#### 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,

R_{L} = 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:-

\[\mu = 15\]

\[ r_P = 2000 Ω = 2 k\Omega\]