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For a BJT, IC = 5 mA, IB = 50 μA and ICBO = 0.5 μA, then the value of β is
1. 99
2. 91
3. 79
4. 61

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Correct Answer - Option 1 : 99

Given that, IC = 5 mA, IB = 50 μA and ICBO = 0.5 μA

Collector current IC = β IB + (1 + β) ICBO

⇒ 5 × 10-3 = β (50 × 10-6) + (1 + β) (0.5 × 10-6)

⇒ 5 × 10-3 = 10-6 (50β + 0.5 + 0.5β)

⇒ β = 99

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