**8.**

For a transistor with α=0.95\alpha = 0.95 and IE=1 mAI_E = 1 ext{ mA}, find IBI_B. The symbols used here are in their conventional sense.

Question

**8.**

For a transistor with α=0.95\alpha = 0.95 and IE=1 mAI_E = 1 	ext{ mA}, find IBI_B. The symbols used here are in their conventional sense.

Accepted Answer

**AI-তৈরি নমুনা উত্তর**

Given:
$\alpha = 0.95$
$I_E = 1	ext{ mA}$

We know,
$I_C = \alpha I_E = 0.95 	imes 1	ext{ mA} = 0.95	ext{ mA}$

Also,
$I_E = I_B + I_C$
$\Rightarrow I_B = I_E - I_C = 1	ext{ mA} - 0.95	ext{ mA} = 0.05	ext{ mA} = 50	ext{ }\mu	ext{A}$

Ans: $0.05	ext{ mA}$ (or $50	ext{ }\mu	ext{A}$)

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