# A certain n-p-n transistor has the common emitter output characteristics as shown in the figure.(a) Find the emitter current at$V_{CE} = 12.5 \: V \: and\: I_{b} = 60\: \mu A$, and (b) Current gain $'\beta '$ at this point.

We have given;

$V_{CE}=12.5\: v$

$I_{b}=60\; \mu A$

$I_{c}=6\; mA\;$   (from the given graph)

We know,

$I_{e}=I_{b}+I_{c}$

Where, $I_{e}$ = emitter current

$I_{b}$ = base current

$I_{c}$ = Collector current

$\therefore I_{e}=60\; \mu A+6\; mA$

$\therefore I_{e}=60\times 10^{-6}A+6\times 10^{-3}A$

$I_{e}= 6.060 \; mA$

Calculating the current gain

$(\beta )=\frac{I_{c}}{I_{b}}=\frac{6mA}{60\mu A}$

$\beta =\frac{6\times 10^{-3}}{6\times 10^{-6}}=\frac{6}{60}\times 10^{-3}\times 10^{6}$

$\beta = 0.1\times 10^{3}$

$\beta = 100$

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