MHTCET 26 Apr 2025 Shift 1 Paper

Concept:In a common emitter amplifier, the output voltage is found using the input current, AC current gain, and load resistance.Explanation:Given: $R_{in} = 1.8 \mathrm{k}\Omega$, $R_L = 9 \mathrm{k}\Omega$, $\beta_{ac} = 70$, $V_{in} = 6 \mathrm{mV}$.First, find the AC input current:$$I_{in} = \frac{V_{in}}{R_{in}} = \frac{6 \times 10^{-3}}{1.8 \times 10^{3}} = 3.33 \times 10^{-6} \mathrm{A}$$Next, calculate the AC output current using the current gain:$$I_{out} = \beta_{ac} \cdot I_{in} = 70 \times 3.33 \times 10^{-6} = 2.33 \times 10^{-4} \mathrm{A}$$Finally, find the output voltage across the load:$$V_{out} = I_{out} \cdot R_L = 2.33 \times 10^{-4} \times 9 \times 10^{3} \approx 2.1 \mathrm{V}$$Answer:The output voltage will be $2.1 \mathrm{V}$, which corresponds to option C.