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What will be the current flowing through the $6 \mathrm{k} \Omega$ resistor in the circuit shown, where the breakdown voltage of the Zener is $6 \mathrm{V} ?$
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The correct answer is:
$\frac{2}{3} m A$
In the given circuit, the zener diode is used as a voltage regulating device. The voltage across $6 \mathrm{k} \Omega$ resistance is $(10-6) \mathrm{V}=4 \mathrm{V}$
Current through $6 \mathrm{k} \Omega$ resistor, $I=\frac{4 \mathrm{V}}{6 \mathrm{k} \Omega}=\frac{4}{6 \times 10^{3}}=\frac{2}{3} \times 10^{-3}=\frac{2}{3} \mathrm{mA}$
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