A) -5.25V
B) -0.75V
C) 6V
D) zero
Correct Answer: A
Solution :
Using Kirchhoff s voltage law, \[{{V}_{CE}}=9-{{I}_{C}}{{R}_{C}}\] \[=9-(1.5)\,\,(2)=6\,V\] \[{{I}_{B}}=\frac{{{I}_{C}}}{\beta }\] \[=\frac{1.5\times {{10}^{-3}}}{100}=15\,\mu A\] Also, \[{{V}_{BE}}=9-{{I}_{B}}{{R}_{B}}\] \[=9-(15\times {{10}^{-6}})\,\,(550\times {{10}^{3}})\] \[=0.75\,V\] \[{{V}_{BC}}={{V}_{BE}}-{{V}_{CE}}\] \[=-5.25\,V\].
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