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A Zener diode having break-down voltage $5.6 \mathrm{V}$ is connected in reverse bias with a battery of emf $10 \mathrm{V}$ and a resistance of $100 \Omega$ in series. The current flowing through the Zener diode is
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The correct answer is:
$44 \mathrm{mA}$
Breakdown voltage of Zener diode $=5.6 \mathrm{V}$
Voltage of source $=10 \mathrm{V}$
Potential difference $\Delta V=(10-5.6) \mathrm{V}=4.4 \mathrm{V}$
Resistance of circuit $=100 \Omega$
The current passing through the Zener diode
$$
\begin{aligned}
l &=\frac{\Delta V}{R} \\
&=\frac{4.4}{100} \\
&=4.4 \times 10^{-2} \mathrm{A} \\
&=44 \times 10^{-3} \mathrm{A} \\
&=44 \mathrm{mA}
\end{aligned}
$$
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