Question

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

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

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.

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

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) \(400 \Omega\).

Key concepts

Voltage Drop

When operating an LED, understanding the concept of voltage drop is key. A voltage drop is the difference in electrical potential between two points in a circuit. For an LED, this is the amount of voltage required to turn it on. This ensures that the LED emits light at a desired brightness.

In our exercise, the LED has a voltage drop of 2V. This means, the LED needs 2V to function correctly. If connected to a voltage source, the LED will 'consume' 2V, leaving the excess to be handled by other components like a resistor, in the circuit.

It's crucial because not all of the supply voltage can be used by the LED. The circuit must compensate and control this using other components.

Current Through LED

The current through the LED is essential for its operation. Like most electronic components, LEDs have specific current requirements that must be met to function efficiently and safely.

LEDs typically use a small amount of current. In this exercise, the LED requires a current of 10mA. If the current exceeds this limit, there is a risk of damaging the LED. Conversely, too little current and the LED might not light up properly.

Controlling the current is achieved using a resistor. By setting a specific amount of resistance in the circuit, we can manage how much current flows through the LED.

Limiting Resistor Calculation

Calculating the correct value for a limiting resistor is a crucial step in designing circuits with LEDs. The purpose of the limiting resistor is to ensure the LED receives an appropriate current level to prevent overheating or failure.

Using Ohm's Law, which states that the resistance (R) is equal to the voltage (V) divided by the current (I) \[ R = \frac{V}{I} \], we can find the required resistance.

In our example:

• Voltage across the resistor is 4V.
• The current through the resistor is 10mA or 0.01A.

By calculating, \[ R = \frac{4V}{0.01A} = 400 \Omega \].

Hence, a 400Ω resistor is needed to limit the current through the LED.

Voltage Division

Voltage division is a principle used to determine the voltage across individual components in a linear circuit. When multiple components are in series, the total voltage is divided among them.

For the LED reported here, we had a total battery voltage of 6V. The LED uses 2V, leaving the remaining 4V across the resistor. This means:

• Battery voltage = 6V
• LED voltage drop = 2V
• Resistor voltage = 4V (calculated as 6V - 2V)

This calculation ensures that each component in the circuit gets the proper voltage it requires.

LED Operation with Battery

Operating an LED with a battery requires meeting its voltage and current specifications to maintain efficiency and safety. When using a 6V battery to power an LED with a 2V drop, we ensure that remaining voltage after the LED operates some other circuit component which in this case is a resistor.

To protect the LED, along with ensuring its correct functionality, a limiting resistor is employed. It tailors the surplus voltage, allowing only the necessary current to flow through the LED. This makes sure the LED operates without any damage.

Understanding this setup helps in designing effective LED circuits powered by batteries, ensuring all components are safely within their operating levels.