Q. The linear portions of the characteristic curves of a triode valve give the following readings $V_{g}(volt)$ 0 $- 2$ $- 4$ $- 6$ $I_{p}(mA)$ for $V_{p} = 150\ volts$ 15 12.5 10 7.5 $I_{p}(mA)$ for $V_{p} = 120\ volts$ 10 7.5 5 2.5 The plate resistance is
A) $2000\ ohms$
B) 4000 $ohms$
C) 8000 $ohms$
D) 6000 $ohms$
Explanation
(d) $g_{m} = \left( \frac{\mathrm{\Delta}i_{p}}{\mathrm{\Delta}V_{g}} \right)_{V_{p} = constant} = \frac{(15 - 10) \times 10^{- 3}}{0 - ( - 4)} = 1.25 \times 10^{- 3}\Omega$
$\mu = \left( \frac{\Delta V_{p}}{\Delta V_{g}} \right)_{I_{p} = constant} = \frac{150 - 120}{0 - ( - 4)} = 7.5$
$\therefore r_{p} = \frac{\mu}{g_{m}} = \frac{7.5}{1.25 \times 10^{- 3}} = 6000\ ohms$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.