Originally posted by beckyj43@May 10 2005, 02:45 PM
could someone please help with this question?
Assume a device draws 50 mA of current and a 150 vdc across it . what is the resistance?
[post=7594]Quoted post[/post]
V=IROriginally posted by beckyj43@May 10 2005, 02:45 PM
could someone please help with this question?
Assume a device draws 50 mA of current and a 150 vdc across it . what is the resistance?
[post=7594]Quoted post[/post]
HiOriginally posted by beckyj43@May 10 2005, 02:45 PM
could someone please help with this question?
Assume a device draws 50 mA of current and a 150 vdc across it . what is the resistance?
[post=7594]Quoted post[/post]
No, the term *resistance* of the device is correct.Originally posted by legac@May 11 2005, 04:53 AM
Hi
I think the term INTERNAL RESISTANCE of the DEVICE is more precise. For a circuit we use the term IMPEDANCE.
Cheers
Legac
[post=7610]Quoted post[/post]
Originally posted by pebe@May 11 2005, 01:01 AM
No, the term *resistance* of the device is correct.
*Impedance* is the resistance of a device or circuit to the flow of AC.
[post=7612]Quoted post[/post]
Yes. I was going to say "Impedance is the *effective* resistance..." implying it could loosely be called that. But changed my mind.Originally posted by Brandon@May 11 2005, 05:49 PM
Heh, got to get in on this.
Actually, isn't impedance the combined Resistance (real) + Reactance (img) of a component?
[post=7628]Quoted post[/post]
5mA is 0.005AOriginally posted by susi@May 12 2005, 08:08 PM
^man its so easy...just have tofind the resistance by Ohm's Law
Data:
Current=I=5mA => 0.05Amp
Voltage=V=150vdc
Resistance=R=??
Solution:
By using Ohm's Law:
V=IR
R=V/I
R=150/0.05=3000 Ω ------Answer/
[I hope I solved it right..lol..check the conversion of mA into Amp]
[post=7651]Quoted post[/post]