SteveB
Active member
- Joined
- Mar 9, 2009
- Messages
- 26
- Fluid Motion Model
- C-24 C SE
- Vessel Name
- PugTug
Hi everyone! I have a couple of questions about measuring/planning for 12 volt electricity consumption on my boat. In all cases, assume the engine is off and electricity is supplied only by the house battery (75 amp-hours)
Suppose I have a 12 volt lamp that is rated at 6 watts. Dividing WATTS by VOLTAGE (6 divided by 12), I get a result of 0.5. Is is correct that this lamp will consume .5 (one-half of one) amp-hours when running off of my house battery?
(I think that's correct)
Next, I have a 300 watt inverter connected to my boat's 12v accessory plug.
These are hypothetical questions:
Looking at a standard 110 volt AC-to-DC adapter I have here for a radio, it says "Input 110 volts 60hz : Output 8V --- 2A". How much current will this draw from the house battery when I'm running it off of the inverter? Is it 2 amp-hours or do I have to use another way to compute the current draw?
Last question. I have a regular household AC desk lamp with a 40 watt bulb. Which voltage do I use to compute the draw this will place on my house battery when used through an inverter?
Is it 40 divided by 12 volts (house battery power), for 3.3 amp-hours
-or-
Is it 40 divided by 110 volts (A/C from the inverter) for .36 amp-hours? This seems unlikely.
Or is it something completely different?
Thanks for any help
-Steve
Suppose I have a 12 volt lamp that is rated at 6 watts. Dividing WATTS by VOLTAGE (6 divided by 12), I get a result of 0.5. Is is correct that this lamp will consume .5 (one-half of one) amp-hours when running off of my house battery?
(I think that's correct)
Next, I have a 300 watt inverter connected to my boat's 12v accessory plug.
These are hypothetical questions:
Looking at a standard 110 volt AC-to-DC adapter I have here for a radio, it says "Input 110 volts 60hz : Output 8V --- 2A". How much current will this draw from the house battery when I'm running it off of the inverter? Is it 2 amp-hours or do I have to use another way to compute the current draw?
Last question. I have a regular household AC desk lamp with a 40 watt bulb. Which voltage do I use to compute the draw this will place on my house battery when used through an inverter?
Is it 40 divided by 12 volts (house battery power), for 3.3 amp-hours
-or-
Is it 40 divided by 110 volts (A/C from the inverter) for .36 amp-hours? This seems unlikely.
Or is it something completely different?
Thanks for any help
-Steve