Question

r1 =7 kω vcc = 12 v r2 =7 kω rc = 4.7 kω β = 15 re = 6.3 kω

Explanation:

Step1: Find Vb (Base Voltage)

The base voltage \( V_b \) is the voltage at the junction of \( R1 \) and \( R2 \), which form a voltage divider. The formula for \( V_b \) is \( V_b = V_{cc} \times \frac{R2}{R1 + R2} \).
Given \( V_{cc} = 12 \, V \), \( R1 = 7 \, k\Omega \), \( R2 = 7 \, k\Omega \).

$$ V_b = 12 \times \frac{7}{7 + 7} = 12 \times \frac{7}{14} = 6 \, V $$

Step2: Assume Vbe = 0.7 V (for silicon transistor) to find Ve (Emitter Voltage)

For a transistor, \( V_{be} = V_b - V_e \), so \( V_e = V_b - V_{be} \). Assuming \( V_{be} = 0.7 \, V \):

$$ V_e = 6 - 0.7 = 5.3 \, V $$

Step3: Find Ie (Emitter Current)

The emitter current \( I_e \) is given by \( I_e = \frac{V_e}{R_e} \). Given \( R_e = 6.3 \, k\Omega \):

$$ I_e = \frac{5.3}{6.3 \times 10^3} \approx 0.841 \, mA $$

Step4: Find Ic (Collector Current) using \( \beta \)

For a transistor, \( I_c \approx I_e \) when \( \beta \) is considered, but also \( I_c = \beta \times I_b \), and \( I_e = I_c + I_b \). However, since \( I_e \approx I_c \) (because \( I_b \) is small), or using \( I_c = \frac{\beta}{\beta + 1} I_e \). Let's use \( I_c \approx I_e \) for approximation (or more accurately \( I_c = \frac{\beta}{\beta + 1} I_e \)).
Given \( \beta = 15 \):

$$ I_c = \frac{15}{15 + 1} \times 0.841 \approx \frac{15}{16} \times 0.841 \approx 0.789 \, mA $$

Step5: Find Vc (Collector Voltage)

The collector voltage \( V_c = V_{cc} - I_c \times R_c \). Given \( R_c = 4.7 \, k\Omega \):

$$ V_c = 12 - 0.789 \times 10^{-3} \times 4.7 \times 10^3 \approx 12 - 3.708 \approx 8.292 \, V $$

Step6: Find Vce (Collector - Emitter Voltage)

\( V_{ce} = V_c - V_e \):

$$ V_{ce} = 8.292 - 5.3 = 2.992 \, V \approx 3 \, V $$

(Note: If the problem was to find a specific parameter like \( I_c \), \( V_{ce} \), etc., the final answer would be adjusted. Assuming we found \( V_{ce} \) as an example.)

Answer:

If the question was to find \( V_{ce} \), the answer is approximately \( \boldsymbol{3 \, V} \) (or based on the specific parameter asked, adjust accordingly).