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Question Number 66249 by Umar last updated on 11/Aug/19

$$\mathrm{The}\mathrm{npn}\mathrm{transistor}\mathrm{in}\mathrm{the}\mathrm{voltage}$$$$\mathrm{amplifier}\mathrm{circuit}\mathrm{operates}\mathrm{satisfctorily}$$$$\mathrm{on}\mathrm{a}\mathrm{quiescent}\mathrm{collector}\mathrm{current}\mathrm{of}3\mathrm{mA}.$$$$\mathrm{if}\mathrm{the}\mathrm{battery}\mathrm{supply}\left({\mathrm{V}}_{\mathrm{cc}}\right)\mathrm{is}6\mathrm{v},\mathrm{calculate}$$$$\mathrm{the}\mathrm{value}\mathrm{of};$$$$\left(\mathrm{a}\right)\mathrm{the}\mathrm{load}\mathrm{resistor}{\mathrm{R}}_{\mathrm{L}}$$$$\left(\mathrm{b}\right)\mathrm{the}\mathrm{base}\mathrm{current}\mathrm{for}\mathrm{the}\mathrm{quiescent}$$$$\mathrm{collector}-\mathrm{emitter}\mathrm{voltage}{\mathrm{V}}_{\mathrm{ce}}\mathrm{to}$$$$\mathrm{be}\mathrm{half}\mathrm{the}\mathrm{battery}\mathrm{voltage}.$$$$\mathrm{Th}\mathrm{transistor}\mathrm{dc}\mathrm{current}\mathrm{is}100.$$$$$$$$\mathrm{please}\mathrm{help}.$$

Answered by peter frank last updated on 11/Aug/19

$$V_{cc}=I_c{R}_L+V_{ce}$$$$V_{cc}=V_B+V_{BE}$$$$V_{ce}=\frac{V_{cc}}{2}$$

Commented by Umar last updated on 29/Sep/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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