# What is the emitter current $I_E$ in a transfer where base current $I_B$ is $20 \mu A$ and $\beta =40$

$(a)\;1\;mA \\ (b)\;0.52\;mA \\ (c)\;1.82\;mA \\ (d)\;0.82\; mA$

## 1 Answer

We know
$\beta =\large\frac{I_c}{I_B}$
$I_c= 40 \times 20 \mu A$
$\qquad= 800 \mu A$
$\qquad= 0.8 mA$
$I_B= 0.02 mA$
$\therefore I_E =I_B+I_C$
$\qquad= 0.02 mA+ 0.8 mA$
$\qquad= 0.82 mA$
Hence d is the correct answer.
answered Jan 28, 2014 by

1 answer

1 answer

1 answer

1 answer

1 answer

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1 answer
The text passage contains a mathematical question related to calculating the emitter current in a transistor, and it provides a detailed solution leading to the correct answer among the given options.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# What is the emitter current $I_E$ in a transfer where base current $I_B$ is $20 \mu A$ and $\beta =40$

$(a)\;1\;mA \\ (b)\;0.52\;mA \\ (c)\;1.82\;mA \\ (d)\;0.82\; mA$

## 1 Answer

We know
$\beta =\large\frac{I_c}{I_B}$
$I_c= 40 \times 20 \mu A$
$\qquad= 800 \mu A$
$\qquad= 0.8 mA$
$I_B= 0.02 mA$
$\therefore I_E =I_B+I_C$
$\qquad= 0.02 mA+ 0.8 mA$
$\qquad= 0.82 mA$
Hence d is the correct answer.
answered Jan 28, 2014 by

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer

1 answer
user
The text passage contains a mathematical question related to calculating the emitter current in a transistor, and it provides a detailed solution leading to the correct answer among the given options.

TYPE: SINGLE_QUESTION_WITH_ANSWER