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(B) Given, the breakdown voltage of the Zener diode is $V_z = 3.0 \, V$.The total voltage supplied is $V = 10 \, V$ and the series resistance is $R_s

Vedclass · Accepted Answer

$5.5$

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(B) Given, the breakdown voltage of the Zener diode is $V_z = 3.0 \, V$.The total voltage supplied is $V = 10 \, V$ and the series resistance is $R_s = 1 \, k\Omega = 1000 \, \Omega$.The current $I$ flowing through the series resistor is given by:$I = \frac{V - V_z}{R_s} = \frac{10 \, V - 3 \, V}{1000 \, \Omega} = \frac{7 \, V}{1000 \, \Omega} = 7 \, mA$.The load resistor is $R_L = 2 \, k\Omega = 2000 \, \Omega$. The current $I_L$ flowing through the load resistor is:$I_L = \frac{V_z}{R_L} = \frac{3 \, V}{2000 \, \Omega} = 1.5 \, mA$.Applying Kirchhoff's Current Law at node $A$, we have:$I = I_z + I_L$$I_z = I - I_L = 7 \, mA - 1.5 \, mA = 5.5 \, mA$.Thus, the value of $I_z$ is $5.5 \, mA$.

In the given circuit, the breakdown voltage of the Zener diode is $3.0 \, V$. What is the value of $I_z$ (in $ \, mA$)?

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