Ohms Law My Guitar amp has 16 ohm or 8 ohm outputs.?

I don't know where the previous contributor got that answer from but talk about overdoing it. You don't need Ohm's law to work this one out, (Ohm's Law is all to do with the relationship between resistance, voltage and current) you need the following equation:

1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ... 1/Rn (where n is the number of resistors)

So for your question it would be:

Internal Impedance: 1/Rtotal = 1/8 + 1/8 = 4 ohms as stated.

You have the following possibilities, internal speakers in series or in parallel, connected with the external speakers, which I assume is a parallel connection. Assuming you are not going to rewire the external cab, you have:

Internal speakers in Series: 1/Rtotal = 1/16 + 1/8 = 5.3 ohms

Internal speakers in parallel: 1/Rtotal = 1/4 + 1/8 = 2.67 ohms

So if the output from your amp is 4 ohms leave the internal speakers alone. If the out put is 2 ohms, rewire to parallel.

As a rule of thumb, connecting resistance in parallel always produces a resistance that is smaller than the smallest lowest resistance.

See http://www.1728.com/resistrs.htm

Hello there, I am a little confused. Does this amp only have one channel going out. I did not think if was stereo. If it is mono, how can you be running two lines out to the same cab. I am a little lost on that. Also, if you have your cab wired for a 16 ohm load why are you running your amp's output at 8 ohm. An impedance mismatch on a tube amp is dangerous. May trash your amp doing that. Not sure how you can change a 16 ohm load cab into an 8 ohm cab without changing speakers. Lets say you have the cab loaded with four 8 ohm speakers. If you wire the two of the speakers in series you have a 16 ohm load. Then if you wire the two pairs in parallel you have an 8 ohm load. However, if you wire the two pairs in series you have a 32 ohm load. Granted, the 1960 is normally loaded with 16 ohm Celestrions. If you wire the speakers in pairs in series and then wire the two pairs in parallel you have a 16 ohm load. How do you get 8 ohm out of yours? I think some of your tone problem is that you may not understand how to figure the impedance load of speakers and that you are trying to run an impedance mistmatch between the head and the speakers. As for getting to the break up point in your speakers without an attenuator where you cannot crank the amp up, I suggest using an overdrive pedal to get you there. Something like the tubescreamer should do the trick.. Alternatively, you can swap out your speakers for lower power speakers that will get to the break up point at a lower power. The problem with that is you cannot crank the volume back up without blowing your speakers. That is a one way mod. So unless you are only using your amp at home, I don't suggest that one. Go with the overdrive pedal. Forget the impedance mistmatch idea. That is a big risk. If you need more information about the damage you can do to a tube amp trying to run an impedance mismatch, let me know, I can give you some links. You can get by with certain impedance mismatches running solid state without damaging the amp. Not so with tubes. Man, don't do that to a tube amp. Later,

Yes your ohmic calcs at the start are correct.

Not sure what u mean by 8x12 ext cable?

The max power transfer theorem states the the impedance of the sources should match that of the load.

Mind you the ohmic values are for DC, as the frequency of the AF can vary form say 20 Hz to 20kHz on a good qual amp, then the impedance of the load {speakers{ is given by the vector addition Real number / j notation / trig }or the resistance R and inductance, XL, where XL = 2 Pi FL {canny get the Pi symbol up here !}.

So basally, unless u overload the amp, you should have no real prods.

, the maximum power (transfer) theorem states that, to obtain maximum power from a source with a fixed internal resistance, the resistance of the load must be made the same as that of the source. It is claimed that Moritz von Jacobi was first to discover the maximum power (transfer) theorem which is referred to as "Jacobi's law".

The theorem applies only when the source resistance is fixed. If the source resistance were variable (but the load resistance fixed), maximum power would be transferred to the load simply by setting the source resistance to zero. Raising the source impedance to match the load would, in this case, reduce power transfer. This is the case when driving a load such as a loudspeaker with a modern amplifier. In this case, the load presented by the loudspeaker is fixed (typically, 8 ohms for home audio) and maximum power occurs with an impedance bridging connection. This type of connection also serves to maximize control of the speaker cone (due to high damping factor), which serves to lower distortion.

Hope this helps.

There is not a simple answer though.

then of course you have left and right signals and bass, mid and high freq speakers usually in both L and R speakers, each with a different impedance and frequency response.

I could go on . . . . . . .

Paul

{B. Eng Hons}