I've created the following circuit for educational purposes. It's basically 2 LEDs in parallel. I've used my multimeter to measure the current and voltage of the various components and noted everything down. The idea was to come up with the formula to calculate the current flowing through the 2 LEDs. I know that the total current will be divided between the 2 LEDs but I'm having trouble calculating everything.

enter image description here

I'm able to use Ohms law for a simple 1 resistor/LED circuit but I'm having trouble understanding how to calculate the current flowing through the 2 LEDs in this circuit. Here's what I did :

Current for the green LED :

Here I attempted to use the basic formula to calculate the current through the green LED, but that obviously doesn't take into account that there is a another LED in parallel.

I = V/R = (5V - 2,065V drop green LED) / 270 = 10.9mA

So this is obviously wrong. I don't know what I calculated here. I guess this would have been valid if only the green LED was present, and the voltage drop would have been 2,065V.

Current for the red LED :

Here I tried to visualize the current flow, so I asssumed that the current flowing through the red LED would have to pass the 2 resistors in series (270 + 100) and used Ohms law again

I = V/R = (5V - 1.863V drop red LED) / 370 = 8,47mA

This is again incorrect but seems to correspond to the current flowing through the green LED. Don't know if this is coincidence.

I've chosen the resistor values randomly just to ensure that the 2 LEDs would be lit. This circuit is just for educational purposes.

The logic behind calculating the current is something I'm currently missing. Any help would be great.

  • \$\begingroup\$ Are the values on the diagram measured values or calculated values? \$\endgroup\$
    – pjc50
    Nov 5, 2013 at 11:13
  • \$\begingroup\$ The resistor on the right absorbs the voltage difference. The rest is Ohm's Law and KCL. \$\endgroup\$ Nov 5, 2013 at 11:19
  • \$\begingroup\$ Picture on the right are all measured values using the multi-meter. \$\endgroup\$
    – ddewaele
    Nov 5, 2013 at 11:22

2 Answers 2


The big problem is that you didn't draw the schematic properly. A good schematic always makes the things clear:


simulate this circuit – Schematic created using CircuitLab

The current is \$I = \frac{U}{I}\$

I0 = I1+I2

I0 = (5V-2.065V)/270 = 10.87037037mA

I1 = (2.065V - 1.863V)/100 = 2.02mA

I2 = I0 - I1 = 8.85037037mA

  • \$\begingroup\$ I'm very new to all of this and noticed different people draw things differently. Yours is indeed clearer. Don't know if this is making sense but You've derived I2 from I1 and I0 (correct as it needs to add up). Is it possible to calculate I2 directly without deriving it from I1 ? There's no resistor value on the I2 branch and voltage gets dropped to 0V. How does the formulate look like I2 = (2.065V - ?) / ? = 8.85037037mA. How do you calculate that ? Or is that not possible ? \$\endgroup\$
    – ddewaele
    Nov 5, 2013 at 12:20
  • \$\begingroup\$ @ddewaele if you look at the data sheet for the diode, you'll see a relationship between voltage drop and current. If this relationship is linear in the region you're operating in, then you could calculate the current by treating the diode as a voltage source and series resistor, where the voltage is where the tangent to I/V plot crosses the I=0 axis and the resistance in the gradient of the tangent. \$\endgroup\$ Nov 5, 2013 at 14:03
  • \$\begingroup\$ @ddewaele - I2 can't be easy determined by the diode voltage, because the diodes are highly non-linear devices. The Ohms law is not valid for such devices. \$\endgroup\$
    – johnfound
    Nov 5, 2013 at 14:27
  • \$\begingroup\$ en.wikipedia.org/wiki/False_precision \$\endgroup\$
    – markrages
    Nov 5, 2013 at 19:55
  • \$\begingroup\$ @markrages - if you are talking about the precision in my computations - it was intentional. \$\endgroup\$
    – johnfound
    Nov 5, 2013 at 20:01

The improper characterization of parallel and series connected circuit elements needs to be addressed particularly if this is an educational circuit.

It's basically 2 LEDs in parallel

The two LEDs are not in parallel. If the two LEDs were parallel connected, the LEDs would have identical voltages across. What you have is a green LED in parallel with a series connected resistor and red LED

This is explicitly expressed by the KVL equation around that loop:

$$V_{green} = V_{100} + V_{red} $$

so I asssumed that the current flowing through the red LED would have to pass the 2 resistors in series

The two resistors are not in series. For two series connected resistors, all of the current exiting one resistor enters the other. In your circuit, the current through the 270 Ohm resistor splits between the green LED and the 100 Ohm resistor.

This is explicitly expressed by the KCL equation at that node:

$$I_{270} = I_{green} + I_{100} $$


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