Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
Share

# 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

ConceptSemiconductor Diode

#### 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:-

$\mu = 15$

$r_P = 2000 Ω = 2 k\Omega$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [2]

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.
S