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(B) The voltage across the zener diode is $V_{Z} = 10 \ V$. Since the zener diode is in parallel with resistor $R_{2} = 1500 \ \Omega$,the voltage acr

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$3.3$

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(B) The voltage across the zener diode is $V_{Z} = 10 \ V$. Since the zener diode is in parallel with resistor $R_{2} = 1500 \ \Omega$,the voltage across $R_{2}$ is $V_{R_{2}} = V_{Z} = 10 \ V$.The current through $R_{2}$ is:$I_{R_{2}} = \frac{V_{R_{2}}}{R_{2}} = \frac{10 \ V}{1500 \ \Omega} = \frac{1}{150} \ A \approx 6.67 	imes 10^{-3} \ A = 6.67 \ mA$.The voltage across $R_{1} = 500 \ \Omega$ is:$V_{R_{1}} = V_{source} - V_{Z} = 15 \ V - 10 \ V = 5 \ V$.The total current through $R_{1}$ is:$I_{R_{1}} = \frac{V_{R_{1}}}{R_{1}} = \frac{5 \ V}{500 \ \Omega} = 0.01 \ A = 10 \ mA$.Applying Kirchhoff's Current Law at the junction,the current through the zener diode $I_{Z}$ is:$I_{Z} = I_{R_{1}} - I_{R_{2}} = 10 \ mA - 6.67 \ mA = 3.33 \ mA \approx 3.3 \ mA$.

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

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