question_answer

A Zener diode having breakdown voltage equal to 15 V, is used in a voltage regulator. Circuit as shown in figure, the current through the diode is

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 \[{{l}_{2}}=\,\,l-l'\,=\,\,\left( 20-15 \right)mA=\,\,5mA\]