For a transistor $\alpha$ and $\beta$ are given as $\alpha=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{E}}}$ and $\beta=\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}$. Then the correct relation between $\alpha$ and $\beta$ will be:

(A) $\beta=\frac{\alpha}{1-\alpha}$

(C) $\alpha=\frac{\beta}{1-\beta}$

(D) $\alpha=\frac{1-\beta}{\beta}$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Semiconductors

For a transistor

α

and

β

are given as

α = \frac{I _{C}}{I _{E}}

and

β = \frac{I _{C}}{I _{B}}

. Then the correct relation between

α

and

β

will be:

Answer: (A)

For a transistor,

I_{E} = I_{B} + I_{C}

. From

α = \frac{I _{C}}{I _{E}}

, we get

I_{C} = α I_{E}

. From

β = \frac{I _{C}}{I _{B}}

, we get

I_{B} = \frac{I _{C}}{β}

. Substituting into

I_{E} = I_{B} + I_{C}

I_{E} = \frac{I _{C}}{β} + I_{C} = I_{C} (\frac{1}{β} + 1) = I_{C} \frac{1 + β}{β}

. Therefore,

α = \frac{I _{C}}{I _{E}} = \frac{β}{1 + β}

, which gives

β = \frac{α}{1 - α}

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option