A Zener diode with voltage V_z = 6 V is used | vedclass

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(A) Given: $E = 9 \ V$,$V_z = 6 \ V$,$R_L = 1000 \ \Omega$,and $R_s = 100 \ \Omega$.The potential drop across the series resistor $R_s$ is $V_R = E -

Vedclass · Accepted Answer

$0.144$

Vedclass · Answer

(A) Given: $E = 9 \ V$,$V_z = 6 \ V$,$R_L = 1000 \ \Omega$,and $R_s = 100 \ \Omega$.The potential drop across the series resistor $R_s$ is $V_R = E - V_z = 9 \ V - 6 \ V = 3 \ V$.The total current flowing through the series resistor $R_s$ is $I = \frac{V_R}{R_s} = \frac{3 \ V}{100 \ \Omega} = 0.03 \ A$.The current flowing through the load resistor $R_L$ is $I_L = \frac{V_z}{R_L} = \frac{6 \ V}{1000 \ \Omega} = 0.006 \ A$.The current through the Zener diode is $I_z = I - I_L = 0.03 \ A - 0.006 \ A = 0.024 \ A$.The power dissipated in the Zener diode is $P_z = V_z 	imes I_z = 6 \ V 	imes 0.024 \ A = 0.144 \ W$.

$A$ Zener diode with voltage $V_z = 6 \ V$ is used to maintain a constant voltage across a load resistance $R_L = 1000 \ \Omega$. $A$ series resistor $R_s = 100 \ \Omega$ is used. If the source $e.m.f$ is $E = 9 \ V$,then the power dissipated in the Zener diode is .......... $W$.

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