A Zener diode having a Zener voltage of 8, V | vedclass

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(C) The power dissipation rating of the Zener diode is given by $P = V_z \cdot I_z$,where $V_z = 8\, V$ and $P = 0.5\, W$.Calculating the maximum curr

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$192$

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(C) The power dissipation rating of the Zener diode is given by $P = V_z \cdot I_z$,where $V_z = 8\, V$ and $P = 0.5\, W$.Calculating the maximum current $I_z$ through the Zener diode:$I_z = \frac{P}{V_z} = \frac{0.5}{8} = \frac{1}{16}\, A$.In the circuit,the total voltage $E = 20\, V$ is divided between the protective resistance $R_p$ and the Zener diode.The voltage drop across the protective resistance $R_p$ is $V_{R_p} = E - V_z = 20\, V - 8\, V = 12\, V$.Using Ohm's law for the protective resistance $R_p = \frac{V_{R_p}}{I_z}$:$R_p = \frac{12}{1/16} = 12 	imes 16 = 192\, \Omega$.

$A$ Zener diode having a Zener voltage of $8\, V$ and a power dissipation rating of $0.5\, W$ is connected across a potential divider circuit as shown in the diagram. The value of the protective resistance $R_p$ is $....\, \Omega$.

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