Question

In common emitter configuration of transistor amplifier, \(r_i\), \(R_L\) and \(β\) represent the input resistance, load resistance and the a.c. current gain respectively. The voltage gain \(A_V\) and power gain \(A_P\) are represented in magnitude respectively by

• A. \(β(\frac{r_i}{R_L}),β^2(\frac{r_i}{R_L})\)
• B. \(β(\frac{R_L}{r_i}),β(\frac{R_L}{r_i})^2\)
• C. \(β(\frac{R_L}{r_i}),β^2(\frac{R_L}{r_i})\)
• D. \(β(\frac{r_i}{R_L}),β(\frac{r_i}{R_L})^2\)

Hint:

A forward biased ideal diode acts as a wire and a reverse biased one acts as an open switch.

Answer 1

The Correct Option is C

Step 1: Conductance method:
Forward: conductances add, $\frac1{30} + \frac1{30} = \frac{1}{15}$, so $R_1 = 15\ \Omega$.

Step 2: Reverse:
One branch is switched off, leaving $R_2 = 30\ \Omega$. The ratio is $\frac{15}{30} = \frac12$ (C).

Final Answer:
$\frac12$. \[ \boxed{\frac{1}{2}} \]