Help me calculate the total voltage loss of circuit - DC

I am trying to anticipate the total voltage loss in my circuit.

Useful info: I have a power supply (12V, 60A, 720w) Wire used has a cross section of 2.5mm2 Each 6A represent the power consumption of an LED striplight of 5m Each segment (shown with length in meters) will start and finish at the power supply.

• Is it a homework question? – dirac16 Jun 20 '17 at 13:32
• Nop, i am planning a 35m RGB strip light for my home. – Reuben Chetcuti Jun 20 '17 at 14:07
• What is the resistance per meter for your wire? – Tyler Jun 20 '17 at 14:32
• wire resistance is 0.00741 per meter – Reuben Chetcuti Jun 20 '17 at 15:48

You have five circuits so you want to calculate the voltage loss in each individual circuit not "the total voltage loss in my circuit".

2.5 mm² copper has a resistance of 7.50 mΩ/m. You have your lengths (but don't forget to double the lengths to include the return leg). Now you can work out R for each circuit.

You already have I so V = IR will get you the voltage drop.

Your voltage drop is 12V for this to be a complete circuit. If you plan to have additional loads after the strip lighting then you can consider the voltage drop across your lighting circuit shown.

If you put 2 strips in series you will cut the current flowing through them in half causing them to be dim. You should daisy chain your lighting (place in parallel).

Based on your circuit assuming you are using copper wire and the lengths provided are the lengths of wire not including the lengths of the LEDs I estimate you will get the following results:

• since i have 7 sets of leds (each consuming up to 72W) should the total W be around 504W? – Reuben Chetcuti Jun 20 '17 at 15:43
• If you were to place them all in parallel then ideally yes if you do not consider the reduction in load current due to the resistance of your wiring. But since you have 2 paths with LED strips in series you will consume less power and those will be dim. Consider Ohms Law (V=IR or V/R = I). If you increase your load R then you decrease your current. Power is VI or I^2*R. If your current drops, your power consumption also drops. – Steve771 Jun 20 '17 at 15:49

First you lookup the resistance per meter of the wire you are using. There are many on-line resources for this. If it is a pure copper, solid wire then the resistance is 0.0075 ohms per meter.

For each loop, multiply your total wire length in meters by the 0.0075 ohms per meter. This will give you the resistance of the wire for that total length. Make certain you include the wire length from both sides of the LED strip.

Since the wire resistance will be much smaller than the load resistance, it is a matter of applying Ohm's law for a series circuit to get a good approximation of voltage drop across the wire. You know the total current in each loop (6 amps according to your drawing). You know the resistance of the wire in that loop from the earlier calculation. Voltage = I*R so multiply the current times the wire resistance and you know how much voltage you are dropping in each of the wire segments. For example, in your 18 meter loop (assuming it is 36 meters in total), the voltage drop in the wire will be about 6 amps * 0.0075 ohms/meter * 36 meters = 1.6 volts.

You should note that you have LED strips in series connected to the supply and others are by themselves connected to the supply. Unless these two strips are rated differently, you cannot do that. You will either burn out the single strips or the series strips will be too dim. Can you post a link to the datasheet for the LED strips for additional assistance with this?

• 0.0034 ohms per meter is the resistance per length of 2.5 mm diameter solid wire, not 2.5 mm² cross-sectional area. 0.0075 ohms per meter as listed below by Transistor is a more appropriate value. – davidmneedham Jun 20 '17 at 14:56
• Well i looked at the specs if the cable and say its 7.41ohms / km (0.00741). Regarding the led specs, all they say its 12V, 72W every 5m. I was planning to use a mini amplifier with each segment to increase the brightness. – Reuben Chetcuti Jun 20 '17 at 15:46