A) 5 mA
B) 10 mA
C) 15 mA
D) 20 mA
Correct Answer: A
Solution :
The voltage drop across \[1\,k\Omega \,\,=\,\,{{V}_{2}}=15\,V\] The current through \[1 k\Omega \] is \[l'=\frac{15\,V}{1\times {{10}^{-\,3}}\,\Omega }\] \[=\,\,\,15\times 1{{0}^{-\,3}}\,A=\,\,15\,\,mA\] The voltage drop across \[2500 = 20 V\,-15V\]\[=\,\,5\,V\] The current through \[250\,\Omega \] is \[l=\frac{5\,V}{250\,\Omega }=0.02\,A\]\[=\,\,20mA\] The current through Zener diode is \[{{l}_{2}}=\,\,l-l'\,=\,\,\left( 20-15 \right)mA=\,\,5mA\]You need to login to perform this action.
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