You need to tell us the lamp name or show intended circuit. You dont even have a picture frame.
Here is a calculation to light the corners of your missing pic frame at the corners with white 4.5Vfwd LEDs 20 ma each from 12VDC
We will put two LEDs in series twice. Two LEDs = 4.5Vf + 4.5Vf = 9.0Vf Resistor drop = 12VDC - 9Vf = 3.0VR R value = VR / I = 3Vr / 20ma =3 / 0.02A = 150 ohms Power of R = V x I = 3VR x .02A = 0.06 Watt
So you can light two LEDs in series with a 1/4W 150 ohm Resistor to 12VDC
Add two more with one more 150 ohm does the four corners.
That is two total resistors using a total of 40 ma from the 12VDC source.
Discussions
7 years ago
I like Re-design's answer below.
9 years ago
Try this calculator.
9 years ago
You need to tell us the lamp name or show intended circuit.
You dont even have a picture frame.
Here is a calculation to light the corners of your missing pic frame
at the corners with white 4.5Vfwd LEDs 20 ma each from 12VDC
We will put two LEDs in series twice.
Two LEDs = 4.5Vf + 4.5Vf = 9.0Vf
Resistor drop = 12VDC - 9Vf = 3.0VR
R value = VR / I = 3Vr / 20ma =3 / 0.02A = 150 ohms
Power of R = V x I = 3VR x .02A = 0.06 Watt
So you can light
two LEDs in series with a 1/4W 150 ohm Resistor to 12VDC
Add two more with one more 150 ohm does the four corners.
That is two total resistors using a total of 40 ma from the 12VDC source.
What say you ?
A