In an NPN transistor the collector current is $24$ mA. If $80 %$ of electrons reach collector its base current in mA is?

36

No worries! We‘ve got your back. Try BYJU‘S free classes today!

26

No worries! We‘ve got your back. Try BYJU‘S free classes today!

16

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $6$
We know that, $I_{e} = I_{c} + I_{b}$
Given 80% of $I_{e} = I_{c} = 24 m A$
$(80 / 100) I_{e} = 24 m A$
$I_{e} = (24 * 100) / 80$
$I_{e} = 30 m A$
$I_{b} = I_{e} - I_{c}$
$= (30 - 24) m A$
$= 6 m A$

Suggest Corrections

Similar questions

In an npn transistor, the collector current is 24 mA. If 80% of electrons reach collector, its base current in mA is

10

90 %

20 mA

80 %

10 mA

90 %

(I_{E})

(I_{B})

In an n – p – n transistor circuit, the collector current is 10 mA. If $90 %$ of the electrons emitted reach the collector. Then

Join BYJU'S Learning Program