Question

In a common-emitter transistor amplifier, the output resistance is 500 kΩ, and the current gain β = 49. If the power gain of the amplifier is 5 × 10⁶, the input resistance is

• A. $325\Omega$
• B. $165\Omega$
• C. $198\Omega$
• D. $240\Omega$

Answer 1

The correct option is D

$240\Omega$

Step 1: Given:

The output resistance $R_0=500K\Omega$

The current gain $\beta =49$

The power gain = $5\times {10}^6$

Step 2: Formula used:

$Powergain={\beta }^2\frac{R_0}{R_i}$

Step 3: Finding input resistance

Substituting the known values,

⇒ $5\times {10}^6={\left(49\right)}^2\frac{500\times {10}^3}{R_i}$

⇒ $R_i=\frac{{\left(49\right)}^2}{10}$

⇒ $R_i=\frac{2401}{10}$

⇒ $R_i\approx 240\Omega$

Hence, the correct answer is option D.