- posted on October 22, 2003, 10:00 pm

if it's 19v & 90w does that means it's 4.7amps output ?

thanks Les

- posted on October 22, 2003, 10:17 pm

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

etc etc

HTH

Nick

- posted on October 22, 2003, 10:28 pm

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

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

Les

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

Les

- posted on October 22, 2003, 10:51 pm

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.

- posted on October 22, 2003, 10:56 pm

it means you were correct

- posted on October 22, 2003, 11:20 pm

as title (c;

Les

Les

- posted on October 22, 2003, 11:28 pm

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

P = Power in Watts V = Volts I = Electrical Current in Amps R = Electrical Resistance in Ohms SQR = Square Root

Sparks...

- posted on October 22, 2003, 10:52 pm

Volts * Amps = Watts

--

geoff

geoff

- posted on October 22, 2003, 11:33 pm

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.

--

*I didn't fight my way to the top of the food chain to be a vegetarian.

Dave Plowman snipped-for-privacy@argonet.co.uk London SW 12

*I didn't fight my way to the top of the food chain to be a vegetarian.

Dave Plowman snipped-for-privacy@argonet.co.uk London SW 12

Click to see the full signature.

- posted on October 23, 2003, 8:47 am

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.

- posted on October 23, 2003, 9:13 am

Christian McArdle wrote:

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

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

- posted on October 23, 2003, 9:12 am

in2minds wrote:

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

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

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