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I came across a device at home which has the specs bellow written on it:

  • Rated Power: 30 W
  • Rated Voltage: DC 12 V
  • Current: 1200 mA

Since P = V × I, P = 12 × 1.2 = 14.4 W

Why on the specs it says 30 W?

What am I missing?

Edit: It is mounted now on the wall, but I found online the specs also: enter image description here

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    \$\begingroup\$ Post a photo. 30 W is probably the max input rating. \$\endgroup\$ – Transistor Mar 30 at 17:47
  • \$\begingroup\$ I updated the post \$\endgroup\$ – Kris Mar 30 at 17:51
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    \$\begingroup\$ 600mA standby current??? That doesn't make sense. Suggests the specs aren't at all trustworthy. \$\endgroup\$ – Brian Drummond Mar 30 at 21:39
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Rated power may be some sort of marketing "audio output power" which, as you calculated, is nonsense. They might justify it by comparison with a similar loudspeaker element mounted in a wet cardboard box rather than their hard plastic horn.

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    \$\begingroup\$ A lot of audio vendors quote "peak audio power" as 2 x the average power for a sinusoidal output (since instantaneous power delivery when voltage and current are in-phase peaks at 2 x the average value). It's a scuzzy practice as old as Hi-Fi. \$\endgroup\$ – Peter Mar 30 at 19:05
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The highest power allowed to flow through siren is 30 W (rated power). Standby current is the current that siren draws when when it is off. Now, to find out current for rated power is simple:

  • P = V * I ---> eq.(1)
  • 30 W = 12 VDC × I
  • I = 30 / 12
  • I = 2.5 amps ---> eq.(2)
  • Substitute value of eq.(2) into eq.(1)
  • P = 12 V × 2.5 amps
  • P = 30 W
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    \$\begingroup\$ Welcome to EE.SE. Did you notice the specification is 600 mA on standby and 1200 mA when in alarm? All you've done is calculated the current that should be drawn and then multiplied it by 12 to get back where you started. OP (original poster) is aware that the current is wrong and wants to know why and you haven't answered that. You can edit your answer to improve it. \$\endgroup\$ – Transistor Mar 30 at 22:30

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