Question
Download Solution PDFयदि किसी उत्सर्जक धारा में 4 mA का परिवर्तन होने पर संग्राही धारा में 3.5 mA का परिवर्तन होता है, तो β का मान क्या होगा?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFस्पष्टीकरण:
⇒ IE = IB+ IC 
,
⇒ \(\Delta I_{E} = \Delta I_{B} + \Delta I_{C}\)
दिया गया है:
उत्सर्जी धारा में परिवर्तन = 4 mA
संग्राही धारा में परिवर्तन = 3.5 mA
\(I_{E} = I_{B}+ I_{C}\)
\(\Delta I_{E} = \Delta I_{B} + \Delta I_{C}\)
4 = \( \Delta I_{B} +\) 3.5
∴ आधार धारा में परिवर्तन, \(\Delta I_{B} = 0.5~mA\)
हम यह भी जानते हैं कि, प्रवर्धन कारक, \(β = \frac{\Delta I_{c}}{\Delta I_{B}}\)
\( ⇒ β = \frac{3.5}{0.5} \) => β = 7
अतः विकल्प A सही है।
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