When a diode is forward biased, it has a voltage drop of $0.5 \mathrm{~V}$. The safe limit of current through the diode is 10 $\mathrm{mA}$. If a battery of emf $1.5 \mathrm{~V}$ is used in the 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 :
Correct Option: , 3
(3) According to question, when diode is forward biased,
$V_{\text {diode }}=0.5 \mathrm{~V}$
Safe limit of current, $I=10 \mathrm{~mA}=10^{-2} \mathrm{~A}$
$R_{\min }=?$
Voltage through resistance
$V_{R}=1.5-0.5=1 \mathrm{volt}$
$i R=1\left(=V_{R}\right)$
$\therefore R_{\min }=\frac{1}{i}=\frac{1}{10^{-2}}=100 \Omega$
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