An n-p-n transistor in a common - emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The positive terminal of a 8 V battery is connected to the collector through a load resistance ` R_L ` and to the base through a resistance ` R_B ` . The collector - emitter voltage ` V_(CE) = 4 V` , the base - emitter voltage ` V_(BE) = 0.6 V` and the current amplification factor ` beta = 100 ` . Calculate the values of ` R_L ` and `R_B` .

Explanation

To solve the problem, we need to calculate the values of the load resistance \( R_L \) and the base resistance \( R_B \) for the given n-p-n transistor in common-emitter mode.

### Step 1: Calculate Load Resistance \( R_L \)

1. **Given Values**:
- Collector current \( I_C = 4 \, \text{mA} = 4 \times 10^{-3} \, \text{A} \)
- Collector-emitter voltage \( V_{CE} = 4 \, \text{V} \)
- Supply voltage \( V_{CC} = 8 \, \text{V} \)

2. **Apply Kirchhoff's Voltage Law (KVL)** in the collector loop:
\[ V_{CE} = V_{CC} - I_C \cdot R_L \]
Rearranging gives:
\[ I_C \cdot R_L = V_{CC} - V_{CE} \]
Substituting the known values:
\[ I_C \cdot R_L = 8 \, \text{V} - 4 \, \text{V} = 4 \, \text{V} \]

3. **Solve for \( R_L \)**:
\[ R_L = \frac{4 \, \text{V}}{I_C} = \frac{4 \, \text{V}}{4 \times 10^{-3} \, \text{A}} = 1000 \, \Omega \]

### Step 2: Calculate Base Resistance \( R_B \)

1. **Calculate Base Current \( I_B \)** using the current amplification factor \( \beta \):
\[ \beta = \frac{I_C}{I_B} \implies I_B = \frac{I_C}{\beta} \]
Substituting the known values:
\[ I_B = \frac{4 \times 10^{-3} \, \text{A}}{100} = 4 \times 10^{-5} \, \text{A} = 40 \, \mu\text{A} \]

2. **Apply KVL in the base loop**:
\[ V_{BE} = V_{CC} - I_B \cdot R_B \]
Rearranging gives:
\[ I_B \cdot R_B = V_{CC} - V_{BE} \]
Substituting the known values:
\[ I_B \cdot R_B = 8 \, \text{V} - 0.6 \, \text{V} = 7.4 \, \text{V} \]

3. **Solve for \( R_B \)**:
\[ R_B = \frac{7.4 \, \text{V}}{I_B} = \frac{7.4 \, \text{V}}{40 \times 10^{-6} \, \text{A}} = 185000 \, \Omega = 1.85 \times 10^5 \, \Omega \]

### Final Answers:
- Load Resistance \( R_L = 1000 \, \Omega \)
- Base Resistance \( R_B = 1.85 \times 10^5 \, \Omega \)