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(D) The Zener diode is connected in parallel with resistor $R_{2} = 1500 \, \Omega$. Since the Zener breakdown voltage is $V_{z} = 10 \, V$,the voltag

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$3.33$

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(D) The Zener diode is connected in parallel with resistor $R_{2} = 1500 \, \Omega$. Since the Zener breakdown voltage is $V_{z} = 10 \, V$,the voltage across $R_{2}$ is fixed at $V_{R_{2}} = 10 \, V$.The current through $R_{2}$ is:$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10 \, V}{1500 \, \Omega} = \frac{1}{150} \, A \approx 6.67 	imes 10^{-3} \, A = 6.67 \, mA$.The voltage drop across resistor $R_{1} = 500 \, \Omega$ is:$V_{R_{1}} = V_{source} - V_{z} = 15 \, V - 10 \, V = 5 \, V$.The total current supplied by the source through $R_{1}$ is:$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5 \, V}{500 \, \Omega} = 10^{-2} \, A = 10 \, mA$.Applying Kirchhoff's Current Law at the node,the current through the Zener diode $(I_{z})$ is:$I_{z} = I_{R_{1}} - I_{R_{2}} = 10 \, mA - 6.67 \, mA = 3.33 \, mA$.

In the circuit given,the current through the Zener diode is ....... $mA$.

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