Question

In a transistor amplifier $\beta =62,R_L=5000\Omega$ and internal resistance of the transistor is $500\Omega$. Its power amplification will be

• A. $25580$
• B. $33760$
• C. $38440$
• D. $55760$

Answer 1

The correct option is C $38440$

Given that:
Current gain, $\beta =62$
$R_L=5000\Omega$
$R_I=500\Omega$

Hence, resistance gain = $\frac{R_L}{R_i}$

Resistance gain = $\frac{5000}{500}=10$

The power amplification, i.e., power gain is
Power gain = (current gain)${}^2\times$ resistance gain
Power gain = $(62{)}^2\times 10=38440$