Question
Download Solution PDFIf an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of β will be:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
⇒ IE = IB+ IC 
,
⇒ \(\Delta I_{E} = \Delta I_{B} + \Delta I_{C}\)
Given :
Change in emitter current = 4 mA
Change in collector current = 3.5 mA
\(I_{E} = I_{B}+ I_{C}\)
\(\Delta I_{E} = \Delta I_{B} + \Delta I_{C}\)
4 = \( \Delta I_{B} +\) 3.5
∴ Change in base current, \(\Delta I_{B} = 0.5~mA\)
Also we know that, the amplification factor, \(β = \frac{\Delta I_{c}}{\Delta I_{B}}\)
\( ⇒ β = \frac{3.5}{0.5} \) => β = 7
Hence, Option A is correct.
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