A) 7 mA
B) 17 Ma
C) 10 mA
D) 15mA
Correct Answer: C
Solution :
\[9={{V}_{Z}}+{{V}_{{{R}_{1}}}}\] \[{{V}_{Z}}=5.6V\] \[{{V}_{{{R}_{1}}}}=9-5.6\] \[{{V}_{{{R}_{1}}}}=3.4\] \[{{I}_{{{R}_{1}}}}=\frac{{{V}_{{{R}_{1}}}}}{R}=\frac{3.4}{200}\] \[{{I}_{{{R}_{1}}}}=17mA\] \[{{V}_{Z}}={{V}_{{{R}_{2}}}}={{I}_{{{R}_{2}}}}({{R}_{2}})\] \[\frac{5.6}{800}={{I}_{R{{ & }_{2}}}}\] \[{{I}_{R{{ & }_{2}}}}=7mA\] \[{{I}_{Z}}=(17-7)mA\] \[=10mA\]You need to login to perform this action.
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