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Chapter 19: Problem 2604

A light emitting diode has a voltage drop of \(2 \mathrm{~V}\) across it when \(10 \mathrm{~mA}\) current is passed. If this \(\mathrm{LED}\) is to be operated with \(6 \mathrm{~V}\) battery the value of limiting resistor would be (A) \(400 \Omega\) (B) \(4000 \Omega\) (C) \(40 \mathrm{k} \Omega\) (D) \(300 \Omega\)

Short Answer

Expert verified
The value of the limiting resistor should be 400Ω to operate the LED with a 6V battery. So, the correct option is (A) \(400 \Omega\).

Step by step solution

01

Write down given information and values

We are given the LED voltage drop, which is 2V, and the operating current, which is 10mA. We are also given the battery voltage, which is 6V.
02

Find the voltage across the limiting resistor

In order to find the value of the limiting resistor, find the voltage across it. Since we know the voltage drop across the LED, we can use the concept of voltage division. The voltage across the limiting resistor, V_R, will be the difference between the battery voltage and the LED voltage drop: V_R = V_battery - V_LED = 6V - 2V = 4V
03

Use Ohm's Law to find the value of the limiting resistor

Now that we know the voltage across the limiting resistor (4V) and the current through it (10mA), we can use Ohm's Law to calculate the resistor value: Ohm's Law: V = IR R = V / I Here: V = 4V (Voltage across the limiting resistor), I = 10mA = 0.01A (Operating current) R = (4V) / (0.01A) = 400Ω The value of the limiting resistor should be 400Ω to operate the LED with a 6V battery. So, the correct option is (A) \(400 \Omega\).

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