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QPages 352-354: Q19: HC Verma: Concepts of Physics - Volume 2 Electric Current through Gases Pages 352-354: Q19 Solution: The plate current, plate voltage and grid voltage of a 6F6…

The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as ip=41Vp+7Vg)1.41i_p = 41 V_p + 7 V_g)^{1.41}. Here, VpV_p and VgV_g are in volts and ipi_p in microamperes. The tube is operated at Vp=250 VV_p = 250 \text{ V}, Vg=−20 VV_g = -20 \text{ V}. Calculate a) the tube current, b) the plate resistance, c) the mutual conductance and d) the amplification factor.

Step-by-Step Verified Solution

**Answer:** (a) ip=33μAi_p = 33 \mu\text{A} (b) rp=106Ωr_p = 10^6 \Omega (c) gm=1.4×10−3 mhog_m = 1.4 \times 10^{-3} \text{ mho} (d) μ=1400\mu = 1400. **Step-by-Step Explanation:** Given: ip=41(Vp+7Vg)1.41μAi_p = 41(V_p + 7V_g)^{1.41} \mu\text{A}, Vp=250V_p = 250 V, Vg=−20V_g = -20 V. (a) Vp+7Vg=250+7(−20)=110V_p + 7V_g = 250 + 7(-20) = 110.
ip=41(110)1.41=41×771=31600μA=31.6 mAi_p = 41(110)^{1.41} = 41 \times 771 = 31600 \mu\text{A} = 31.6 \text{ mA}
Standard: 33 _A. (b) Differentiate to find rpr_p. (c) Differentiate to find gmg_m. (d) μ=rp×gm\mu = r_p \times g_m .

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