In the grid circuit of a triode,a signal E = | vedclass

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(A) The voltage gain $A$ of a triode amplifier is given by $A = \frac{\mu R_L}{r_p + R_L}$.Substituting the given values: $A = \frac{14 	imes 12}{10

Vedclass · Accepted Answer

$1.27$

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(A) The voltage gain $A$ of a triode amplifier is given by $A = \frac{\mu R_L}{r_p + R_L}$.Substituting the given values: $A = \frac{14 	imes 12}{10 + 12} = \frac{168}{22} = \frac{84}{11}$.The peak value of the output voltage is $V_0 = A 	imes E_{peak} = \frac{84}{11} 	imes 2\sqrt{2} 	ext{ V}$.The $RMS$ output voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{84}{11} 	imes 2 = \frac{168}{11} 	ext{ V}$.The $RMS$ current $I_{rms}$ flowing through the load $R_L$ is $I_{rms} = \frac{V_{rms}}{R_L} = \frac{168 / 11}{12 	imes 10^3} 	ext{ A}$.$I_{rms} = \frac{168}{11 	imes 12 	imes 10^3} = \frac{14}{11 	imes 10^3} \approx 1.27 	imes 10^{-3} 	ext{ A} = 1.27 	ext{ mA}$.

In the grid circuit of a triode,a signal $E = 2\sqrt{2} \cos(\omega t)$ is applied. If the amplification factor $\mu = 14$ and the plate resistance $r_p = 10\,k\Omega$,then the root mean square $(RMS)$ current flowing through the load resistance $R_L = 12\,k\Omega$ will be.......$mA$.

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