Question

\(A_I\) in the given circuit is

• A. \(50\)
• B. \(100\)
• C. \(120\)
• D. \(121\)

Hint:

For a transistor, \[ I_e=I_c+I_b. \] Since \[ I_c=\beta I_b, \] we get \[ I_e=(\beta+1)I_b. \] Thus many current-gain questions directly reduce to using \[ \beta+1. \]

Answer 1

The Correct Option is D

Concept: Current gain is defined as \[ A_I=\frac{I_o}{I_i} \] For a common-emitter transistor amplifier, \[ I_c=h_{fe}I_b \] and \[ I_e=I_c+I_b. \] Hence, \[ I_e=(h_{fe}+1)I_b. \] Since the input current is essentially the base current and the output current is the collector-side current delivered through the collector resistor, the current gain of the circuit becomes approximately \[ A_I=h_{fe}+1. \]

Step 1: Write the given transistor parameter.
From the figure, \[ h_{fe}=120. \]

Step 2: Apply the current gain relation.
For this CE amplifier, \[ A_I=h_{fe}+1. \] Substituting the value, \[ A_I=120+1. \] \[ A_I=121. \]

Step 3: State the result.
Therefore, \[ \boxed{A_I=121} \] Hence the correct option is \[ \boxed{(D)\;121} \]