Given the information appearing in Fig. 4.119, determine: a) IC, b) RC, c) RB

$$I_C = 3200 \times 10^{-6} \, \text{A} = 3.2 \, \text{mA}$$

The collector current is $\boxed{\text{3.2 mA}}$.

b) $R_C$

$$V_{R_C} = V_{CC} - V_C$$ $$V_{R_C} = 12 \, \text{V} - 6 \, \text{V} = 6 \, \text{V}$$

$$R_C = 1875 \, \Omega = 1.875 \, \text{k}\Omega$$

The collector resistor is $\boxed{\text{1.875 k}\Omega}$.

c) $R_B$

Since $V_E = 0 \, \text{V}$, $V_B = V_{BE}$.

$$V_B = 0.7 \, \text{V}$$

$$V_{R_B} = V_{CC} - V_B$$ $$V_{R_B} = 12 \, \text{V} - 0.7 \, \text{V} = 11.3 \, \text{V}$$

$$R_B = 282500 \, \Omega = 282.5 \, \text{k}\Omega$$

The base resistor is $\boxed{\text{282.5 k}\Omega}$.

d) $V_{CE}$

$$V_{CE} = 6 \, \text{V}$$

The collector-emitter voltage is $\boxed{\text{6 V}}$.

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: Given the information appearing in Fig. 4.120, determine:
Given: $I_B = 20 \, \mu\text{A}$, $R_C = 2.2 \, \text{k}\Omega$, $V_{CE} = 7.2 \, \text{V}$, $I_E = 4 \, \text{mA}$. The emitter is grounded, so $V_E = 0 \, \text{V}$.

a) $I_C$

$$I_C = (4 \times 10^{-3} \, \text{A}) - (20 \times 10^{-6} \, \text{A})$$ $$I_C = 0.004 \, \text{A} - 0.00002 \, \text{A}$$

$$I_C = 0.00398 \, \text{A} = 3.98 \, \text{mA}$$

The collector current is $\boxed{\text{3.98 mA}}$.

b) $V_{CC}$

Since $V_E = 0 \, \text{V}$.

$$V_C = V_{CE} + V_E$$ $$V_C = 7.2 \, \text{V} + 0 \, \text{V} = 7.2 \, \text{V}$$

$$V_{CC} = 7.2 \, \text{V} + (3.98 \times 10^{-3} \, \text{A}) \times (2.2 \times 10^3 \, \Omega)$$ $$V_{CC} = 7.2 \, \text{V} + 8.756 \, \text{V}$$

$$V_{CC} = 15.956 \, \text{V}$$

The supply voltage is $\boxed{\text{15.956 V}}$.

c) $\beta$

$$\beta = \frac{I_C}{I_B}$$

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