That makes lots of sense but what will happen when the 50 w bulb is off, they only load then will be the 0.1 ohm resistor. The power will be over 5 w and burn the resistor out.
|
Reading current with pic
That makes lots of sense but what will happen when the 50 w bulb is off, they only load then will be the 0.1 ohm resistor. The power will be over 5 w and burn the resistor out.
|
If the bulb is off then the circuit is broken. No current to flow.
Dave
Always wear safety glasses while programming.
|
Never mind. Was not thinking straight.
Last edited by rmteo; - 22nd November 2008 at 00:21.
|
In most cases, the best way to measure current is with a Hall-effect sensor. No voltage loss, no heat, perfect isolation. Check out the Allegro devices.
http://www.allegromicro.com/en/Produ...rentsensor.asp
Charles Linquist
|
Maybe a low ohmic resistor 0.01ohm 5W is suitable in the application.
The hall effect sensor normally is very expensive in my country.
Let say the total amp flows is 15A,:
Voltage dropped = 15 X 0.01 = 0.15V
Wattage dissipated = 15 X 0.15 = 2.25W
I have been using this with the volt meter to measure the voltage up to 22A, and it is working fine. The heat will not be a issue for the resistor. But it is a bit expensive.
|
You mean 15x15x0.15=33.75WWattage dissipated = 15 X 0.15 = 2.25W
Al.
All progress began with an idea
|
P = I*V = 15 x .15 = 2.25W
Members who have read this thread : 0You do not have permission to view the list of names.
Posting Permissions
Bookmarks