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QPages 352-354: Q17: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q17 Solution: The plate resistance of a triode is and the transconductance is …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert transconductance from millimhos to mhos ($S$) or plate resistance from kilohms to ohms ($\Omega$) before applying the linear approximation formulas, leading to magnitude errors. Always ensure $g_m$ is in $S$ (or $A/V$) and $r_p$ is in $\Omega$ when calculating $\Delta I_p$ and $\Delta V_g$.

The plate resistance of a triode is 8 kΩ8 \text{ k}\Omega and the transconductance is 2.5 millimho. a) If the plate voltage is increased by 48 V and the grid voltage is kept constant, what will be the increase in the plate current? b) With plate voltage kept constant at this increased value, by how much should the grid voltage be decreased in order to bring the plate current back to its initial value?

Step-by-Step Verified Solution

**Answer:** (a) ΔIp=6 mA\Delta I_p = 6 \text{ mA} (b) ΔVg=2.4 V\Delta V_g = 2.4 \text{ V}. **Step-by-Step Explanation:** (a) rp=8 kΩ=8000Ωr_p = 8 \text{ k}\Omega = 8000 \Omega, ΔVp=48\Delta V_p = 48 V.
ΔIp=488000=0.006 A=6 mA\Delta I_p = \frac{48}{8000} = 0.006 \text{ A} = 6 \text{ mA}
(b) To bring current back, the change in grid voltage must produce an equal and opposite effect.
gm=2.5 millimho=0.0025 mhog_m = 2.5 \text{ millimho} = 0.0025 \text{ mho}
ΔVg=ΔIpgm=0.0060.0025=2.4 V\Delta V_g = \frac{\Delta I_p}{g_m} = \frac{0.006}{0.0025} = 2.4 \text{ V}

All Questions in this Chapter

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