![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ6LvaMOg6_gGtExiDirjBrcTdpBSY8q_joxx2x9otvWO44DHcl67vRcj20WNcmFGA3P_KFX1JCYTJZUvXmh0DdysU8_kbxWxu105xNdbu2epnn-hxuKHKgq6MAtkTp7gH4gdvpsnOE2xK/s320/image.jpeg)
Introduction:
- Using a bench-top power supply as our 9V
- LED1: 5V and 22.75 mA
- LED2: 2V and 20 MA
Using V=IR we find the values of the resistance in the LEDs:
- LED1: 219.78 Ohms
- LED2: 100.0 Ohms
In order to find R1 and R2 required we use the concept that parallel branches have the same voltage. To find the power we use the power equation from the previous blog. (Values shown later)
We choose the available resistors closest to the calculated value. and set up the following circuit:
The currents and voltages were measured in 3 loops
- all components active
- LED 1 active
- LED 2 active
Data:
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