# Voltage loss on DC circuits

I want to connect 4 bulbs in parallel. The bulbs need 12V DC and rated 18 watts. P=V*I, that means I=1.5 A on each bulb. From that the resistance of the bulbs are 8 Ohm (R=12/1.5). The circuit will look like this:

simulate this circuit – Schematic created using CircuitLab

But in real life wires have resistance too. The resistance of my wire is 12 mOhm/m. I need around 12 meters of this wire. The bulbs will be 3 meters from each other. If I calculate this, on the last bulb only 11V will fall and 1.375A will flow through. Which is only around 15Watts.

simulate this circuit

If I want to connect more bulbs (lets say 8 or even more) there won't be enough Voltage across the bulb and it wont work. Did i calculate it wrong? How is this problem problem solved in real life?

• This problem is solved by using higher voltages and lower currents. – Hearth Jul 3 at 10:34
• Do they have to be wired in parallel? – HandyHowie Jul 3 at 10:40
• No, they dont have to. But is there any other way? – Kaleb Jul 3 at 11:15
• At the cost of more wire, feed them at the midpoint. You have 54 mOhm from feed common point to outside bulbs and half the current in each half so Vdrop difference is substantially lower. | For "perfect" results you could use a number of feeds optimised for R and Ls (and more wire). | If you don't care about power and only about relative brightness, an eg LM317 gegulator at each node will allow you more even V but need a higher feed voltage. (LM317 Imax varies - select to suit). |A TL431 and MOSFET / lamp allows customisable results. (Probably need a jellybean bjt and few Rs as well per lamp. – Russell McMahon Jul 3 at 11:29

You have come across the reason why high voltages are used in the home (230V, 110V), even higher are used to transmit power over longer distances (10kV and higher). This is also the reason many cars are moving away from 12V power to 48V, and lorries/trucks use 24V rather than 12V.

As you mention, Power is the current times the voltage, P=IV. If you remember Ohm's law: V=IR, you can put the two equations together: P=Ix(IxR), P=I^2R. Looking at the wires, they have a fixed resistance, other than increasing the diameter of the wirse(expensive and additional weight) or using some very fancy high end superconductor (very very expensive and most are still just not practical for a whole plethora of reasons), the only thing we can change is the current. To get the same power, if you half the current, you need to double the voltage, which isn't too hard to do. But with half the current, the power lost due to resistance is reduced by a factor of 4.

So, to answer your question: how is it solved in real life, we increase the voltage.

Other ideas you may want to consider for your application: Reduce wire length Increase wire diameter (reduce wire resistance) Check any connectors, which often have higher resistance than the wires

As mentioned in the comments, a "star" configuration could also improve matters. The reason for this is pretty obvious, as using a "star" (each load being powered by its own wire) decreases the load per wire. But you are having to spend more money and space on wire, which may not be suitable for your application.

If those aren't useful for whatever reason, increase input voltage.

Increasing input voltage may mean you need to drop the voltage back down at the point of load to make things work. But that can be done with a simple buck converter. There are lots of things to consider when designing the power connection system, but that is beyond what I can do in a quick answer on here.

• Thank you very much! These are the things that came to my mind but i wanted some validation. – Kaleb Jul 3 at 11:22
• Another option is to change the wiring layout to a "star" topology, with individual lamps, or small groups of lamps, connected back to the battery, independently of the others. That way, the first stretch of wire doesn't have to carry all the current. – Simon B Jul 3 at 11:22
• @Kaleb if this has answered the question, you can accept it by clicking the tick next to the answer. – Puffafish Jul 4 at 7:18
• @SimonB indeed you're correct, I've added at note about that in the answer. – Puffafish Jul 4 at 7:19

If you care about matching brightness (which I assume you do based on the nature of your question) then what you want to shoot for is equal current through each bulb.

So there’s two ways of solving that problem. One is to wire the bulbs in series and use a boost LED DC-DC driver, and let Kirchhoff’s Law do the rest. This is arguably the cheapest and most efficient way, with the downside that when one bulb fails the whole string goes down.

The other is to regulate current locally at each bulb using a current sense regulator.

What is the application? - it probably matters.
Do you care about volts. If so why?
You may. Or not.

Brightness of about 2:1 can just be discerned by eye for most people.
Tungsten bulb resistance rises with voltage so 18V:15V difference will be less than 1.2:1 brightness - but IF incandescent, colour change may be unacceptable.

Use LEDs, assuming "bulb" <> LED. FAR less current than incandescent per lux out. If these ARE LEDs then a current regulator per LED leg may suit.
A bjt jellybean and a FET and a few resistors maketh a current source with about 1V min headroom.

At the cost of more wire, feed them at the midpoint.
You then have 54 mOhm from feed common point to outside bulbs and half the current in each half so Vdrop difference is substantially lower.

For "perfect" results you could use a number of feeds optimised for R and Ls (and more wire).

If you don't care about power and only about relative brightness, an eg LM317 regulator at each node will allow you more even V but need a higher feed voltage. (LM317 Imax varies - select to suit). |A TL431 and MOSFET / lamp allows customisable results. (Probably need a jellybean bjt and few Rs as well per lamp.)

Circuits can be drawn for any of the above if of interest.