Let's analyze the circuit step by step to determine the values of $R_C$, all node voltages, and branch currents.

Given Data:

• $\beta = 75$ (Current gain of the transistor)
• $I_C = 0.5 \, \text{mA}$ (Collector current provided by the current source)
• $V_C = 1 \, \text{V}$ (Voltage at the collector)
• Power supply voltages: $+5 \, \text{V}$ and $-5 \, \text{V}$

1. Calculate $I_B$ (Base current):

The transistor's collector current is related to the base current by: $I_C = \beta \cdot I_B$

Rearranging for $I_B$: $I_B = \frac{I_C}{\beta}$

Substitute $I_C = 0.5 \, \text{mA}$ and $\beta = 75$: $I_B = \frac{0.5}{75} \, \text{mA} = 0.00667 \, \text{mA} = 6.67 \, \mu\text{A}$

2. Calculate $V_B$ (Base voltage):

The base voltage is determined by the voltage divider formed by the $25 \, \text{k}\Omega$ resistor and the base-emitter junction of the transistor.

For a silicon transistor, the base-emitter voltage $V_{BE}$ is approximately $0.7 \, \text{V}$. Therefore: $V_B = V_E + V_{BE}$

The emitter current is given by: $I_E = I_C + I_B$

Substitute $I_C = 0.5 \, \text{mA}$ and $I_B = 0.00667 \, \text{mA}$: $I_E = 0.5 + 0.00667 \, \text{mA} = 0.50667 \, \text{mA}$

The voltage across the $25 \, \text{k}\Omega$ resistor is: $V_E = I_E \cdot 25 \, \text{k}\Omega$

Substitute $I_E = 0.50667 \, \text{mA}$: $V_E = 0.50667 \cdot 25 = 12.67 \, \text{V}$

Thus, the base voltage is: $V_B = V_E + V_{BE} = 12.67 + 0.7 = 13.37 \, \text{V}$

3. Calculate $R_C$:

The voltage across $R_C$ is: $V_{RC} = V_{CC} - V_C$

Substitute $V_{CC} = 5 \, \text{V}$ and $V_C = 1 \, \text{V}$: $V_{RC} = 5 - 1 = 4 \, \text{V}$

The collector current $I_C = 0.5 \, \text{mA}$. Using Ohm's law, the value of $R_C$ is: $R_C = \frac{V_{RC}}{I_C}$

Substitute $V_{RC} = 4 \, \text{V}$ and $I_C = 0.5 \, \text{mA}$: $R_C = \frac{4}{0.5 \times 10^{-3}} = 8 \, \text{k}\Omega$

4. Summary of Results:

1. $R_C = 8 \, \text{k}\Omega$
2. Base current: $I_B = 6.67 \, \mu\text{A}$
3. Emitter current: $I_E = 0.50667 \, \text{mA}$
4. Node voltages:
   • $V_B = 13.37 \, \text{V}$
   • $V_C = 1 \, \text{V}$
   • $V_E = 12.67 \, \text{V}$

Let me know if you would like further clarification or additional steps explained in detail. Here are some follow-up questions to deepen your understanding:

1. How does changing $\beta$ affect the values of $I_B$ and $R_C$?
2. Why is the $V_{BE}$ drop approximately $0.7 \, \text{V}$ for silicon transistors?
3. How would the circuit change if the current source was increased to $1 \, \text{mA}$?
4. What would happen to $R_C$ if $V_C$ was set to $2 \, \text{V}$ instead?
5. How does the voltage divider impact the transistor's base voltage in this circuit?

Tip: Always verify the power dissipation across resistors to ensure they operate within their rated limits.