Q221 / 225 · JEE Main 2004 · Physics

For a transistor amplifier in common emitter configuration having load impedance of $1 \mathrm{k} \Omega\left(\mathrm{h}_{\mathrm{fe}}\right.$ $=50$ and $h_{o e}=25$ ) the current gain is

A
$-5.2$
B
$-15.7$
C
$-24.8$
Correct answer
D
$-48.78$

Solution

In CE configuration, $A_{\mathrm{i}}=\frac{-\mathrm{h}_{\mathrm{fe}}}{1+\mathrm{h}_{0 \mathrm{e}} R_{\mathrm{L}}}$ $=\frac{-50}{1+25 \times 10^{-6} \times 1 \times 10^3}=-48.78$

Hence, the correct answer is option: 'D'.

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