Question 38 of 209

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V. For a $400 \Omega$ resistor connected in series, the zener current is found to be 4 times load current. The load current $\left(I_L\right)$ and load resistance $\left(R_{\mathrm{L}}\right)$ are :

A$\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$B$\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \Omega$C$\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega$D$\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$

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Answer: (D) $\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega$

From the circuit diagram,
$\begin{aligned}
& 5 \mathrm{i}=\frac{20}{400}=\frac{1}{20} \mathrm{~A} \\
& \therefore \mathrm{i}=\frac{1}{100} \mathrm{~A}=10 \mathrm{~mA}=\text { Load current }
\end{aligned}$
Also, $\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}$
$\therefore \mathrm{R}_{\mathrm{L}}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega$