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(C) The Zener diode is used for voltage regulation. The power dissipation $P_D$ of the Zener diode is given by $P_D = V_Z 	imes i$,where $V_Z$ is the

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$125$

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(C) The Zener diode is used for voltage regulation. The power dissipation $P_D$ of the Zener diode is given by $P_D = V_Z 	imes i$,where $V_Z$ is the breakdown voltage and $i$ is the current through the diode.Given $P_D = 0.4 \ W$ and $V_Z = 10 \ V$,we have:$0.4 = 10 	imes i$$i = \frac{0.4}{10} = 0.04 \ A$The protective resistance $R$ is connected in series with the Zener diode. The voltage drop across the resistance $R$ is $V_R = V_{source} - V_Z = 15 \ V - 10 \ V = 5 \ V$.Using Ohm's law,$V_R = i 	imes R$,we get:$R = \frac{V_R}{i} = \frac{5 \ V}{0.04 \ A} = 125 \ \Omega$.

$A$ voltage regulating circuit consisting of a Zener diode,having a breakdown voltage of $10 \ V$ and a maximum power dissipation of $0.4 \ W$,is operated at $15 \ V$. The approximate value of the protective resistance in this circuit is . . . . . . $\Omega$.

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