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Power calculations
AC power, which includes audio power, is best measured as an average power. This is the accurate method. It is based on this formula
[2]:
For a purely resistive load (not a speaker), a simpler equation can be used:
In the case of a steady sinusoidal tone (not music) into a purely resistive load, this can be calculated from the peak voltage and the resistance:
Though a speaker is
not purely resistive, these equations can be used to approximate power measurements for such a system, as follows.
An ideal (100% efficient)
class AB amplifier with a 12-volt
peak-to-peak supply can drive a signal with a
peak amplitude of 6 V. In an 8
ohm (see
impedance) loudspeaker this would deliver:
Ppeak = (6 V)2 / 8 Ω = 4.5 watts peak instantaneous.
[3] If this signal is
sinusoidal, its RMS value is 6 V × 0.707 = 4.242 V(RMS). This voltage into a speaker load of 8 Ω gives a power of:
Pavg = (4.242 V)2 / 8 Ω = 2.25 watts average
[4] Thus the output of an inexpensive car audio amplifier is limited by the voltage of the battery. In most actual car systems, the amplifiers are connected in a
bridge-tied load configuration, and speakers are no higher than 4Ω. High-power car amplifiers use a DC-to-DC converter to generate a higher supply voltage.
The true power output of an amplifier can be estimated by examining the input current. Linear amplifiers tend to be about 60% efficient at best. A switch-mode amplifier (known as class D) can achieve much higher efficiency, sometimes as high as 95%. A linear car amplifier labeled "500 W PMPO" but fitted with a 5-amp fuse can, at most, deliver an average power of 5 A × 14.4 V × 60%, or about 43 watts.
