If the voltage across the Zener diode is 6, V | vedclass

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(D) The total voltage of the source is $V_{in} = 30\, V$.The voltage across the Zener diode is $V_z = 6\, V$.Since the Zener diode is in parallel with

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(D) The total voltage of the source is $V_{in} = 30\, V$.The voltage across the Zener diode is $V_z = 6\, V$.Since the Zener diode is in parallel with the resistor $R$, the voltage across the resistor $R$ is $V_R = V_z = 6\, V$.The voltage across the $1\, k\Omega$ series resistor is $V_s = V_{in} - V_z = 30\, V - 6\, V = 24\, V$.The current flowing through the series resistor is $I_s = \frac{V_s}{1\, k\Omega} = \frac{24\, V}{1\, k\Omega} = 24\, mA$.Let $I_z$ be the current through the Zener diode and $I_R$ be the current through the resistor $R$. By Kirchhoff's Current Law, $I_s = I_z + I_R$.For the resistance $R$ to be maximum, the current $I_R$ must be minimum. The minimum current through the Zener diode is $I_{z,min} = 0$.Thus, $I_R = I_s = 24\, mA$.Using Ohm's law for resistor $R$, $R = \frac{V_R}{I_R} = \frac{6\, V}{24\, mA} = 0.25\, k\Omega$.Wait, re-evaluating the circuit: The Zener diode is in parallel with $R$. The voltage across the series resistor $1\, k\Omega$ is $V_{series} = 30 - 6 = 24\, V$. The current through the series resistor is $I = \frac{24\, V}{1\, k\Omega} = 24\, mA$. This current splits into the Zener diode and the resistor $R$. To maximize $R$, we minimize the current through it. If we assume the Zener diode takes all excess current, the minimum current through $R$ is not well-defined without a minimum Zener current. However, if the question implies the Zener diode is just at the breakdown point, then $I_R = 24\, mA$, giving $R = 0.25\, k\Omega$. Given the options, let's re-read the circuit. If $R$ is in series with the Zener, $R = (30-6)/I_z$. If $I_z = 6\, mA$ (as per diagram label), then $R = 24/6 = 4\, k\Omega$.

If the voltage across the Zener diode is $6\, V$, then find out the value of the maximum resistance $R$ in this condition. (in $k\Omega$)

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