In the given circuit diagram,a 5 ,V Zener dio | vedclass

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(C) The Zener diode is connected in parallel with the load resistance $R_L$. The voltage across the Zener diode is constant at $V_Z = 5 \,V$.The volta

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$500$

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(C) The Zener diode is connected in parallel with the load resistance $R_L$. The voltage across the Zener diode is constant at $V_Z = 5 \,V$.The voltage drop across the series resistor $R$ is $V_R = V_i - V_Z = 50 \,V - 5 \,V = 45 \,V$.The total current $I$ flowing through the series resistor $R$ is the sum of the Zener current $I_Z$ and the load current $I_L$,so $I = I_Z + I_L$.To find the minimum value of $R$,we use the relation $R = \frac{V_R}{I}$. Since $V_R$ is constant,$R$ is minimum when the total current $I$ is maximum.The current $I$ is maximum when the Zener current $I_Z$ is at its maximum value $(90 \,mA)$ and the load current $I_L$ is zero (which occurs when the load is disconnected or $R_L ightarrow \infty$).Thus,$I_{max} = I_{Z,max} + 0 = 90 \,mA = 90 	imes 10^{-3} \,A$.Therefore,$R_{min} = \frac{45 \,V}{90 	imes 10^{-3} \,A} = \frac{45000}{90} \,\Omega = 500 \,\Omega$.

In the given circuit diagram,a $5 \,V$ Zener diode along with a series resistance $R$ is connected across a $50 \,V$ power supply. If the maximum Zener current is $90 \,mA$,the minimum value of the resistance $R$ required will be $..... \,\Omega$.

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