# The Plate Current, Plate Voltage and Grid Voltage of a 6f6 Triode Tube Are Related as Ip = 41 (Vp + 7 Vg)1.41 Here, Vp And Vg Are in Volts And Ip In Microamperes. - Physics

Sum

The plate current, plate voltage and grid voltage of a 6F6 triode tube are related as ip = 41 (Vp + 7 Vg)1.41

Here, Vp and Vg are in volts and ip in microamperes.
The tube is operated at Vp = 250 V, Vg = −20 V. Calculate (a) the tube current, (b) the plate resistance, (c) the mutual conductance and (d) the amplification factor.

#### Solution

Given:-

Plate voltage, VP = 250 V

Grid voltage, VG = -20V

(a) As given in the question, plate current varies as,

$i_p = 41( V_p + 7 V_g )^{1 . 41}$

$i_p = 41(250 - 140 )^{1 . 41}$

$i_p = 41 \times (110 )^{1 . 41}$

$i_p = 30984 . 71 \mu A = 30 . 98 \text{ mA}$

(b) $i_p = 41( V_p + 7 V_G )^{1 . 41}$

Differentiating this equation, we get:-

$d i_P = 41 \times 1 . 41 \times ( V_p + 7 V_g )^{0 . 41} \times (d V_p + 7d V_g ) ................(1)$

Plate resistance is defined as:-

$r_p = \left( \frac{d V_p}{d i_p} \right)_{V_g = \text{Constant}}$

From equation (1),

$\frac{d V_p}{d i_p} = \frac{1 \times {10}^6}{41 \times 1 . 41 \times {110}^{0 . 41}}$

$\frac{d V_p}{d i_p} = {10}^6 \times 2 . 51 \times {10}^{- 3}$

$\frac{d V_p}{d i_p} = 2 . 5 \times {10}^3 \Omega = 2 . 5 K\Omega$

(c) From above:-

As $d I_{{}_P} = 41 \times 1 . 41 \times (250 + 7 \times ( - 20) )^{0 . 41} \times 7d V_{g,}$

$g_m = (\frac{d I_p}{d V_g} )_{V_P =\text{ Constant}}$

From equation (1),

$\frac{1}{7} \left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 41 \times 1 . 41 \times (250 + 7 \times ( - 20) )^{0 . 41}$

$\left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 41 \times 1 . 41 \times (110 )^{0 . 41} \times 7\text{ mho}$

$\left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 41 \times 1 . 41 \times 6 . 87 \times 7\text{ mho}$

$\left( \frac{d I_p}{d V_g} \right)_{V_P =\text{ Constant}} = 2 . 78\text{ milli mho}$

(d) Amplification factor,

$\mu = r_p \times g_m$

$= 2 . 5 \times {10}^3 \times 2 . 78 \times {10}^{- 3}$

$= 6 . 95 \approx 7$

Concept: Triode
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 19 Electric Current through Gases
Q 19 | Page 353