John Howe Posted April 29, 2003 Share Posted April 29, 2003 No doubt due to the fact that my father was a spark and we never got on, I have a complete lack of intelligence to all things electical. Can someone kindly suggest a rule of thumb by which I can establish when one should add a relay, rather than a straight connection, in to a circuit? ... and what are the yardsticks to the "size" of relays. JH Deliveries by Saffron, the yellow 222bhp Sausage delivery machine Link to comment Share on other sites More sharing options...
V7 SLR Posted April 29, 2003 Share Posted April 29, 2003 It's a case of current draw. Rules of thumb for using a relay: 100W headlights Air horns (motors draw lots of current) Radiator fan (motor again) Fuel pump (motor) Any electrical heater If you post what you want to do I might be able to help further. Worcs L7 club joint AO.//Membership No. 4379//Azure Blue SLR No. 0077//Se7ens List Tours Link to comment Share on other sites More sharing options...
Chris W Posted April 29, 2003 Share Posted April 29, 2003 John Basically, 3 reasons for using a relay. 1. If the current the device will draw is larger than the switch's rating. 2. If using a switch directly means that you have to run long lengths of cable back to the dash. Siting a relay nearer to the device means that shorter lengths of cable can be used for the main current, while the long lengths go back to the switch and only have to power the relay's coil (a small current so no appreciable volts drop). Long lengths may cause an appreciable voltage drop in the cable at higher currents and will result in the device not running as effectively as it would with full current. For some devices this doesn't matter of course but for others it is critical. For example, when all the lights are on, the current draw is around 20 amps. Since volts drop (by Ohms Law) is current x resistance, then even if the cable resistance is only 1/20 ohm, (ie: 1/10 ohm to the switch and back) you will lose 2v in the cable. This may not seem much, but power is proportional to the volts squared so if you drop from 12v to 10v (ie: 83% of the original voltage) the power in the lights drops to 0.83 squared, ie: 69% of its original value. So the lights would be noticably dimmer. The switch on the dash would also be have to be fairly chunky to handle 20 amps continuously. Whereas, if you use a typical car relay whose operating current is less than 100mA, then the volts drop for the same resistance in the switch cable is only 1 millivolt, which will have no adverse effect on the operation of the relay. 3. If you want to switch a couple of separate circuits with one physical switch where you need to keep the circuits electrically isolated from each other. (A double pole relay) A long answer to an apparently straightforward question. Just shows there are no dumb questions cheers Chris 1.8K SV 140hp see it here Link to comment Share on other sites More sharing options...
John Howe Posted April 30, 2003 Author Share Posted April 30, 2003 Nigel / Chris - thanks for your replies JH Deliveries by Saffron, the yellow 222bhp Sausage delivery machine Link to comment Share on other sites More sharing options...
J.R. Posted May 1, 2003 Share Posted May 1, 2003 Chris W. If the cable resistance was 1/10 ohm to the switch and 1/10 ohm back then the total resistance would be 1/5 of an ohm not 1/20th. Current draw on main beam circuit is 5 amps per light (each light has seperate feed cable, but even using your figures of 20 amps and a resistance of 1/20 ohm would only result in a voltage drop of 1 volt. Electrical power (watts) is directly proportional to voltage, not squared. Watts = amps * volts. So using your figures the reduction in power would only be 8.33%. If you check you will see that the resistance of the lighting cable is lower by several orders of magnitude than 1/20th of an ohm, more likely to be less than 100th of an ohm per metre. Also the current flow does not go to "the switch and back" the return is via the nearby earthtag to the chassis return & to the battery via the earth strap. Your calculations would hold true (excepting the voltage drop squared) for a bad earth or switch connection. The easiest way to measure volt drop if you are worried is to use an accurate multi-meter set to the lowest range with the lights switched on, one probe at the switch connection the other at the headlight connection. The fractional voltage displayed will be the voltage drop, this divided by the battery voltage and multiplying by 100 will give the percentage reduction in power from the voltage drop in the circuit. Link to comment Share on other sites More sharing options...
Chris W Posted May 1, 2003 Share Posted May 1, 2003 JR You're totally wrong I'm afraid You're talking through your 🙆🏻 Power IS proportional to voltage squared. You're quite correct that Power = current x voltage but remember that current = voltage/resistance, (ie: current is a function of voltage applied) therefore Power = voltage squared/resistance. You can of course rearrange everything to show that Power is also equal to current squared x resistance. What you're overlooking is that if the voltage drops by half, then the current ALSO drops by half and, as you say, power = current x voltage. So the power will therefore drop to 25% of what it was. If you could half the voltage and miraculously keep the current the same, you would be correct. But unfortunately you're not. A Blue Peter experiment.....connect a couple of fresh AA batteries in series to a resistor (say 1K ohms). Measure the current through the resistor and the voltage across it. Now take away one of the AA batteries and measure again. Assuming both batteries were reasonably identical to commence with, the voltage across the resistor and the current through it will have both halved. So if the power in the resistor the first time = v x i (where v = volts across the resistor and i = the current through it) then the power in the resistor the second time = v/2 x i/2 = vi/4 Thus the power will be 1/4 of what it was the first time. Re the 1/10th of an ohm on the cables etc, I said in my posting 1/20th of an ohm actually if you read it correctly, ie: 1/10th of an ohm there and back (ie: total). I accept that one side will be earthed to the chassis but I was merely using the example to illustrate the point that, at high currents, cable resistance is very critical. I agree that real cables do have much lower resistances than I used in the calculations. And you can see why, you would lose so much power if it were otherwise. Chris 1.8K SV 140hp see it here Edited by - Chris W on 1 May 2003 20:18:40 Link to comment Share on other sites More sharing options...
