In the grid circuit of a triode a signal $E = 2\sqrt{2}\cos{\omega t}$ is applied. If $\mu = 14$ and $r_{p} = 10\ k\Omega$ then root mean square current flowing through $R_{L} = 12\ k\Omega$ will be

Question

Dr. Abubacker · Accepted Answer

(a) $A = \frac{\mu R_{L}}{r_{p} + R_{L}} = \frac{14 	imes 12}{10 + 12} = \frac{84}{11}$. Peak value of output signal $V_{0} = \frac{84}{11} 	imes 2\sqrt{2}V$
$\Rightarrow V_{rms} = \frac{V_{0}}{\sqrt{2}} = \frac{84 	imes 2}{11}V$
$\Rightarrow r.m.s.$ value of current through the load
$= \frac{84 	imes 2}{11 	imes 12 	imes 10^{3}}A = 1.27\ mA$

Q. In the grid circuit of a triode a signal $E = 2\sqrt{2}\cos{\omega t}$ is applied. If $\mu = 14$ and $r_{p} = 10\ k\Omega$ then root mean square current flowing through $R_{L} = 12\ k\Omega$ will be

Explanation

Reviewed by Dr. Abubacker

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