Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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The Gain Factor of an Amplifier in Increased from 10 to 12 as the Load Resistance is Changed from 4 Kω to 8 Kω. Calculate (A) the Amplification Factor and (B) the Plate Resistance. - Physics

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Question

The gain factor of an amplifier in increased from 10 to 12 as the load resistance is changed from 4 kΩ to 8 kΩ. Calculate (a) the amplification factor and (b) the plate resistance.

Solution

We know:-

Voltage gain = \[\frac{\mu}{1 + \frac{r_p}{R_L}} ............(1)\]

When voltage amplification factor, A = 10,

RL = 4 kΩ

\[10 = \frac{\mu}{1 + \frac{r_p}{4 \times {10}^3}}\]

\[\Rightarrow 10=\frac{\mu \times 4 \times {10}^3}{4 \times {10}^3 + r_p}\]

\[ \Rightarrow 4 \times  {10}^4  + 10 r_P    =   4000  \mu .........(2)\]

Now, increased gain,  A = 12

Substituting this value in (1) ,we get:-

\[12 = \frac{\mu}{1 + \frac{r_P}{R_L}}\]

\[12 = \frac{\mu}{1 + \frac{r_p}{8 \times {10}^3}}\]

\[ \Rightarrow 12 = \frac{\mu \times 8000}{8000 + r_P}\]

\[ \Rightarrow 96000 + 12 r_P = 8000 \mu ...........(3)\]

On solving equations (2) and (3), we get:-

\[\mu   =   15\]

\[ r_P    =   2000   Ω  = 2  k\Omega\]

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Solution The Gain Factor of an Amplifier in Increased from 10 to 12 as the Load Resistance is Changed from 4 Kω to 8 Kω. Calculate (A) the Amplification Factor and (B) the Plate Resistance. Concept: Semiconductor Diode.
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