A) 10, 16.7 kW, 0.6 m mho
B) 15, 16.7 kW, 0.06 m mho
C) 20, 6 kW, 16.7 m mho
D) None of these
Correct Answer: A
Solution :
\[{{I}_{p}}=0.004\,{{({{V}_{p}}+10{{V}_{g}})}^{3/2}}\] Þ \[\frac{\Delta {{I}_{p}}}{\Delta {{V}_{g}}}=0.004\,\left[ \frac{3}{2}\,{{({{V}_{p}}+10{{V}_{g}})}^{1/2}}\times 10 \right]\] Þ \[{{g}_{m}}=0.004\times \frac{3}{2}\,{{(120+10\times -2)}^{1/2}}\times 10\] Þ \[{{g}_{m}}=6\times {{10}^{-4}}mho=0.6\,m\,mho\] Comparing the given equation of Ip with standard equation \[{{I}_{p}}=K\,{{({{V}_{p}}+\mu {{V}_{g}})}^{3/2}}\] we get m = 10 Also from m = rp ´ gm Þ \[{{r}_{p}}=\frac{\mu }{{{g}_{m}}}=\frac{10}{0.6\times {{10}^{-3}}}\] Þ \[{{r}_{p}}=16.67\,\times {{10}^{3}}\Omega =16.67\,k\Omega .\]
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