Mick Day Posted May 1, 2003 Share Posted May 1, 2003 No wonder John switched off to all things electrical. Mick Link to comment Share on other sites More sharing options...
J.R. Posted May 2, 2003 Share Posted May 2, 2003 Totally wrong and talking out of my 🙆🏻?? Makes you wonder how I have earned a living in this field for the last 12 years doesnt it? I suppose my algebra must be equally bad because I cant see how you can turn a linear equation - "power = voltage * current" and "current = voltage resistance" into the exponential "power = voltage squared * resistance. But then what would I know Link to comment Share on other sites More sharing options...
Chris W Posted May 2, 2003 Share Posted May 2, 2003 JR Makes you wonder how I have earned a living in this field for the last 12 years doesnt it? My thoughts exactly..... if your bosses find out you know Jack Sh*t about basic electronics, you aren't going to be employed much longer ❗ Your understanding of mathematics is obviously very basic too because Power = voltage x current is NOT a linear equation because current is NOT independent of voltage, it is a direct function of it. ie: P = v x f(v) where f(v) represents a function of voltage (viz: current). Therefore P is proportional to voltage squared. Follow this slowly and you will understand all.... Power = (voltage x current) we agree on this I believe (equation 1) Current = (voltage / resistance) we agree on this too I believe since its Ohms law (equation 2); Therefore, substituting the expression for current from equation 2 into equation 1, we get Power = voltage x (voltage / resistance) ie: Power = voltage squared / resistance (note it's voltage squared DIVIDED by resistance and not multiplied as you also incorrectly wrote in your last post). This is electronics lesson one, first 10 minutes basic Ohms law calculations. Geez i hope you don't repair aircraft or anything else I might travel in Chris 1.8K SV 140hp see it here Edited by - Chris W on 2 May 2003 12:38:21 Link to comment Share on other sites More sharing options...
Chris W Posted May 2, 2003 Share Posted May 2, 2003 Jr Just to finally convince you..... Suppose we have a variable voltage source and we connect this voltage source across a 1 Kohm resistor. From Ohms law we will get the results in the table below for current. Power in the resistor can be simply calculated by Power = voltage x current, as you yourself said previously. Note that if the voltage doubles from say 6v to 12v then power quadruples from 36mW to 144mw. Again, if the volts go from 2v to 8v (factor of 4 increase) the power goes from 4mW to 64mW (a factor of 16 increase) ie: Power is proportional to voltage squared. QED resistance = 1 Kohm volts (v)_____________current (mA)___________Power (mW) 0___________________ 0_____________________ 0 1___________________ 1_____________________ 1 2___________________ 2_____________________ 4 3___________________ 3_____________________ 9 4___________________ 4_____________________ 16 5___________________ 5_____________________ 25 6___________________ 6_____________________ 36 7___________________ 7_____________________ 49 8___________________ 8_____________________ 64 9___________________ 9_____________________ 81 10___________________10____________________100 11___________________11_____________________121 12___________________12_____________________144 Chris Edited by - Chris W on 2 May 2003 13:12:48 Link to comment Share on other sites More sharing options...
John Howe Posted May 2, 2003 Author Share Posted May 2, 2003 Well you guys have lost me... but it makes for riveting reading. Can't wait for the next exciting instalment... and such a shame dualling has been banned. Can just imagen it - Avo's at dawn! JH Deliveries by Saffron, the yellow 222bhp Sausage delivery machine Link to comment Share on other sites More sharing options...
J.R. Posted May 2, 2003 Share Posted May 2, 2003 Chris I stand corrected. Thankfully being self employed my boss is as ignorant as I am. No I dont repair aircraft but if you are worried perhaps you might want test the aircraft mechanic on ohms law before you fly. Just goes to show that you are never too old to learn the basics of electrical theory, mathematics or politeness Link to comment Share on other sites More sharing options...
Chris W Posted May 2, 2003 Share Posted May 2, 2003 JR I admire your strength in recognising your error and your courtesy in acknowledging it. rgds Chris 1.8K SV 140hp see it here Link to comment Share on other sites More sharing options...
