Power is a product of voltage and current - i.e. Power (in Watts) equals
Voltage (Volts) times current (Amps). For reasons that I can't remember,
although voltage is conventionally represented in the formula by the symbol
V, and power (watts) by W, current in amperes is represented by I rather
than 'A' as might be expected. So in the previous reply, VI=W means volts x
amps = watts. You can transpose the equation to solve for any third
parameter if you know the other two.
So the answer to your original question is "yes" (presuming the "it" to
which you refer is a DC power source specified as delivering 90Watts at
19Volts). 90W/19V = 4.7 Amps approximately (!)
I haven't tried it, but I suspect a quick Google for "Ohms Law" would turn
up a lot more information if you're interested.
Ok, let me try to explain!
To work out Amperage, you do the following: -
Watts divided by Voltage
To work out Wattage, you do the following: -
Amperage multiplied by Voltage
To work out the voltage, you do the following: -
Wattage divided by Amperage
BASE FORMULAS P=I*V V=I*R
TO FIND VOLTAGE V=P/I V=I*R V=SQR(P*R)
TO FIND CURRENT I=P/V I=V/R I=SQR(P/R)
TO FIND POWER P=I*V P=V2/R P=I2*R
TO FIND RESISTANCE R=V2/P R=V/I R=P/I2
P = Power in Watts
V = Volts
I = Electrical Current in Amps
R = Electrical Resistance in Ohms
SQR = Square Root
Only if it is a resistive load (or DC rather than AC). If there is an
uncorrected capacitive or inductive element to the impedence (called
reactance), then the current may be greater.
Resistive loads would include incandescent lighting and heating elements.
Reactive loading (in addition to the resistive) would be produced by
uncorrected fluorescent lighting and motors.
Reactive loading makes the supply less efficient, as it increases the
current (and hence transmission losses) for no benefit. Large premises such
as factories are required to correct their reactive loads to reduce these
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