For a transistor, the current amplification factor is $0.8$ . The transistor is connected in common emitter configuration, the change in collector current when the base current changes by $6 m A$ is

6 m A

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

4.8 m A

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

24 m A

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

8 m A

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

Open in App

Solution

The correct option is C $24 m A$
Given that:
$α = 0.8$
$I_{b} = 6 m A$
we have
$β = \frac{α}{1 - α}$
$β = \frac{0.8}{1 - 0.8} = 4$
But,
$β = \frac{I_{c}}{I_{b}}$
$I_{c} = β \times I_{b}$
$I_{c} = 4 \times 6 = 24 m A$

Suggest Corrections

Similar questions

For a transistor, the current amplification factor(α) is 0.8. The transistor is connected in common emitter configuration. The change in collector current, when the base current changes by 6 mA is

Q. For a transistor, the current amplification factor is

0.8

. The transistor is connected in common base configuration, the change in collector current when the base current changes by

6 m A

For a transistor, the value of $α$ is 0.8. The transistor is connected in common-emitter configuration. The change in the collector current when the base current changes by 6 mA is:

Q. A transistor is used in common-emitter configuration. Given its

α = 0.9

, calculate the change in collector current when the base current changes by

2 μ A

Q. A transistor is connected in common emitter (CE) configuration. The collector supply is 8V and the voltage drop across a resistor of

800 Ω

in the collector circuit is 0.8 V. If the current gain factor (

α

) is 0.96, then the change in base current is

Join BYJU'S Learning Program

For a transistor, the current amplification factor is 0.8. The transistor is connected in common emitter configuration, the change in collector current when the base current changes by 6 mA is

The correct option is C 24 mAGiven that:α=0.8Ib=6 mAwe haveβ=α1−αβ=0.81−0.8=4But,β=IcIbIc=β×IbIc=4×6=24 mA

For a transistor, the current amplification factor is $0.8$ . The transistor is connected in common emitter configuration, the change in collector current when the base current changes by $6 m A$ is

The correct option is C $24 m A$
Given that:
$α = 0.8$
$I_{b} = 6 m A$
we have
$β = \frac{α}{1 - α}$
$β = \frac{0.8}{1 - 0.8} = 4$
But,
$β = \frac{I_{c}}{I_{b}}$
$I_{c} = β \times I_{b}$
$I_{c} = 4 \times 6 = 24 m A$