Calculate the output voltage

Using Kirchhoff's Voltage Law (KVL) for the base-emitter loop:

$$V_{in} = I_B R_B + V_{BE}$$

Given $V_{in} = 2.7 \, \text{V}$, $R_B = 10 \, \text{k}\Omega = 10 \times 10^3 \, \Omega$, and $V_{BE} = 0.7 \, \text{V}$.

$$2.7 \, \text{V} = I_B (10 \times 10^3 \, \Omega) + 0.7 \, \text{V}$$ $$I_B (10 \times 10^3 \, \Omega) = 2.7 \, \text{V} - 0.7 \, \text{V}$$ $$I_B (10 \times 10^3 \, \Omega) = 2.0 \, \text{V}$$ $$I_B = \frac{2.0 \, \text{V}}{10 \times 10^3 \, \Omega}$$ $$I_B = 0.2 \times 10^{-3} \, \text{A} = 0.2 \, \text{mA}$$

The collector current is related to the base current by the current gain $\beta$:

$$I_C = \beta I_B$$

Given $\beta = 60$.

$$I_C = 60 \times (0.2 \times 10^{-3} \, \text{A})$$ $$I_C = 12 \times 10^{-3} \, \text{A} = 12 \, \text{mA}$$

The output voltage $V_o$ is the collector-emitter voltage $V_{CE}$. Using KVL for the collector-emitter loop:

$$V_{CC} = I_C R_C + V_o$$ $$V_o = V_{CC} - I_C R_C$$

Given $V_{CC} = 4.5 \, \text{V}$ and $R_C = 200 \, \Omega$.

$$V_o = 4.5 \, \text{V} - (12 \times 10^{-3} \, \text{A}) \times (200 \, \Omega)$$ $$V_o = 4.5 \, \text{V} - 2.4 \, \text{V}$$ $$V_o = 2.1 \, \text{V}$$

The output voltage is $\text{2.1 V}$.