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Question

Question: In the circuit given, the current through the zener diode is. <img src="https://cdn.pureessence.tec...

In the circuit given, the current through the zener diode is.

A

10 mA

B

6.676.67 Ma

C

5 mA

D

3.333.33 mA

Answer

3.333.33 mA

Explanation

Solution

:

The voltage drop across R2R_{2}is

VR2=VZ=V_{R_{2}} = V_{Z} =10V

The current through R2R_{2}is

IR2=VR2R2=10V1500Ω=0.667×102AI_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10V}{1500\Omega} = 0.667 \times 10^{- 2}A

The voltage drop across R1R_{1}is ()

}{= 15V - 10V = 5V}$$ The current through $R_{1}$is $$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5V}{500\Omega} = 10^{- 2}A = 10 \times 10^{- 3}A = 10mA$$ The current through the zener diode is $${I_{Z} = I_{R_{1}} - I_{R_{2}} }{= (10 - 6.67)mA = 3.33mA}$$