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QPages 352-354: Q23: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q23 Solution: The figure shows two identical triode tubes connected in…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume that the amplification factor $\mu$ doubles in parallel connection; however, since $\mu = r_p g_m$, the halving of plate resistance and doubling of mutual conductance cancel out, leaving $\mu$ unchanged.

The figure shows two identical triode tubes connected in parallel. The anodes are connected together and the cathodes are connected together. Show that the equivalent plate resistance is half the individual plate resistance, the equivalent mutual conductance is double the individual mutual conductance and the equivalent amplification factor is the same as the individual amplification factor.

Step-by-Step Verified Solution

**Answer:** Equivalent plate resistance is half, mutual conductance is double, amplification factor is the same. **Step-by-Step Explanation:** For two identical triodes in parallel: - Plate current doubles for the same plate voltage: rp,eq=rp/2r_{p,eq} = r_p/2. - Change in plate current doubles for the same grid voltage: gm,eq=2gmg_{m,eq} = 2g_m. - μeq=rp,eq×gm,eq=(rp/2)×2gm=rp×gm=μ\mu_{eq} = r_{p,eq} \times g_{m,eq} = (r_p/2) \times 2g_m = r_p \times g_m = \mu.

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