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 the 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)

$198\Omega$

Power gain = Current gain $\times$ volume gain .....(1)
Voltage gain $A_v=\beta \frac{R_2}{R_1}$

Output resistance $R_2=500\times {10}^3\Omega$
Also, current gain $\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{1-0.98}=49$
$\therefore A_v=\left(49\right)\left(\frac{500\times {10}^3}{R_1}\right)$
Power gain $=6.0625\times {10}^6$ (Given)

From (1), we get
$6.0625\times {10}^6$ $=49\times \left(\frac{500\times {10}^3}{R_1}\right)\times 49$
$\Rightarrow R_1=198\Omega$