Dave L. Posted May 2, 2003 Share Posted May 2, 2003 JR, You give up too easily v=ir p=vi p=i*i*r i.e. p is proportional to i squared (which Chris's chart also proves !) v=ir i=v/r p=vi p=v*v/r i.e. p is also proportional to v squared Primary school electronics (and algebra for that matter). And JR is also right when he says that 1/10 ohm plus 1/10 ohm in series, as per your inference "to the switch and back", is indeed 1/5 ohm. Chris, you're thinking of resistor in parallel, when it would be 1/20 ohm. A little knowledge ... That's your company car !?! Link to comment Share on other sites More sharing options...
Dave L. Posted May 2, 2003 Share Posted May 2, 2003 Oh, and another thing - JR is quite right that using Chris's figures of 20 amps flowing through a resistance of 1/20 ohm results in a drop of 1 Volt not 2. 20 times 1/20 results in 1 in anybody's maths ! Except for the GF's when it would result in 42.3 thingummijigs That's your company car !?! Link to comment Share on other sites More sharing options...
Chris W Posted May 2, 2003 Share Posted May 2, 2003 Dave Read my original post and you will see I said 1/20th of an ohm per cable. ie: a total of 1/10th of an ohm there and back. Therefore 20 amps flowing through 1/10th ohm results in a voltage drop of 2v which is what I said in the first place. So I'm not thinking of resistors in parallel. I know how to do these simple sums thank you. And indeed as I put in my first post.... power is proportional to voltage squared OR current squared. That's why the cables on electricity pylons run at such huge voltages (33,000 v) so that the same amount of power can be delivered but at a much lower current (since p = v x i). This dramatically reduces the power losses in the cable due to current squared x r losses. Just read what was written Dave. Chris 1.8K SV 140hp see it here Link to comment Share on other sites More sharing options...
Dave L. Posted May 2, 2003 Share Posted May 2, 2003 Chris, The magic words were "per cable" which did not appear in your original post. Subsequently armed with those I apologise for mis-interpreting the bit in brackets, but I'll be damned if I can find anywhere in your first post, even now, where it says power is proportional to current squared. It does however appear in your second post - pedantic, moi ? So it seems my reading lessons have paid off after all, and I do know how to read these simple things, thank you. That's your company car !?! Link to comment Share on other sites More sharing options...
Chris W Posted May 2, 2003 Share Posted May 2, 2003 Dave The words in my first post were: then even if the cable resistance is only 1/20 ohm, (ie: 1/10 ohm to the switch and back) ........so it couldn't be clearer. Anyway that wasn't the point of the post. The point was to show that power is proportional to volts squared which I guess everyone agrees we have now done. This was to show why cable resistance has to be kept really low with high currents, so the volts at the end device remain high. I don't understand your belabouring the point about power being proportional to current squared as if I somehow disagreed with you. I agree with you. I have never had a doubt about it. It was others who were disagreeing. Not me. It wasn't relevant to the first post because we were considering the volts drop in cables (v^2/r), not the heating effect of the current (i^2*r). BTW, you couldn't really use a cable with a resistance of 1/10th ohm at 20Amps as, besides the 2v drop in the cable, the power (heating effect) in the cable would be 20^2 x 0.1 = 40 Watts. At least the engine bay would be illuminated! As I said earlier, cables actually must have far lower resistances but its always interesting to see why. Hope everything is now clear and we can get on with life. cheers Chris 1.8K SV 140hp see it here Link to comment Share on other sites More sharing options...
Ken elle Posted May 2, 2003 Share Posted May 2, 2003 When I started reading this post I knew nothing about electrics, now I think I know even less 😬 Edited by - ken elle on 2 May 2003 18:41:37 Link to comment Share on other sites More sharing options...
Dave L. Posted May 2, 2003 Share Posted May 2, 2003 Hey Ken, Wait 'til we get started on AC power theory ... eek That's your company car !?! Link to comment Share on other sites More sharing options...
Chris W Posted May 2, 2003 Share Posted May 2, 2003 For that we'll start doing some Fourier and Laplace Transforms or maybe Maxwell's Equations would be a good start. 1.8K SV 140hp see it here Link to comment Share on other sites More sharing options...
Tony C Posted May 2, 2003 Share Posted May 2, 2003 As far as I know, Maxwell's left hand didn't know what his right hand was doing 😬 Or was that Fleming? BRG SV 😬 Edited by - Tony C on 3 May 2003 19:58:08 Link to comment Share on other sites More sharing options...
EFA Posted May 2, 2003 Share Posted May 2, 2003 How are you on Helmholtz theorem, Chris?? Fat Arn Visit the K2 RUM website See the Lotus Seven Club 4 Counties Area Website here Link to comment Share on other sites More sharing options...
Chris W Posted May 3, 2003 Share Posted May 3, 2003 You hum it Arnie, I'll play it 1.8K SV 140hp see it here Link to comment Share on other sites More sharing options...
neilsjuke Posted May 3, 2003 Share Posted May 3, 2003 cut the cra? rule of *thumbup*if it draws over 8 amps fit a relay Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now