Let us analyze and solve the questions from the provided circuit diagrams and descriptions.

Question 8: Voltage Gain with Load Resistor and Capacitor

The voltage gain ($A_v$) for the circuit in Figure 2 with a load resistor ($R_L = 100 \, \Omega$) and capacitor ($C_3 = 1 \, \mu\text{F}$) can be calculated using the formula:

$A_v = - \frac{\beta \cdot R_L}{r_e}$

Where:

• $\beta = 150$ (given transistor gain)
• $R_L = 100 \, \Omega$
• $r_e = \frac{V_T}{I_E}$ is the intrinsic emitter resistance, with $V_T \approx 26 \, \text{mV}$ at room temperature and $I_E$ is the emitter current.

To calculate $r_e$, we need $I_E$, which depends on the circuit's biasing. If the collector current ($I_C$) is approximately equal to $I_E$, we estimate:

$r_e = \frac{26 \, \text{mV}}{I_E}$

Substitute values once $I_E$ is known, calculate $r_e$, and then substitute in the gain equation to find $A_v$.

Question 9: Effect of Load Resistance on Gain

Adding a load resistor ($R_L$) decreases the overall voltage gain because the effective load resistance $R_{eff}$ seen by the transistor becomes:

$R_{eff} = \left( \frac{R_C \cdot R_L}{R_C + R_L} \right)$

Here:

• $R_C = 3.6 \, \text{k}\Omega$ (collector resistor)
• $R_L = 100 \, \Omega$

Since $R_L$ is much smaller than $R_C$, $R_{eff}$ significantly reduces, which reduces $A_v$.

Question 10: Effect of Removing Capacitor $C_2$

$C_2$ bypasses the emitter resistor to stabilize the gain at high