The reverse breakdown voltage of a Zener diod | vedclass

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(A) The total voltage supplied is $V = 9\, V$. The Zener diode is in parallel with the load resistor $R_L = 800\, \Omega$,so the voltage across the lo

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(A) The total voltage supplied is $V = 9\, V$. The Zener diode is in parallel with the load resistor $R_L = 800\, \Omega$,so the voltage across the load is $V_z = 5.6\, V$.The voltage drop across the series resistor $R_s = 200\, \Omega$ is $V_{R_s} = V - V_z = 9 - 5.6 = 3.4\, V$.The total current flowing through the series resistor $R_s$ is $I = \frac{V_{R_s}}{R_s} = \frac{3.4}{200} = 0.017\, A = 17\, mA$.The current flowing through the load resistor $R_L$ is $I_L = \frac{V_z}{R_L} = \frac{5.6}{800} = 0.007\, A = 7\, mA$.Applying Kirchhoff's current law at the junction,the current through the Zener diode is $I_z = I - I_L = 17\, mA - 7\, mA = 10\, mA$.

The reverse breakdown voltage of a Zener diode is $5.6\, V$ in the given circuit. The current $I_z$ through the Zener is......$mA$

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