Easy - The lower the resistance, the more current flow can occur. Think about resistance as a water hose. The less resistance the more flow. When you have more current flow, there is more heat generated because conductors are not perfect, they also have a resistance. The amp runs hotter and requires bigger cooling heatinks, bigger caps to store all those electrons, wider electrical traces, bigger posts, etc.
More resistance is less load??
Hi, can someone explain, in "ohms for dummies" language, why a 4 ohm speaker, which has half the resistance of an 8 ohm speaker, is said to be more demanding on the amp? And the other way, why a 16 ohm speaker, with twice the resistance, is less demanding?
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Expanding on the relationship Audioman pointed to... The relationships between resistance (ohms), current and voltage are defined by Ohm’s Law: V/I = R where V is voltage (volts); I is current (amps); R is resistance (ohms). See link here.... https://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.html At constant V, if R increases, I will decrease. At constant V, if R decreases, I will increase. As a simple example (I’ll defer to Almarg or Atmasphere for their input on a better voltage value or range to use in discussing an operating amplifier).... At 120V if R = 8 ohm, I = 15 At 120V if R = 4 ohm, I = 30 At 120 V if R = 16 ohm, I = 7.5 So you can see how greatly current requirements are affected by resistance. It’s important to remember that a speaker’s resistance might not be constant across the frequency range. Some portions of the frequency range might be associated with low resistance and consequently run up against the current limitations of the amp reproducing sound in that frequency range. Hope this helps. |
Simply, the lower the resistance of a speaker, the closer it is to a short (zero ohms). EG: Like a car going up hill, the steeper the hill harder the engine has to work, if almost vertical it could blow if not built for it. In speakers v amp this means that amp has to supply current more and more the lower the speakers impedance (resistance). Cheers George |
Good responses by the others above, except that the reference to E/IR should be E = IR (meaning E equals I multiplied by R). (E is commonly used in the context of Ohm’s Law to denote volts, and means the same thing in that context as V). So when Ghosthouse referred to V/I = R, he was correct. By simple algebra V/I = R is equivalent to V (or E) = I x R = IR. Regarding the voltages that are typically provided to speakers, for a resistive load: Power = (I squared) x R = (E squared)/R where, for example, P is expressed in watts, I in amps, R in ohms, and E in volts. Therefore E = (Square root (P x R)). So for example 100 watts into an 8 ohm resistive load corresponds to: Square root (100 x 8) = 28.28 volts. Regards, -- Al |
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