Question

1)
In a N P N transistor about 10^10 electrons enter the emitter in 2meu s when it
is connected to a battery. Then I =
(A) 200
(B) 400
(C) 800
(D) 1600​

Answer 1

Answer:

200

Explanation:

200

200

200

200

200 200 200

Answer 2

Answer:

The number of electrons entering the emitter is 1010 electrons.

The time taken by the electrons is 10−6 s.

Current gain common emitter 

β=IBIC

The emitter current is given as:

IE=tq

⇒10−61010×1.6×10−19

IE=1.6 mA

Now, 5% of the electrons recombine in the base region. So the base current will be:

IB=5%×1.6 mA

IB=0.08 mA

We know that the emitter current is the sum of the base current and collector current.

IE=IB+IC

So, IC=1.6−0.08 = 1.52 mA

⟹β=0.081.52

=19

Explanation:

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