A light emitting diode has a voltage drop of 2V across it when 10mA current is passed. If this LED is to be operated with 6V battery the value of limiting resistor would be
To find the value of the limiting resistor ( $R$ ) for an LED, we use Ohm's law. Given the parameters: the voltage drop across the LED ( $V_{LED} = 2V$ ), the current through the LED ( $I = 10 mA$ ), and the supply voltage ( $V_{supply} = 6V$ ). The resistor value can be calculated as follows:
$$R = \frac{V_{supply} - V_{LED}}{I}$$
Substituting the values:
$$R = \frac{6V - 2V}{10 mA}$$
$$R = \frac{4V}{10 mA}$$
$$R = 400 \Omega$$
Thus, the value of the limiting resistor is $400 \Omega$ .