A Zener diode is connected to a battery and a | vedclass

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(D) The circuit is given as shown in the figure.Potential drop across the load resistance $R_L$ is $V_L = V_Z = 10 	ext{ V}$.Potential drop across th

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$12.5 	ext{ mA}, 7.5 	ext{ mA}, 5 	ext{ mA}$

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(D) The circuit is given as shown in the figure.Potential drop across the load resistance $R_L$ is $V_L = V_Z = 10 	ext{ V}$.Potential drop across the $4 	ext{ k}\Omega$ series resistance is $V_R = 60 	ext{ V} - 10 	ext{ V} = 50 	ext{ V}$.Current through the $4 	ext{ k}\Omega$ resistance is $I = \frac{V_R}{R} = \frac{50 	ext{ V}}{4 	imes 10^3 \Omega} = 12.5 	imes 10^{-3} 	ext{ A} = 12.5 	ext{ mA}$.Current through the load resistance $I_L$ is $I_L = \frac{V_L}{R_L} = \frac{10 	ext{ V}}{2 	imes 10^3 \Omega} = 5 	ext{ mA}$.Using Kirchhoff's current law at the node, $I = I_Z + I_L$, therefore $I_Z = I - I_L = 12.5 	ext{ mA} - 5 	ext{ mA} = 7.5 	ext{ mA}$.Thus, the currents are $I = 12.5 	ext{ mA}$, $I_Z = 7.5 	ext{ mA}$, and $I_L = 5 	ext{ mA}$.

$A$ Zener diode is connected to a battery and a load resistance as shown below. The currents $I, I_Z$ and $I_L$ respectively are:

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