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Physics Question on Semiconductor electronics: materials, devices and simple circuits

A Zener diode is connected to a battery and a load as shown below : The currents I,IzI, I_z and ILI_L are respectively

A

15 mA,5 mA,10 mA15\, mA, 5 \,mA, 10\, mA

B

15 mA,7.5 mA,7.5 mA15\, mA, 7.5\, mA, 7.5 \,mA

C

12.5 mA,5 mA,7.5 mA12.5\,mA, 5 \,mA, 7.5\, mA

D

12.5 mA,7.5 mA,5 mA12.5 \,mA, 7.5\, mA, 5 \,mA

Answer

12.5 mA,7.5 mA,5 mA12.5 \,mA, 7.5\, mA, 5 \,mA

Explanation

Solution

IL=10V2kΩ=5mAI_{L} = \frac{10V}{2k\Omega} = 5mA
I=(60−10)v4kΩ=504kΩ=12.5 mAI = \frac{\left(60-10\right)v}{4k\Omega} = \frac{50}{4k\Omega} = 12.5\,mA
IZ=I−IL=(12.5−5) mA=7.5 mAI_{Z} = I-I_{L} = \left(12.5-5\right)\,mA = 7.5\,mA