Step 1: Understanding the Concept:
A transistor amplifier increases the amplitude of a weak input signal.
The voltage gain is a measure of this amplification, defined as the ratio of output voltage to input voltage.
It can be calculated directly from the current gain and the ratio of output (load) resistance to input resistance.
Step 2: Key Formula or Approach:
Voltage gain ($A_v$) formula: $A_v = \frac{V_{\text{out}}}{V_{\text{in}}}$.
Using Ohm's law, $V_{\text{out}} = I_{\text{out}} R_{\text{out}}$ and $V_{\text{in}} = I_{\text{in}} R_{\text{in}}$.
So, $A_v = \frac{I_{\text{out}}}{I_{\text{in}}} \times \frac{R_{\text{out}}}{R_{\text{in}}}$.
The ratio $\frac{I_{\text{out}}}{I_{\text{in}}}$ is the AC current gain, denoted by $\beta$.
Therefore, $A_v = \beta \times \frac{R_L}{R_{\text{in}}}$, where $R_L$ is load resistance.
Step 3: Detailed Explanation:
Given values from the problem:
AC current gain, $\beta = 64$.
Load (output) resistance, $R_L = 5400 \Omega$.
Input resistance, $R_{\text{in}} = 540 \Omega$.
Substitute these values into the voltage gain formula:
\[ A_v = \beta \times \left(\frac{R_L}{R_{\text{in}}}\right) \]
\[ A_v = 64 \times \left(\frac{5400}{540}\right) \]
Calculate the resistance ratio:
\[ \frac{5400}{540} = 10 \]
Multiply by the current gain:
\[ A_v = 64 \times 10 = 640 \]
Step 4: Final Answer:
The voltage gain is 640.