volts - watts - amps

V/I=R VI=W or V=W/I or I=W/V

etc etc

HTH

Nick

"V/I=R VI=W or V=W/I or I=W/V" [..]

well that makes no sense to me at all (c;

Les

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

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.

In message , in2minds writes

Volts * Amps = Watts

it means you were correct

as title (c;

Les

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

OHM'S LAW 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

Sparks...

The output or input would be strictly speaking 90 watts. If the voltage it requires is 19, then the current flowing would be 4.7 amps.

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 losses.

Christian.

Or should that be 4.7amps input, and 90W output...? :-)

The guy is having enough truble with his sums without introducing complex numbers and vector arithmetic.