The Zener diode has a breakdown voltage V_{z} | vedclass

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(B) The total voltage supplied is $V_{in} = 90\, V$ and the series resistance is $R_{s} = 4\, k\Omega = 4000\, \Omega$.The Zener diode maintains a con

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$9$

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(B) The total voltage supplied is $V_{in} = 90\, V$ and the series resistance is $R_{s} = 4\, k\Omega = 4000\, \Omega$.The Zener diode maintains a constant voltage of $V_{z} = 30\, V$ across it.The voltage drop across the series resistor $R_{s}$ is $V_{R} = V_{in} - V_{z} = 90\, V - 30\, V = 60\, V$.The total current $i$ flowing from the source through the series resistor is $i = \frac{V_{R}}{R_{s}} = \frac{60\, V}{4000\, \Omega} = 0.015\, A = 15\, mA$.The load resistor is $R_{L} = 5\, k\Omega = 5000\, \Omega$. The current $i_{1}$ flowing through the load resistor is $i_{1} = \frac{V_{z}}{R_{L}} = \frac{30\, V}{5000\, \Omega} = 0.006\, A = 6\, mA$.Applying Kirchhoff's Current Law at the junction,the current through the Zener diode $i_{z}$ is given by $i_{z} = i - i_{1} = 15\, mA - 6\, mA = 9\, mA$.

The Zener diode has a breakdown voltage $V_{z} = 30\, V$. The current passing through the diode for the following circuit is $\ldots \ldots \ldots \,mA.$

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