a ce amplifier is driving a 10 kω load. if rc=2.2 kω and re=10 ω, the voltage gain is approximately\nselect…

a ce amplifier is driving a 10 kω load. if rc=2.2 kω and re=10 ω, the voltage gain is approximately\nselect one:\n○ a. 10\n○ b. 220\n○ c. 180\n○ d. 1000

a ce amplifier is driving a 10 kω load. if rc=2.2 kω and re=10 ω, the voltage gain is approximately\nselect one:\n○ a. 10\n○ b. 220\n○ c. 180\n○ d. 1000

Answer

Answer:

c. 180

Explanation:

Step1: Find the parallel resistance

First, we need to find the parallel resistance of ( R_C ) (2.2 kΩ) and the load resistance (10 kΩ). The formula for parallel resistance ( R_{parallel} ) is ( \frac{R_C \times R_{load}}{R_C + R_{load}} ). Plugging in the values: ( \frac{2.2 \times 10}{2.2 + 10} ) kΩ ( = \frac{22}{12.2} \approx 1.803 ) kΩ ( = 1803 ) Ω.

Step2: Calculate voltage gain

The formula for the voltage gain ( A_v ) of a CE amplifier is ( A_v = -\frac{R_{parallel}}{r'_e} ). We know ( r'e = 10 ) Ω and ( R{parallel} \approx 1803 ) Ω. So, ( A_v \approx \frac{1803}{10} \approx 180 ).