Question

When a diode is forward biased, it has a voltage drop of $0.5\text{V}.$ The safe limit of current through the diode is $10\text{mA}$. If a battery of emf $1.5\text{V}$ is used in a circuit. The value of minimum resistance to be connected in series with the diode so that the current does not exceed the safe limit is-

• A. $300\Omega$
• B. $50\Omega$
• C. $100\Omega$
• D. $200\Omega$

Answer 1

The correct option is C $100\Omega$

According to the question, when diode is forward biased,

$V_{Diode}=0.5\text{V}$
Safe limit of current $=10\text{mA}={10}^{-2}\text{A}$


$E=V_{Diode}+V_R$

$V_R=E-V_{Diode}=1.5-0.5=1\text{V}$

$\because V_R=i{R}_{min}$

$\therefore R_{min}=\frac{V_R}{i}=\frac{1}{{10}^{-2}}=100\Omega$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.