A) 0.5 mA ; 6 mA
B) 0.5 mA ; 8.5 Ma
C) 1.5 mA ; 8.5 mA
D) 1 mA ; 8.5 mA
Correct Answer: B
Solution :
\[At\,{{V}_{B}}=8V\] \[{{i}_{L}}=\frac{6\times {{10}^{-3}}}{4}=1.5\times {{10}^{-3}}A\] \[{{i}_{R}}=\frac{8-6\times {{10}^{-3}}}{1}=2\times {{10}^{-3}}A\] \[\therefore \]\[{{i}_{zener\,diode}}={{i}_{R}}-{{i}_{load}}\] \[=0.5\times {{10}^{-3}}A\] \[At\,{{V}_{B}}=16V\] \[{{i}_{L}}=1.5\times {{10}^{-3}}A\] \[{{i}_{R}}=\frac{(16-6)\times {{10}^{-3}}}{1}=10\times {{10}^{-3}}A\] \[\therefore \]\[{{i}_{zener}}diode={{i}_{R}}-{{i}_{L}}\] \[=8.5\times {{10}^{-3}}A\]You need to login to perform this action.
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