In common emitter amplifier, input resistance is 1000 Ω, peak value of input signal voltage is 5 mV and β = 60. The peak value of output current is (a) 3 × 10⁻⁴ A (b) 0.5 × 10⁻⁴ A (c) 2 × 10⁻⁵ A (d) 1 × 10⁻⁵ A

Answer: (a). Peak input (base) current IB = (Vi)/(Ri) = 5 × 10⁻³1000 = 5 × 10⁻⁶ A. Peak output (collector) current IC = β IB = 60 × 5 × 10⁻⁶ = 3 × 10⁻⁴ A. So the answer is (a) 3 × 10⁻⁴ A.

In common emitter amplifier, input resistance is 1000 Ω, peak value… | MHT CET 2020 (16 Oct, Shift 1)

Q78MHT CET 2020 · 16 OCT · SHIFT 1Medium

Junction Transistor

In common emitter amplifier, input resistance is $1000\ \Omega$ , peak value of input signal voltage is $5\text{ mV}$ and $\beta = 60$ . The peak value of output current is

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Answer: (a)

Peak input (base) current $I_B = \dfrac{V_i}{R_i} = \dfrac{5 \times 10^{-3}}{1000} = 5 \times 10^{-6}$ A.
Peak output (collector) current $I_C = \beta I_B = 60 \times 5 \times 10^{-6} = 3 \times 10^{-4}$ A.
So the answer is (a) $3 \times 10^{-4}$ A.

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