In a previous exercise, we applied Ohm’s law using a circuit with a single resistor as the load. In this lesson, we will look at a load that consists of two components: a resistor and an LED.

The circuit above has an LED with a forward voltage (V_f) of 1.4V which is smaller than the source voltage (V) of 9V. To address this difference in voltage, a resistor is placed in the circuit to share some of the applied voltage.

The goal now is to find the value of the resistor that will share the right amount of voltage and induce the necessary current.

Before we get into the math, we first need to cover two rules of this circuit:

• Rule 1: The sum of the voltages in the load will equal the voltage source. In the example circuit, the voltage across the resistor (V_r) added to the forward voltage (V_f) of the LED will equal 9V.
• Rule 2: The current through all the components will be equal. Remembering the resistance water analogy, the smaller portion of the pipe will dictate the water flow for the entire pipe. In the example circuit, the resistor will dictate the goal current which is equal to the forward current of the LED, 0.02A.

Now let’s plug in the numbers using rule 1: The sum of the voltages in the load will equal the voltage source

$V = V_{r} + V_{f}$

$9V = V_{r} + 1.4V$

$V_{r} = 9V - 1.4V$

$V_{r} = 7.6V$

Using this answer from the previous equation and circuit rule 2 we now have enough information to apply Ohm’s Law:

1. The voltage across the resistor is 7.6V
2. The goal current through the circuit is equal to the LED forward current, 0.02A

   $R = V_{r} ÷ I$

   $R = 7.6V ÷ 0.02A$

   $R = 380Ω$

   Picking a resistor to go along with an LED is an important multi-step task. Move on to the instructions to give it a try on a new circuit.