In the study of transistor as an amplifier if α = (IC)/(IE) = 0.98 and β = (IC)/(IB) = 49, where IC, IB and IE are collector, base and emitter current respectively then ((1)/(α) - (1)/(β)) is equal to (a) zero (b) 1/2 (c) 2 (d) 1

Answer: (d). IE = IB + IC. Divide by IC: (1)/(α) = (1)/(β) + 1. So (1)/(α) - (1)/(β) = 1. Check: 1/0.98 - 1/49 = 1.0204 - 0.0204 = 1 So the answer is (d) 1.

In the study of transistor as an amplifier if α = (IC)/(IE) = 0.98… | MHT CET 2023 (11 May, Shift 2)

Q43MHT CET 2023 · 11 MAY · SHIFT 2Easy

Junction Transistor

In the study of transistor as an amplifier if $\alpha = \frac{I_C}{I_E} = 0.98$ and $\beta = \frac{I_C}{I_B} = 49$ , where $I_C, I_B$ and $\text{I}_\text{E}$ are collector, base and emitter current respectively then $\left(\frac{1}{\alpha} - \frac{1}{\beta}\right)$ is equal to

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Answer: (d)

$I_E = I_B + I_C$ . Divide by $I_C$ : $\dfrac{1}{\alpha} = \dfrac{1}{\beta} + 1$ .
So $\dfrac{1}{\alpha} - \dfrac{1}{\beta} = 1$ .
Check: $\dfrac{1}{0.98} - \dfrac{1}{49} = 1.0204 - 0.0204 = 1$
So the answer is (d) 1.

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