between the input and output signals.Therefore,the equation for the output voltage is $v_o = 2 \sin(1000t + \pi) 	ext{ V}$.

The input voltage applied to a $CE$ amplifier is $v_i = 0.2 \sin(1000t) \text{ V}$. If the voltage gain of the amplifier is $10$,the equation for the output voltage is ....... $\text{V}$.

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