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Assertion : In a common-emitter amplfier, the load resistance of the output circuit is 1000 times the load resistance of the input circuit. If $\alpha=0.98$, then voltage gain is $49 \times 10^3$.
Reason : $\alpha=\frac{\beta}{1-\beta}$ (symbols have their usual meaning).
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Reason : $\alpha=\frac{\beta}{1-\beta}$ (symbols have their usual meaning).
Solution:
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The correct answer is:
If assertion is true but reason is false
We know that
$\alpha=\frac{\Delta i_C}{\Delta i_E}$ and $\beta=\frac{\Delta i_C}{\Delta i_B}$
Also $\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49$
$\therefore$ Voltage gain $=\beta=\frac{R_2}{R_1}=49 \times 1000=49 \times 10^3$
$\alpha=\frac{\Delta i_C}{\Delta i_E}$ and $\beta=\frac{\Delta i_C}{\Delta i_B}$
Also $\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49$
$\therefore$ Voltage gain $=\beta=\frac{R_2}{R_1}=49 \times 1000=49 \times 10^3$
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