A common emitter amplifier circuit is shown in the figure below. For the transistor used in the circuit the current amplification factor, $β_{d c} = 100$ . Other parameters are mentioned in the figure.
We find that :

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$V_{B E} = + 20.7 V, V_{B C} = + 3 . 75 V a n d a m p l i f i e r i s n o t w o r k i n g .$

$V_{B E} = + 18.2 V, V_{B C} = = - 3.45 V a n d a m p l i f i e r i s w o r k i n g .$

$V_{B E} = + 18.5 V, V_{B C} = + 2.85 V a n d a m p l i f i e r i s n o t w o r k i n g .$

$V_{B E} = + 21.5 V, V_{B C} = - 2 . 75 V a n d a m p l i f i e r i s w o r k i n g .$

answer is D.

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Detailed Solution

$\begin{array}{l} β_{d c} = 100, β_{d c} = \frac{i_{C}}{i_{B}} \Rightarrow i_{B} = \frac{i_{c}}{100} - - - - (1) \end{array}$

$\begin{array}{l} V_{C C} = + i_{C} R_{L} + V_{B C} + V_{B E} - - - - - - (2) \\ - i_{B} R_{B} + V_{B C} + i_{C} R_{L} = 0 - - - - - - (3) \\ S o l v i n g w e g e t V_{B E} = + 20.7 V, V_{B C} = - 3.75 V \end{array}$

NOTE: According to the question none of the options are matching. 4 th option is nearest.

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