To determine the values of \(I_C\) (collector current) and \(V_{CE}\) (collector-emitter voltage) for the given emitter-bias network, we need to follow these steps:

Given Values:

• \(\beta = 100\)
• \(V_{BE} = 0.7 \, V\)
• Supply Voltage \(V_{CC} = 15 \, V\)
• Resistor values: \(R_1 = 330 \, k\Omega\), \(R_2 = 2 \, k\Omega\), \(R_E = 1 \, k\Omega\)

Step 1: Determine \(I_E\) (Emitter Current)

Using the approximation \(I_E \approx I_C\) (since \(\beta \approx 100\)), we find \(I_E\).

Assuming a voltage across \(R_E\) due to emitter current:

\(V_{E} = I_{E} \times R_{E}\)

Using Kirchhoff's Voltage Law (KVL) in the input loop:

\(V_{CC} = I_{E} \times R_{E} + V_{BE} + I_B \times R_1\) \(15 = I_{E} \times 1k + 0.7 + \frac{I_{E}}{\beta} \times 330k\)

Assuming \(I_B\) is much smaller than \(I_E\), the voltage drop across \(R_1\) can be small, we approximate:

\(I_E \approx \frac{14.3}{331} \approx 43.2 \, mA\)

Step 2: Relate \(I_E\) to \(I_C\)

Since \(I_C \approx \beta I_B\):

\(I_C = \frac{\beta}{\beta+1} \times I_E \approx \frac{100}{101} \times 43.2 \, mA \approx 43.1 \, mA\)

Step 3: Determine \(V_{CE}\)

Using KVL for the outer loop:

\(V_{CC} = I_C \times R_2 + V_{CE} + I_E \times R_E\)

Rearrange and solve for \(V_{CE}\):

\(15 = 43.1 \, mA \times 2k + V_{CE} + 43.2 \, mA \times 1k\) \(V_{CE} = 15 - 86.2 - 43.2 \approx 5.04 \, V\)

Conclusion

The calculated values are \(I_C \approx 43.2 \, mA\) and \(V_{CE} \approx 5.04 \, V\). Matching with the options provided, the correct answer is:

• 3.32 mA and 5.04 V