Question

A light emitting diode (LED) has a voltage drop of \[2\,{\text{V}}\] across it and passes a current of \[{\text{10 mA}}\]. When it operates with a \[{\text{6 V}}\] battery through a limiting resistor \[{\text{R}}\], the value of \[{\text{R}}\] is
A. \[200\,\Omega \]
B. \[400\,\Omega \]
C. \[40\,{\text{k}}\Omega \]
D. \[4\,{\text{k}}\Omega \]

Answer 1

Hint:We use the given information about voltage drop and amount of current flowing through the LED and solve it with the help of voltage resistance formula as given by Ohm’s law. Ohm’s law states the relationship between electric current and potential difference. The current that flows through most conductors is directly proportional to the voltage applied to it.

Complete step by step solution:
From the given information we know that a forward, biased P-N junction that emits light is a light emitting diode.
The voltage drop across it \[ = 2\,{\text{V}}\]
Battery voltage, \[ = 6\,{\text{V}}\]
So, the total voltage is given by:
\[{\text{V}} = 6 - 2 = 4\,{\text{V}}\]
Since an LED in the forward biased region has zero resistance.
So, the total resistance can be written as:
$R = 0 + r \\
\Rightarrow R = r \\$
Here, current,
$I = 10\,{\text{mA}} \\
\Rightarrow I = 10 \times {10^{ - 3}} \\
\Rightarrow I = 0.01\,{\text{A}} \\$
We know that,
Resistance equals voltage divided by current,
\[R = \dfrac{V}{I}\]
$\Rightarrow r = \dfrac{4}{{0.01}} \\
\therefore r = 400\,\Omega \\$
Hence, the required answer is \[400\,\Omega \] .The correct option is B.

Additional information:
A semiconductor device that emits light when an electric current is passed through it is a light-emitting diode (LED). Light is created when inside the semiconductor material, the particles carrying the current (known as electrons and holes) join together.

Resistor: A resistor is a passive electrical two-terminal component that as a circuit element implements electrical resistance. Resistors are used, among other applications in electronic circuits to decrease current flow, change signal levels, separate voltages, bias active components, and terminate transmission lines.

Note:While solving the problem, most of the students forget to convert the unit of the current from milliamperes to amperes. Due to this reason, they get wrong and irrelevant answers. Again, most of the students add the voltage drop of the LED and the battery to get the net drop in the voltage, which is wrong.