Question

The input resistance of a common emitter transistor amplifier, if the output resistance is 500 k$\Omega$, the current gain $\alpha$ = 0.98 and power gain is 6.0625 $\times 10^6$, is

• A. 198 $\Omega$
• B. 300$\Omega$
• C. 100$\Omega$
• D. 400$\Omega$

Answer 1

Answer: (A)

Voltage gain= $A_V =\beta \frac{R_2}{R_1}$
Also,current gain $\beta =\frac{\alpha}{1-\alpha} =\frac{0.98}{1-0.98}=49$
$ \, \, \, \, \, \, \, \, \, \, A_V =(49) \bigg[\frac{500 \times 10^3}{R_1}\bigg]$
Power gain
$ \, \, \, \, =6.0625 \times 10^6 =49 \times \bigg[\frac{500 \times 10^3}{R_1}\bigg] \times 49$
$\therefore \, \, \, \, \, \, \, R_1 =198 \Omega$