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

• A

  $88 \mathrm{mA}$
• B

  $0.88 \mathrm{mA}$
• C

  $4.4 \mathrm{mA}$
• Correct answer
  D

  $44 \mathrm{mA}$

Solution

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 $$ \begin{aligned} l &=\frac{\Delta V}{R} \\ &=\frac{4.4}{100} \\ &=4.4 \times 10^{-2} \mathrm{A} \\ &=44 \times 10^{-3} \mathrm{A} \\ &=44 \mathrm{mA} \end{aligned} $$

Hence, the correct answer is option: 'D'.