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STD 12 - 14. Semicondutor electronics — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 14. Semicondutor electronics4 Marks
MCQ
A $npn$ transistor is connected in common emitter configuration in a given amplifier. A load resistance of $800 \,\,\Omega$ is connected in the collector circuit and the voltage drop across it is $0.8\,\, V.$ If the current amplification factor is $0.96$ and the input resistance of the circuit is $192 \,\,\Omega$, the voltage gain and the power gain of the amplifier will respectively be
  • A
    $3.69,3.84$
  • B
    $4,4$
  • C
    $4,3.69$
  • ✓
    $4,3.84$
✓

Answer

Correct option: D.
$4,3.84$
d
$\text { Here, } R_{0}=800 \Omega, R_{i}=192 \Omega$

current gain, $\beta=0.96$

Voltage gain $=$ Current gain $\times$ Resistance gain

                         $=0.96 \times \frac{800}{192}=4$

Power gain $=$ [Current gain ] $\times$ [ Voltage gain ]

                  $=0.96 \times 4=3.84$

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