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  • In the grid circuit of a triode a signal cos ωt is applied. If <img alt="\mu =14" src="http

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

Option: 1

\mathrm{1.27\ mA}


Option: 2

\mathrm{10\ mA}


Option: 3

\mathrm{1.5\ mA}


Option: 4

\mathrm{12.4\ mA}


Answers (1)

best_answer

\mathrm{A=\frac{\mu R_L}{r_p+R_L}=\frac{14 \times 12}{10+12}=\frac{84}{11},} Peak value of output signal \mathrm{V_0=\frac{84}{11} \times 2 \sqrt{2} V}
\mathrm{\Rightarrow V_{r m s}=\frac{V_0}{\sqrt{2}}=\frac{84 \times 2}{11} \mathrm{~V}}
\mathrm{\Rightarrow r.m.s.} value of current through the load
\mathrm{=\frac{84 \times 2}{11 \times 12 \times 10^3} A=1.27 \mathrm{~mA}}

Posted by

sudhir kumar

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