In an n - p - n transistor circuit- the collector current is 10 mA- If 90 - of the electrons emitted reach the collector- ThenA- The emitter current will be 9 mAB- The base current will be 9 mAC- The base current will be 0-1 MaD- The emitter current will be 11 mA

The correct option is D The emitter current will be 11 mA I_c=10mA I_c = 90/100I_E ⇒ I_E=10/9× 10mA = 11.1mA I_B=I_E I_C =100/T 10=10/9 =1.1mA ...

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Standard XII

Physics

Transistor

In an n - p -...

Question

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

The emitter current will be 9 mA

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The emitter current will be 11 mA

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The base current will be 9 mA

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The base current will be 0.1 Ma

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Solution

The correct option is B
The emitter current will be 11 mA

$I_{c} = 10 m A I_{c} = \frac{90}{100} I_{E} \Rightarrow I_{E} = \frac{10}{9} \times 10 m A = 11.1 m A I_{B} = I_{E} - I_{C} = \frac{100}{T} - 10 = \frac{10}{9} = 1.1 m A$

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

The correct option is B The emitter current will be 11 mA Ic=10mAIc=90100IE⇒ IE=109×10mA=11.1mAIB=IE−IC=100T−10=109=1.1mA

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

The correct option is B
The emitter current will be 11 mA

$I_{c} = 10 m A I_{c} = \frac{90}{100} I_{E} \Rightarrow I_{E} = \frac{10}{9} \times 10 m A = 11.1 m A I_{B} = I_{E} - I_{C} = \frac{100}{T} - 10 = \frac{10}{9} = 1.1 m A$