A) 0.42mA
B) 0.49mA
C) 0.29 mA
D) 0.35 mA
Correct Answer: C
Solution :
Key Idea: Current gain in common-base amplifier is ratio of change in collector current to change in emitter current. Current gain of transistor is \[_{Z}{{X}^{A}}{{\xrightarrow{{}}}_{Z-2}}{{Y}^{A-4}}{{+}_{2}}H{{e}^{4}}\] where, \[\alpha \] is change in collector current, and \[~{{\lambda }_{m}}T=constant\] is change in emitter current. Given, \[\lambda \] \[\therefore \] Also \[\int{E.dA=\frac{q}{{{\varepsilon }_{0}}}}\] \[=\frac{1}{6}\times \frac{q}{{{\varepsilon }_{0}}}=\frac{q}{6{{\varepsilon }_{0}}}\] \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-2Q)}{r}\] \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(Q)}{r}\]
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