In the circuit diagram, the current through the Zener diode is :

10 m A

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3.33 m A

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6.67 m A

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0 m A

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Solution

The correct option is B $3.33 m A$
The voltage drop across $R_{2}$ is
$V_{R_{2}} = V_{z} = 10 V$
The current through $R_{2}$ is
$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10}{1500} = 0.667 \times 10^{- 2} A$
$= 6.67 \times 10^{- 3} A = 6.67 m A$
The voltage drop across $R_{1}$ is
$V_{R_{1}} = 15 V - V_{R_{2}} = 15 V - 10 V = 5 V$
The current through $R_{1}$ is
$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5 V}{500 Ω} = 10^{- 3} A = 10 \times 10^{- 3} = 10 m A$
The current through Zener diode is
$I_{z} = I_{R_{1}} - I_{R_{2}} = (10 - 6.67) m A = 3.33 m A$

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