Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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# The Plate Resistance and Amplification Factor of a Triode Are 10kω and 20. the Tube is Operated at Plate Voltage 250 V and Grid Voltage −7.5 V. the Plate Current is 10 Ma. - Physics

Sum

The plate resistance and amplification factor of a triode are 10kΩ and 20. The tube is operated at plate voltage 250 V and grid voltage −7.5 V. The plate current is 10 mA. (a) To what value should the grid voltage be changed so as to increase the plate current to 15 mA? (b) To what value should the plate voltage be changed to take the plate current back to 10 mA?

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#### Solution

Given:-

Plate resistance,

r_p=10KOmega

Amplification factor,

mu=20

Plate voltage,

V_p=240V

Grid Voltage,

V_G=7.5V

I_p=10"mA"

(a) Transconductance,

$\delta V_G = \frac{\delta I_p}{g_m}$

$\delta V_G = \frac{\left(15 \times {10}^{- 3} - 10 \times {10}^{- 3} \right)}{\mu/ r_p}$

$\delta V_G = \frac{5 \times {10}^{- 3}}{20/10 \times {10}^{- 3}}$

$\delta V_G = \frac{5 \times {10}^{- 3}}{2 \times {10}^{- 3}} = \frac{5}{2} = 2 . 5$

$V_G = + 2 . 5 - 7 . 5 = - 5 . 0 V$

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

$\Rightarrow {10}^4 = \frac{\delta V_p}{\left(10 \times {10}^{- 3} - 15 \times {10}^{- 3} \right)}$

$\Rightarrow \delta V_p = - {10}^4 \times 5 \times {10}^{- 3} = - 50 V$

$\therefore V_p ' - V_p = - 50$

$\Rightarrow V_p ' = - 50 + V_p = 200 V$

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 18 | Page 353
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