In the following transistor amplifier circuit `beta = 50. Vce to` of the transistor is

Options

A. 4 V
B. 6 V
C. 10 V
D. 8 V

Answer

To find the value of \( V_{CE} \) in the given transistor amplifier circuit, we will follow these steps:

### Step 1: Identify Given Values
From the problem, we have the following values:
- \( \beta = 50 \)
- \( I_B = 40 \, \mu A = 40 \times 10^{-6} A \)
- \( R_C = 2 \, k\Omega = 2000 \, \Omega \)
- \( R_B = 245 \, k\Omega = 245000 \, \Omega \)
- \( V_{i} = 10 \, V \)

### Step 2: Calculate Collector Current \( I_C \)
Using the relation between base current \( I_B \) and collector current \( I_C \):
\[ I_C = \beta \times I_B \]
Substituting the values:
\[ I_C = 50 \times 40 \times 10^{-6} = 2 \times 10^{-3} A = 2 \, mA \]

### Step 3: Calculate Voltage Drop Across \( R_C \)
The voltage drop across the collector resistor \( R_C \) can be calculated using Ohm's law:
\[ V_{RC} = I_C \times R_C \]
Substituting the values:
\[ V_{RC} = 2 \times 10^{-3} \times 2000 = 4 \, V \]

### Step 4: Calculate Output Voltage \( V_{out} \)
The output voltage \( V_{out} \) can be calculated as:
\[ V_{out} = V_{i} - V_{RC} \]
Substituting the values:
\[ V_{out} = 10 \, V - 4 \, V = 6 \, V \]

### Step 5: Calculate \( V_{CE} \)
In a common emitter configuration, \( V_{CE} \) can be calculated as:
\[ V_{CE} = V_{out} \]
Thus, we have:
\[ V_{CE} = 6 \, V \]

### Final Answer
The value of \( V_{CE} \) of the transistor is \( 6 \, V \).

---