# Will my chain of LED strips burn out?

I've asked a similar question a few months ago, but I have a few more questions.

Background: I plan to chain multiple LED strips to one another (by soldering) to make one, long LED strip that will be hung up along the outline of my room. I do not currently know whether four LED strips will be enough to surround the entire room, but I purchased five just in case.

LED Strip

Power Supply - Not the exact item, but an identical one rated for 12 V and 25 A.

As you can see, the LED strip is rated at 12 V DC, and consumes 48 W per strip.

I had the opportunity to measure the current drawn from one strip at 12 V, and it reached almost 1 A.

Question: If I were to chain four or five of these strips together, would they be able to handle the amount of current drawn at 12 V? If not, would connecting wires to ground at each chain prevent too much current being drawn?

Note: I plan to supply each side of the long chain (four or five strips) with 12 V and not just one end.

Thank you!

• The LEDs on the middle are definitively going to see a lower voltage. I doubt the strip is going to overheat, but the middle ones will be dimmer than the ones closer to the power input. That said, it sometimes it's not perceptible. There is no way to tell without knowing a bunch of info (such as strip resistance per meter), so it's way easier to just test it out and see if its acceptable. – Wesley Lee Jul 1 '17 at 18:24
• @WesleyLee I will connect more power lines to the middle strips if needed, but I just want to make sure that the whole strip won't burn out if it ends up drawing close to 5 A. – Jacob G. Jul 1 '17 at 18:26
• There's something wrong between the specifications and your measurements. From $P = VI$ you can calculate the expected current as $I = \frac {P}{V} = \frac {48}{12} = 4 \; A$. You only measured 1 A. – Transistor Jul 1 '17 at 18:27
• In my experience it just won't reach high currents simply because the voltage drop will make the LEDs draw less and less current. You should be safe to test and check periodically. – Wesley Lee Jul 1 '17 at 18:28
• The "drawn" current is less than the advertised one because the advertised current does not take into account the resistance of the strip or the wire. – Wesley Lee Jul 1 '17 at 18:28

$(I_{out} + (I_{in} - I_{out})/2) * V_{Drop}$