A light emitting diode (LED) has a voltage dr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The $LED$ is connected in series with a resistor $R$ and a $6\,V$ battery.Since the $LED$ and the resistor are in series,the current flowing throu

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(A) The $LED$ is connected in series with a resistor $R$ and a $6\,V$ battery.Since the $LED$ and the resistor are in series,the current flowing through both is the same,$I = 10\,mA = 10 	imes 10^{-3}\,A$.The voltage drop across the $LED$ is $V_{LED} = 2\,V$.The total voltage of the battery is $V = 6\,V$.According to Kirchhoff's voltage law,the voltage drop across the resistor $R$ is $V_R = V - V_{LED} = 6\,V - 2\,V = 4\,V$.Using Ohm's law,$V_R = I 	imes R$,we get $R = \frac{V_R}{I}$.Substituting the values: $R = \frac{4\,V}{10 	imes 10^{-3}\,A} = 0.4 	imes 10^3\,\Omega = 400\,\Omega$.Since $1\,k\Omega = 1000\,\Omega$,then $400\,\Omega = 0.4\,k\Omega$.

$A$ light emitting diode $(LED)$ has a voltage drop of $2\,V$ across it and passes a current of $10\,mA$ when it operates with a $6\,V$ battery through a limiting resistor $R$. The value of $R$ is $......\,k\Omega$.

Similar Questions

$A$ photodiode is a device

If the energy gap of a semiconductor used for the fabrication of an $LED$ is nearly $1.9 \ eV$, then the color of the light emitted by the $LED$ is

Consider the following statements $A$ and $B$ and identify the correct answer:
$A$. For a solar-cell,the $I-V$ characteristics lies in the $IV$ quadrant of the given graph.
$B$. In a reverse biased $pn$ junction diode,the current measured in $(\mu A)$ is due to majority charge carriers.

With increasing biasing voltage of a photodiode,the photocurrent magnitude

Photodiodes are mostly operated in reverse biased condition because

Vedclass Test Series

Exam Paper Generator

Online Exam Module