andrewsr@u2.rutgers.edu (Rich Andrews) (03/12/91)
Hello All!
I am considering purchasing a TrippLite Line Conditioner Stabilizer
and Brown Out Protector (whew, now there's a mouthfull!)
Q: How many watts do the following pieces of hardware require?
1. Amiga 3000 (w/ 18MB of on board memory)
(And a 200MB SCSI 2ond drive)
2. A1950 monitor
3. JX300 scanner
Q2: For the layman, what is the relationship between watts and amps?
Thanks in advance,
-Rich
--
// Rich | "If there is nothing wrong with me, then maybe there is
\X/ Andrews | something wrong with the universe." - Beverly daveh@cbmvax.commodore.com (Dave Haynie) (03/13/91)
In article <Mar.11.21.27.22.1991.3987@u2.rutgers.edu> andrewsr@u2.rutgers.edu (Rich Andrews) writes: > >Hello All! > >I am considering purchasing a TrippLite Line Conditioner Stabilizer >and Brown Out Protector (whew, now there's a mouthfull!) >Q: How many watts do the following pieces of hardware require? > 1. Amiga 3000 (w/ 18MB of on board memory) > (And a 200MB SCSI 2ond drive) The A3000 power supply is rated for a total of around 135 Watts. A fully loaded A3000 would be expected to draw 135 Watts or less, assuming it wasn't actually an overloaded A3000. An overloaded system might draw more power, or it might just fail. >Q2: For the layman, what is the relationship between watts and amps? It's as easy as pie. P=IE, Power = Current x Voltage, NM Watts = N Volts x M Amps. >-Rich -- Dave Haynie Commodore-Amiga (Amiga 3000) "The Crew That Never Rests" {uunet|pyramid|rutgers}!cbmvax!daveh PLINK: hazy BIX: hazy "What works for me might work for you" -Jimmy Buffett
ptoper@obelix (Andy Nagy) (03/13/91)
In article <Mar.11.21.27.22.1991.3987@u2.rutgers.edu>, andrewsr@u2.rutgers.edu (Rich Andrews) writes: > > Hello All! > > I am considering purchasing a TrippLite Line Conditioner Stabilizer > and Brown Out Protector (whew, now there's a mouthfull!) > > Q: How many watts do the following pieces of hardware require? > 1. Amiga 3000 (w/ 18MB of on board memory) > (And a 200MB SCSI 2ond drive) > 2. A1950 monitor > 3. JX300 scanner > > > Q2: For the layman, what is the relationship between watts and amps? > > Thanks in advance, > -Rich > -- > // Rich | "If there is nothing wrong with me, then maybe there is > \X/ Andrews | something wrong with the universe." - Beverly The simplest answer to your first answer is to lookup the maximum power ratings in your manuals and add them up. This will give you the max amount of power your system could safely handle, not necessarily the actual amount you are using. Answering your second question is easier: power = voltage * current where power is measured in watts (W), voltage in volts (V) and current in amperes (A), both voltage and current are expressed in RMS units (Root Mean Square). For example a 100 W light bulb in a 120 V circuit will have a current draw of current = power / voltage = 100 W / 120 V = 833 mA Because the power line is usually considered constant, the relation- ship between power and current is linear. This means that if one increases the other increases, the inverse is also true. Hope this helps. ------------------------------------------------------------------------------- Andy Nagy (ptoper@asterix.gaul.csd.uwo.ca) The University of Western Ontario, London, Canada "Dee do do do, dee da da da, thats all I want to say to you" -- The Police
kherron@ms.uky.edu (Kenneth Herron) (03/13/91)
In article <Mar.11.21.27.22.1991.3987@u2.rutgers.edu> andrewsr@u2.rutgers.edu (Rich Andrews) writes: >Q2: For the layman, what is the relationship between watts and amps? The way I always thought of it was: Imagine electrons are little trucks carrying the electricty. Amps is the number of trucks. Volts is the amount each truck is carrying. Watts is the total amount of electricity being transported. Ohms (resistance) is the traffic capacity of the highway they're using; in other words the number of trucks that can drive on the road at the same time. -- Kenneth Herron kherron@ms.uky.edu University of Kentucky (606) 257-2975 Department of Mathematics "Never trust gimmicky gadgets" -- the Doctor
ptoper@obelix (Andy Nagy) (03/14/91)
In article <kherron.668836173@s.ms.uky.edu>, kherron@ms.uky.edu (Kenneth Herron) writes: > > In article <Mar.11.21.27.22.1991.3987@u2.rutgers.edu> andrewsr@u2.rutgers.edu (Rich Andrews) writes: > > >Q2: For the layman, what is the relationship between watts and amps? > > The way I always thought of it was: Imagine electrons are little > trucks carrying the electricty. Amps is the number of trucks. Volts > is the amount each truck is carrying. Watts is the total amount of > electricity being transported. Ohms (resistance) is the traffic > capacity of the highway they're using; in other words the number of > trucks that can drive on the road at the same time. > -- > Kenneth Herron kherron@ms.uky.edu > University of Kentucky (606) 257-2975 > Department of Mathematics > "Never trust gimmicky gadgets" -- the Doctor How 'bout this water analogy: Quantity Water Electricity ------------------------------------------------------------------------------- Pressure pascals (Pa) volts (V) pounds per sq inch (psi) Quantity liters per second (l/s) amperes (A) per unit gallons per minute (gpm) time Resistance ??? ohms (Greek lowercase omega) Power watts (W) watts (W) British hourse power (Bhp) Note the ??? for water. I don't remember the proper units for this. (Probably something like Pa*s/l). By the way, coulombs (C), is the measure of electrical quantity not amperes, wattage is the measure of energy or the ability of a system to do work, and ohms gives an indication of energy loss. ------------------------------------------------------------------------------- Andy Nagy (ptoper@asterix.gaul.csd.uwo.ca) The University of Western Ontario, London, Canada "Dee do do do, dee da da da, thats all I want to say to you" -- The Police
navas@cory.Berkeley.EDU (David C. Navas) (03/15/91)
In article andrewsr@u2.rutgers.edu (Rich Andrews) writes: > >Hello All! Hi there. >Q2: For the layman, what is the relationship between watts and amps? Q1 I have no idea about. This one I'll attempt a layman's answer to (if my freezing fingers hold out). Imagine standing under a waterfall [a small one, as you couldn't "stand" under a large one :)]. Now, a certain volume of water is flowing around you over a certain period of time. That is a rough analogy to amperage. Specifically [correct if wrong, please], amperage is the flow of charge per unit time. Also, this same water is hitting you with a certain force. Make the waterfall "taller", and the water will start to hurt you (because it's falling faster) when it hits you. The volume of water hasn't changed, just the force that it has when it impacts. That's wattage. In the same way, force more water to flow over the waterfall (increase the amperage), and the waterfall wil also hurt more, and may eventually reduce ones posture :). Again, this is analogous to an increase of wattage. Voltage is, as you might suspect, the "height" of the waterfall. >Thanks in advance, Hope it helps... David Navas navas@cory.berkeley.edu "Oh, that's an Apple??? I though they just shot themselves in the head..." [Also try c186br@holden, c260-ay@ara and c184-ap@torus]