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

I like Re-design's answer below.

Try this calculator.

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