Parallel Resistance Question - Page 1
Forum section: electronics / beginners

author: PStevenson
author_rank: Regular Contributor
timestamp: January 29, 2013, 05:03:27 pm
content: this may be a bit of a noob question and I assume I know the answer however I thought it best to get a second opinion because this involves my expensive guitar Amp.what I need to do is make a 16(ish) ohm load for the speaker output which I can do by having 2 33r resistors placed in parallel (16.5 is in tolerance of my requirements)however what really matters here is the power rating of the resistor. if I put 2 100watt resistors in parallel do I get 200watt capability or 50watts?I'd assume it would give me 200watts. but as I said - best to get a second opinion.thanks.

author: madires
author_rank: Super Contributor
timestamp: January 29, 2013, 05:13:29 pm
content: 200 Watts, since the current is split between both resistors. If you put both resistors in series you'll get 100W (same current flows through both resistors).

author: PStevenson
author_rank: Regular Contributor
timestamp: January 29, 2013, 05:17:10 pm
content: aye that's what I assumed - a split load.many thanks !

author: ampdoctor
author_rank: Frequent Contributor
timestamp: January 30, 2013, 04:14:25 am
content: As an aside, make sure you evaluate your power rating needs properly. If you're only going to build a dummy load for the purpose of output tube biasing or general purpose troubleshooting you can probably get away with a rating equal to the amps output rating, maybe even a little less. However, if you're going to be doing any work requiring you to run the amplifier at maximum output for any longer than a few minutes you may need a much higher rated load than you think. Under heavy clipping the output will be very nearly a square wave(not exactly but close enough for government work) so the voltage you would use to calculate approximate power dissipation through the load will be much closer to Vpk as opposed to Vpk/sqrt(2) were the output a relatively clean sine wave.

author: IanB
author_rank: Super Contributor
timestamp: January 30, 2013, 04:24:12 am
content: Quote from: madires on January 29, 2013, 05:13:29 pmNot that it actually matters in this case, but it would be 200 W either way.Consider: the 16 ohm load should dissipate 200 W. From the formula P = I2R we find I = 3.54 A.This could be made up from two 32 ohm resistors in parallel, in which case the current would be halved and the power in each resistor would be (3.54/2)2x 32 = 100 W.Or it could be made up from two 8 ohm resistors in series, in which case the power in each resistor would be 3.542x 8 = 100 W.So whether you put them in series or in parallel, two 100 W resistors will make a 200 W resistor.

author: madires
author_rank: Super Contributor
timestamp: January 30, 2013, 01:09:02 pm
content: Quote from: IanB on January 30, 2013, 04:24:12 amYou convinced me ;-)

author: PStevenson
author_rank: Regular Contributor
timestamp: February 01, 2013, 10:59:02 pm
content: aye, I know you need to double the power rating (at least) of the amplifiers top output. as the output isn't DChowever I never use higher than the 100watt mark on my amp (it's a 130watt) I usually use it alot lower, cranking these things to hard tends to be expensive in valves + sounds shit. and it's only really to Direct record it without a speaker.all my other amps are either 30watts, 5 watts...basically lots of sub-100 watt ampsas for the biasing and so on (someone mentioned) it's quite cool in that respect - it's one of these Bugera 6262 amps that allows you to plug your DMM probes into it with a list of voltages written on the top of it next to a trim pot - takes about 2 seconds to do, it also has a switch so you can change the power valves you use from EL34 to 6L6.I'm not a big valve fan but this thing is pretty amazing.