In the grid circuit of a triode, an input vol | vedclass

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

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$1.27$

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$(A)$ The voltage gain $A$ of a triode amplifier is given by the formula: $A = \frac{\mu R_L}{r_p + R_L}$.Substituting the given values: $A = \frac{14 	imes 12}{10 + 12} = \frac{168}{22} = \frac{84}{11}$.The input voltage is $E = 2\sqrt{2} \cos \omega t$, so the peak input voltage is $V_{in, peak} = 2\sqrt{2} 	ext{ V}$.The peak output voltage across the load is $V_{out, peak} = A 	imes V_{in, peak} = \frac{84}{11} 	imes 2\sqrt{2} 	ext{ V}$.The $r.m.s.$ output voltage is $V_{out, rms} = \frac{V_{out, peak}}{\sqrt{2}} = \frac{84}{11} 	imes 2 = \frac{168}{11} 	ext{ V}$.The $r.m.s.$ current through the load $R_L$ is $I_{rms} = \frac{V_{out, rms}}{R_L} = \frac{168 / 11}{12 	imes 10^3} 	ext{ A}$.$I_{rms} = \frac{168}{11 	imes 12} 	imes 10^{-3} 	ext{ A} = \frac{14}{11} 	imes 10^{-3} 	ext{ A} \approx 1.27 	imes 10^{-3} 	ext{ A} = 1.27 	ext{ mA}$.

In the grid circuit of a triode, an input voltage $E = 2\sqrt{2} \cos \omega t$ is applied. If the amplification factor $\mu = 14$ and the plate resistance $r_p = 10 \text{ k}\Omega$, find the $r.m.s.$ current flowing through the load resistance $R_L = 12 \text{ k}\Omega$ in $\text{mA}$.

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