Calculating Series Resistor for LED Current Limiting

This problem involves finding the correct value for a series resistor needed to protect an Light Emitting Diode (LED) from excessive current. We need to ensure the current stays at 20 mA while using a 10-V power supply and considering the LED's forward voltage drop.

Understanding the Circuit Components

• Supply Voltage ($V_S$): The total voltage provided by the power source, which is 10 V.
• LED Forward Voltage ($V_F$): The voltage the LED drops across itself when current flows through it in the forward direction. Here, it's 2.0 V.
• Desired LED Current ($I$): The maximum safe operating current for the LED, specified as 20 mA. We need to convert this to Amperes: $I = 20 \text{ mA} = 0.020 \text{ A}\).
• Series Resistor ($R$): The resistor we need to calculate. It's placed in series with the LED to limit the current.

Applying Ohm's Law

In a simple series circuit like this (power supply -> resistor -> LED -> ground), the total voltage supplied is distributed across the components. The voltage across the resistor ($V_R$) is the difference between the supply voltage and the LED's forward voltage.

The formula is:

Once we know the voltage across the resistor ($V_R$) and the desired current ($I$) flowing through it (which is the same current flowing through the LED), we can use Ohm's Law to find the resistance ($R$):

Step-by-Step Calculation

1. Calculate the voltage across the resistor ($V_R$): Substitute the given values into the formula: $$ V_R = 10 \text{ V} - 2.0 \text{ V} $$ $$ V_R = 8.0 \text{ V} $$ So, the resistor must handle 8.0 volts.
2. Calculate the required resistance ($R$): Now, use Ohm's Law with the calculated $V_R$ and the desired current $I$: $$ R = \frac{8.0 \text{ V}}{0.020 \text{ A}} $$ $$ R = 400 \ \Omega $$ Therefore, a 400 $\Omega$ resistor is required.

Conclusion

Based on the calculations, a 400 $\Omega$ series resistor is needed to limit the current through the LED to 20 mA when connected to a 10-V supply, given the LED's forward voltage drop of 2.0 V. This matches one of the provided options.