A) \[{{E}_{0}}\omega ={{B}_{0}}k\]
B) \[{{E}_{0}}{{R}_{0}}=\omega k\]
C) \[{{E}_{0}}k={{B}_{0}}\omega \]
D) \[r\]
Correct Answer: C
Solution :
Key idea: Base emitter junction is forward biased hence junction resistance is zero. In the given circuit, the input signal is applied across the emitter and base while the amplified output signal is taken across collector and base. Also resistance offered by junction is forward bias is zero. Hence, \[{{f}_{o}}+{{f}_{e}}\] \[\therefore \] \[{{f}_{e}}=\frac{{{f}_{o}}}{|M|}=\frac{100}{50}=2\,cm\] Given, V=7 volt, \[={{f}_{o}}+{{f}_{e}}=100+2=102\,cm\] \[\frac{{{\partial }^{2}}E}{\partial {{x}^{2}}}=\frac{1}{{{c}^{2}}}\frac{{{\partial }^{2}}E}{\partial {{t}^{2}}}\] \[E={{E}_{o}}\,\sin \left( kx-\omega t \right)\] \[B={{B}_{o}}\,\sin \left( kx-\omega t \right)\] \[\frac{{{E}_{o}}}{{{B}_{o}}}=c\]
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