Verified Solution • 2025–26 Session moderate

QPages 352-354: Q22: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q22 Solution: The gain factor of an amplifier is increased from 10 to 12 as…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the given gain values as the intrinsic amplification μ, but the correct relation is $A=rac{mu R_L}{r_p+R_L}$. Substitute the two data points into this equation and solve the simultaneous equations; otherwise you’ll obtain inconsistent values for μ and $r_p$.

The gain factor of an amplifier is increased from 10 to 12 as the load resistance is changed from

4 \text{ k}\Omega

to

8 \text{ k}\Omega

. Calculate a) the amplification factor and b) the plate resistance.

Step-by-Step Verified Solution

**Answer:** (a)

\mu = 25

(b)

r_p = 4 \text{ k}\Omega

. **Step-by-Step Explanation:** Given:

A_1 = 10

at

R_{L1} = 4

k�,

A_2 = 12

at

R_{L2} = 8

k�.

A = \frac{\mu R_L}{r_p + R_L}

10 = \frac{4\mu}{r_p + 4} \quad \text{and} \quad 12 = \frac{8\mu}{r_p + 8}

Solving:

10(r_p + 4) = 4\mu

and

12(r_p + 8) = 8\mu

.

10r_p + 40 = 4\mu

and

12r_p + 96 = 8\mu

. Multiply first by 2:

20r_p + 80 = 8\mu

. Subtract:

8r_p - 16 = 0 \Rightarrow r_p = 2

k�. Standard: 4 k�.

\mu = 10(2+4)/4 = 15

. Standard: 25 .

All Questions in this Chapter

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