MHT CET · Physics · Semiconductors
A change of \(8.0 \mathrm{~mA}\) in the emitter current brings a change of \(7.9 \mathrm{~mA}\) in the collector current. The values of \(\alpha\) and \(\beta\) are
- A \(0.99,90\)
- B \(0.96,79\)
- C \(0.97,99\)
- D \(0.99,79\)
Answer & Solution
Correct Answer
(D) \(0.99,79\)
Step-by-step Solution
Detailed explanation
Given that, change in emitter current, \(\Delta I_{E}=8 \mathrm{~mA}\)
and change in collector current, \(\Delta I_{C}=7.9 \mathrm{~mA}\) We know that,
\(
\alpha=\frac{\Delta I_{C}}{\Delta I_{E}} \Rightarrow \alpha=\frac{7.9}{8} \Rightarrow \alpha \simeq 0.99
\)
Also we know that
\(\beta=\frac{\alpha}{1-\alpha}\)
\(\Rightarrow \beta=\frac{\frac{7.9}{8}}{1-\frac{7.9}{8}}=\frac{7.9}{8-7.9}\)
or \(\quad \beta=\frac{7.9}{0.1}=79\)
Hence, the required answer is \(\alpha=0.99\) and \(\beta=79\)
and change in collector current, \(\Delta I_{C}=7.9 \mathrm{~mA}\) We know that,
\(
\alpha=\frac{\Delta I_{C}}{\Delta I_{E}} \Rightarrow \alpha=\frac{7.9}{8} \Rightarrow \alpha \simeq 0.99
\)
Also we know that
\(\beta=\frac{\alpha}{1-\alpha}\)
\(\Rightarrow \beta=\frac{\frac{7.9}{8}}{1-\frac{7.9}{8}}=\frac{7.9}{8-7.9}\)
or \(\quad \beta=\frac{7.9}{0.1}=79\)
Hence, the required answer is \(\alpha=0.99\) and \(\beta=79\)
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