(Solved): Consider an NPN transistor having a beta =38, operating with an emitter current of IE=7.5mA. ...

Consider an NPN transistor having a beta $$=38$$, operating with an emitter current of $$I_E=7.5\mathrm{mA}$$. The voltage drops across the emitter-base and collector-base junctions are: $$V_{BE}=0.70\mathrm{V},V_{CB}=2.1\mathrm{V}$$ Find the total power dissipated in the transistor. Express your answer in milliwatts. Hint: Since power is equal to voltage times current, in this case the power dissipated by the transistor is equal to the collector current times the collector to emitter voltage plus the base current times the base emitter voltage: $$P=V_{CE}{I}_C+V_{BE}{I}_B$$.