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Wed, 17 April 2002 15:28

dbeardsl
Messages: 127
Registered: May 2009

Master

The volts are different cause I'm calculating sensitivity at 1 Watt.
thats why it doesn't look right.

---the system of equations
Watts = Current * Volts
Volts = Current * Resistance

---expanded solution
so Watts = Current^2 * Resistance

so Current^2 = Watts / Resistance

so Current = sqrt(Watts / Resistance)

since Volts = Current * Resistance
Volts = sqrt(Watts / Resistance) * Resistance

so Volts = sqrt(Watts * Resistance)

so 2.83V = sqrt(1W * 8) for 8 ohms.
2V = sqrt(1W * 4) for 4 ohms.

If you measure sensitivity at 2.82V/1M, then the Voltage will be the same and Wattage will be different.

		Contemplation on combined Sensitivity(long) By: dbeardsl on Wed, 17 April 2002 14:41
		Re: Contemplation on combined Sensitivity(long) By: Adam on Wed, 17 April 2002 15:07
		but By: dbeardsl on Wed, 17 April 2002 15:28
		Re: but By: Adam on Wed, 17 April 2002 15:30
		(epiphany) By: dbeardsl on Wed, 17 April 2002 16:01
		Re: (epiphany) By: Adam on Wed, 17 April 2002 16:37
		Re: Contemplation on combined Sensitivity(long) By: DanF on Wed, 17 April 2002 15:39
		104 pages of answers By: Wayne Parham on Wed, 17 April 2002 16:21
		Re: Contemplation on combined Sensitivity(long) By: Hodel on Wed, 17 April 2002 17:04
		Re: Contemplation on combined Sensitivity(long) By: mikebake on Wed, 17 April 2002 19:22

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