Question

In a PNP transistor, \(10^{10}\) holes enter the emitter in \(10^{-6}\,s\). If 2% of holes is lost in the base, then the current amplification factor is

• A. 49
• B. 19
• C. 29
• D. 39
• E. 59

Hint:

\(\beta = \frac{\text{carriers reaching collector}}{\text{lost carriers}}\)

Answer 1

The Correct Option is A

Concept: Current gain: \[ \beta = \frac{I_C}{I_B} \]

Step 1: Emitter current. \[ I_E \propto 10^{10} \]

Step 2: Loss in base.
2% recombine → base current: \[ I_B = 0.02 I_E \]

Step 3: Collector current. \[ I_C = 0.98 I_E \]

Step 4: Compute gain. \[ \beta = \frac{0.98 I_E}{0.02 I_E} = \frac{98}{2} = 49 \]

Step 5: Final answer. \[ \boxed{49} \]