Mosvalve 500 - Power supply???
Moderator: Dave Mudgett
- Dag Wolf
- Posts: 995
- Joined: 15 Mar 1999 1:01 am
- Location: Bergen, Norway
Mosvalve 500 - Power supply???
Can someone tell me how much watts a mosvalve 500 consume from the power outlet? I need to know how many watts my new power supply will have to be. I`m changing from 110volts to 220volts.
Thanks,Dag
Thanks,Dag
- Jack Stoner
- Posts: 22136
- Joined: 3 Dec 1999 1:01 am
- Location: Kansas City, MO
-
- Posts: 708
- Joined: 3 Dec 1999 1:01 am
- Location: Santa Maria, CA.,
Dag, at 220 VAC, producing 1400 watts would require 6.36 Amperes maximum load on the mains circuit.
This "maximum power consumed" condition would only exist with the amp at full tilt, on a sine-wave test signal, output unclipped.
I'll bet that would be LOUD!<FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Rich Paton on 22 May 2000 at 05:29 PM.]</p></FONT><FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Rich Paton on 22 May 2000 at 09:46 PM.]</p></FONT>
This "maximum power consumed" condition would only exist with the amp at full tilt, on a sine-wave test signal, output unclipped.
I'll bet that would be LOUD!<FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Rich Paton on 22 May 2000 at 05:29 PM.]</p></FONT><FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Rich Paton on 22 May 2000 at 09:46 PM.]</p></FONT>
-
- Posts: 708
- Joined: 3 Dec 1999 1:01 am
- Location: Santa Maria, CA.,
Dag, the consumed power (actually called Volt-Amperes because of the transformer's inductive qualities) is a function of the voltage (E) multiplied by the current (I).
BTW, watts of electrical exist power only exist in a restive load such as an incandescent lamp or resistive heating element.
Anyway, back to the 220 Volt issue,
we'll call it watts for now...
1400 "Watts" equals 220 volts X 6.36 amps,
by using one of the formulas:
P = I x E (power equals Current "I" times Volts "E",
P = I squared X R (power equals the current squared times the load resistance "R"),
or P = E squared / R (power equals the voltage "E" squared, divided by the load resistance "R"
Since we know the power and voltage, use
1400 "watts" divided by 220 Volts. The answer would be 1400/220 = 6.36 Amps.
BTW, watts of electrical exist power only exist in a restive load such as an incandescent lamp or resistive heating element.
Anyway, back to the 220 Volt issue,
we'll call it watts for now...
1400 "Watts" equals 220 volts X 6.36 amps,
by using one of the formulas:
P = I x E (power equals Current "I" times Volts "E",
P = I squared X R (power equals the current squared times the load resistance "R"),
or P = E squared / R (power equals the voltage "E" squared, divided by the load resistance "R"
Since we know the power and voltage, use
1400 "watts" divided by 220 Volts. The answer would be 1400/220 = 6.36 Amps.