A) \[\frac{1}{20}\mu A\]
B) \[\frac{1}{5}\mu A\]
C) \[\frac{1}{20}mA\]
D) \[\frac{1}{24}mA\]
Correct Answer: D
Solution :
[d]: Here, \[{{R}_{C}}=500\Omega ,{{I}_{C}}{{R}_{C}}=0.5\text{ }V,\] \[{{V}_{CC}}=5V,\alpha =0.96\] As \[{{I}_{C}}{{R}_{C}}=0.5\text{ }V\] \[\therefore \]\[{{I}_{C}}=\frac{0.5V}{{{R}_{C}}}=\frac{0.5V}{500\Omega }=1\times {{10}^{-3}}A=1mA\] The current gains\[\alpha \]and \[\beta \]are related as \[\beta =\frac{\alpha }{1-\alpha }=\frac{0.96}{1-0.96}=24\] The base current is\[{{I}_{B}}=\frac{{{I}_{C}}}{\beta }=\frac{1mA}{24}=\frac{1}{24}mA\]
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