I had a look at another Exchange question, and couldn't quite understand.

I have a set of cheap speakers, 8ohms each with a combined output of 6 watts.

If sqrt(6w/8Ω) = 0.866 v, and the post says line level is approximately 1 volt, what should my amperage be?

How can I calculate for an appropriate resistor?

I have about 10 150Ω resistors at home... will wiring them half and half between speakers attenuate the signal? How many do I need of each?


I want to know how to calculate the resistance needed to reduce a powered signal (6w) to line level.


closed as unclear what you're asking by Trevor_G, Enric Blanco, PeterJ, Dmitry Grigoryev, winny Jun 8 '17 at 10:36

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ I think your calculation's a bit wonky. 0.866V into 8 ohms won't give you 6W. \$\endgroup\$ – Simon B Jun 1 '17 at 13:59
  • \$\begingroup\$ It ought to be \$\sqrt(3W * 8R)\approx5V\$, \$I\approx0.6A\$ per speaker. However, no idea what you are asking re the resistors.....question is unclear. Voting to close. \$\endgroup\$ – Trevor_G Jun 1 '17 at 14:10
  • \$\begingroup\$ You seem to be confusing all sorts of nomenclature. Line level is not used to drive speakers, it is used to send audio signals to devices expecting line-level (nominally 1-2Vpp) inputs, like power amps (used to drive speakers), audio interfaces, or studio effects units. \$\endgroup\$ – Chris M. Jun 1 '17 at 14:57
  • \$\begingroup\$ @ChrisM. Hence my question says reduce to line level, not from line level. I have speakers that are powered, but I want to cut the wires to them and output them to a jack at line level. \$\endgroup\$ – yeeeeee Jun 2 '17 at 6:24
  • \$\begingroup\$ @Trevor, I don't come from any electrical background at all. Don't "vote to close" because of my naivety. \$\endgroup\$ – yeeeeee Jun 2 '17 at 6:25


simulate this circuit – Schematic created using CircuitLab

Something like the circuit shown should work. I'm assuming that the amplifier will be happy driving a 18 ohm load, rather than 8 ohm. Doubling the resistance halves the power dissipated. Even so, R1 needs to be something chunky, such as a 2W resistor.

I chose the relative values of R1 and R3 so that you don't have to run the amplifier at full volume. Turning down the volume a bit significantly reduces the distortion.

If the impedance is important, you'd have to halve the values of both resistors, and use an even bigger resistor for R1, such as 5W rated.


Not the answer you're looking for? Browse other questions tagged or ask your own question.