For a transistor, $\frac{1}{\alpha_{D C}}-\frac{1}{\beta_{D C}}$ is equal to $\left[\alpha_{D C}\right.$ and $\beta_{D C}$ are current amplification factors]

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Aicharya · Accepted Answer

$\begin{aligned} &\frac{1}{\alpha_{D C}}-\frac{1}{\beta_{D C}}=\frac{I_e}{I_c}-\frac{I_b}{I_c}\ &\frac{I_e-I_b}{I_c}=\frac{I_c}{I_c}=1 \end{aligned}$ .

For a transistor, $\frac{1}{\alpha_{D C}}-\frac{1}{\beta_{D C}}$ is equal to $\left[\alpha_{D C}\right.$ and…

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