Step 1: Understanding the Concept:
In an electronic amplifier, power gain is defined as the ratio of output power to input power.
Power can be expressed as the product of voltage and current ($P = V \times I$).
Therefore, power gain is inherently linked to voltage gain and current gain.
Step 2: Key Formula or Approach:
Power Gain ($A_p$) = $\frac{\text{Output Power}}{\text{Input Power}}$
$A_p = \frac{V_{\text{out}} \times I_{\text{out}}}{V_{\text{in}} \times I_{\text{in}}} = \left( \frac{V_{\text{out}}}{V_{\text{in}}} \right) \times \left( \frac{I_{\text{out}}}{I_{\text{in}}} \right)$
$A_p = \text{Voltage Gain } (A_v) \times \text{Current Gain } (A_i)$
Step 3: Detailed Explanation:
The question asks for the ratio of power gain to voltage gain:
\[ \text{Ratio} = \frac{A_p}{A_v} \]
From the relationship derived above:
\[ \frac{A_p}{A_v} = A_i \]
So, the ratio is simply equal to the current gain of the configuration.
The problem specifies a "common emitter configuration".
In a common emitter (CE) configuration, the input current is the base current ($I_b$) and the output current is the collector current ($I_c$).
The current gain in CE mode is denoted by $\beta$ (beta):
\[ A_i = \beta = \frac{I_c}{I_b} \]
(Note: $\alpha$ is the current gain for common base configuration, $\alpha = I_c/I_e$).
Therefore, the ratio $\frac{A_p}{A_v} = \beta$.
Step 4: Final Answer:
The ratio is $\beta$.