NEET Class neet Answered
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Asked by jhajuhi19 | 29 Apr, 2019, 05:16: PM
Expert Answer
Figure shows the path of collector current IC , emitter current IE and base current IB .
we have IE = IB + IC ≈ IC and IC = β IB
If we take the path where IC and IE are flowing, sum of potential drop across 2kΩ ,
potential drop across collector-to-emitter (2V) and potential drop across emitter resistance 1kΩ is 5V.
Hence potential drop across 2kΩ and 1 kΩ is 3V . Hence Collector current IC = 3V/(3kΩ) = 1 mA.
Since β of the transistor is 100, we have IB = IC / β = 10-5 A = 10 μA.
If we take the path where IB is flowing, sum of potential drop across resistance R,
Potential difference across Base-emitter junction ( VBE = 0.7 v) and potential difference across
emitter resistance 1 kΩ is 5V.
Since emitter current is 1 mA, potential across emitter resistance 1 kΩ is 1 V.
Hence potential drop across resistance R = (5 - 0.7 - 1) V / (10-5 )A = 330 kΩ
Answered by Thiyagarajan K | 29 Apr, 2019, 08:43: PM
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