In the circuit diagram shown in the figure,th | vedclass

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(B) The zener diode is in parallel with the load resistor $R_L = 1200 \Omega$. Since the zener breakdown voltage is $15 	ext{ V}$,the voltage across

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$17.5$

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(B) The zener diode is in parallel with the load resistor $R_L = 1200 \Omega$. Since the zener breakdown voltage is $15 	ext{ V}$,the voltage across the load resistor is $V_L = 15 	ext{ V}$.The current through the load resistor is $I_L = \frac{V_L}{R_L} = \frac{15 	ext{ V}}{1200 \Omega} = 0.0125 	ext{ A} = 12.5 	ext{ mA}$.The voltage across the series resistor $R_S = 300 \Omega$ is $V_S = V_{in} - V_L = 24 	ext{ V} - 15 	ext{ V} = 9 	ext{ V}$.The total current supplied by the source is $I = \frac{V_S}{R_S} = \frac{9 	ext{ V}}{300 \Omega} = 0.03 	ext{ A} = 30 	ext{ mA}$.Applying Kirchhoff's Current Law at the junction,the current through the zener diode $I_Z$ is given by $I_Z = I - I_L$.$I_Z = 30 	ext{ mA} - 12.5 	ext{ mA} = 17.5 	ext{ mA}$.

In the circuit diagram shown in the figure,the current through the zener diode is (in $mA$)

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