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(B) The circuit consists of a $16 \ V$ $DC$ source, a series resistor $R_s = 1 \ k\Omega$, a Zener diode with breakdown voltage $V_z = 6 \ V$, and a l

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$4$

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(B) The circuit consists of a $16 \ V$ $DC$ source, a series resistor $R_s = 1 \ k\Omega$, a Zener diode with breakdown voltage $V_z = 6 \ V$, and a load resistor $R_L = 1 \ k\Omega$.First, we calculate the total current $I_s$ flowing from the source through the series resistor:$I_s = \frac{V_{in} - V_z}{R_s} = \frac{16 \ V - 6 \ V}{1 \ k\Omega} = \frac{10 \ V}{1 \ k\Omega} = 10 \ mA$.Next, we calculate the current $I_L$ flowing through the load resistor $R_L$:$I_L = \frac{V_z}{R_L} = \frac{6 \ V}{1 \ k\Omega} = 6 \ mA$.Applying Kirchhoff's Current Law at the junction, the current $I$ through the Zener diode is:$I = I_s - I_L = 10 \ mA - 6 \ mA = 4 \ mA$.

What is the value of the current $I$ flowing through the Zener diode in the given circuit? (in $mA$)

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