Let's solve the problem step by step, where we need to find the resistances $R_C$ and $R_B$ for the given circuit.

Given Data:

• $V_{CC} = 10 \, \text{V}$
• $V_{BB} = 5 \, \text{V}$
• $I_C = 10 \, \text{mA} = 0.01 \, \text{A}$
• $V_{CE} = 5 \, \text{V}$
• $V_{BE} = 0.7 \, \text{V}$
• $\beta = 100$ (current gain)
• Negligible reverse saturation current

Step 1: Finding $R_C$

We can use Kirchhoff's Voltage Law (KVL) in the output loop: $V_{CC} = I_C R_C + V_{CE}$ Rearranging to solve for $R_C$: $R_C = \frac{V_{CC} - V_{CE}}{I_C}$ Substitute the known values: $R_C = \frac{10 \, \text{V} - 5 \, \text{V}}{0.01 \, \text{A}} = \frac{5 \, \text{V}}{0.01 \, \text{A}} = 500 \, \Omega$

So, $R_C = 500 \, \Omega$.

Step 2: Finding $R_B$

First, calculate the base current $I_B$ using the relation between collector current $I_C$ and $I_B$: $I_B = \frac{I_C}{\beta}$ Substitute the known values: $I_B = \frac{0.01 \, \text{A}}{100} = 0.0001 \, \text{A} = 100 \, \mu\text{A}$

Next, apply KVL in the input loop: $V_{BB} = I_B R_B + V_{BE}$ Rearranging to solve for $R_B$: $R_B = \frac{V_{BB} - V_{BE}}{I_B}$ Substitute the known values: $R_B = \frac{5 \, \text{V} - 0.7 \, \text{V}}{0.0001 \, \text{A}} = \frac{4.3 \, \text{V}}{0.0001 \, \text{A}} = 43 \, \text{k}\Omega$

So, $R_B = 43 \, \text{k}\Omega$.

Final Answer:

• $R_C = 500 \, \Omega$
• $R_B = 43 \, \text{k}\Omega$

Would you like further details on the calculations or clarification on any steps?

Here are some related questions to further explore this topic:

1. How would changing the value of $V_{CC}$ affect $R_C$?
2. What happens if $\beta$ (current gain) is doubled?
3. How does the choice of $R_B$ influence the stability of the circuit?
4. Can we calculate the emitter current $I_E$ from the given values?
5. How would the circuit behavior change if $V_{BE}$ is 0.6V instead of 0.7V?

Tip: In BJT circuits, small changes in $\beta$ can significantly affect base current, so stability techniques (like feedback) are often used to minimize this impact.