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QPages 352-354: Q13: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q13 Solution: A triode valve operates at and . The plate current remains…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently compute ΔV_g with the wrong order or ignore that grid voltages are negative, yielding a negative or incorrect magnitude for μ; always subtract the initial grid voltage from the final (ΔV_g = V_{g2}-V_{g1}) and take the absolute ratio ΔV_p/ΔV_g since the amplification factor is defined as a positive quantity.

A triode valve operates at Vp=225 VV_p = 225 \text{ V} and Vg=−0.5 VV_g = -0.5 \text{ V}. The plate current remains unchanged if the plate voltage is increased to 250 V and the grid voltage is decreased to -2.5 V. Calculate the amplification factor.

Step-by-Step Verified Solution

**Answer:** Amplification factor μ=12.5\mu = 12.5. **Step-by-Step Explanation:** Given: Vp1=225V_{p1} = 225 V, Vg1=−0.5V_{g1} = -0.5 V; Vp2=250V_{p2} = 250 V, Vg2=−2.5V_{g2} = -2.5 V. Since plate current remains unchanged:
μ=ΔVpΔVg=250−225(−0.5)−(−2.5)=252=12.5\mu = \frac{\Delta V_p}{\Delta V_g} = \frac{250 - 225}{(-0.5) - (-2.5)} = \frac{25}{2} = 12.5

All Questions in this Chapter

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