ohms law help

Thread Starter

beckyj43

Joined May 10, 2005
1
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?
 

Erin G.

Joined Mar 3, 2005
167
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]​

Resistance = Voltage / Current

R = E/I

R = 150 / .05

R = ???
 

Brandon

Joined Dec 14, 2004
306
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=IR

The Voltage drop across 2 points of measure = The current flowing between the 2 points times the resistance between those 2 points.

plug n solve.
 

n9xv

Joined Jan 18, 2005
329
Resistance = Voltage / Current

Voltage = Current X Resistance

Current = Voltage / Resistance


With any two known quantities, you can find the remaining third quantity.
 

legac

Joined May 4, 2005
54
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]​
Hi
I think the term INTERNAL RESISTANCE of the DEVICE is more precise. For a circuit we use the term IMPEDANCE.
Cheers
Legac
 

pebe

Joined Oct 11, 2004
626
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]​
No, the term *resistance* of the device is correct.

*Impedance* is the resistance of a device or circuit to the flow of AC.
 

Brandon

Joined Dec 14, 2004
306
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]​


Heh, got to get in on this.

Actually, isn't impedance the combined Resistance (real) + Reactance (img) of a component?
 

pebe

Joined Oct 11, 2004
626
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]​
Yes. I was going to say "Impedance is the *effective* resistance..." implying it could loosely be called that. But changed my mind.

Wish I had left it in now :)
 

susi

Joined Jun 4, 2004
31
^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]
 

pebe

Joined Oct 11, 2004
626
Originally 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]​
5mA is 0.005A
 

zhi_yi

Joined May 15, 2005
9
i hope i can explain something to you :) ohm law --> V = I.R
V = Voltage, I = Current, and R = Resistance.
if the resistance is high, and the voltage is constant. so the current will be small, and if the resistance is small, voltage constant, the current will be high.
for example, there is some water inside a bucket, so, the sum of water is current, and the pressure is voltage, and if there is many holes under the bucket, it is the resistance, if the holes present is big, the resistance against the current will be small, so amount of water will be flow out from the bottom of the bucket, so the current on the circuit will be high, and if the holes is small, the resistance against the current will be high, so I = V/R <-- if the resistance is high, the current on that circuit will be low, for example, a battery present 1.5 volt, if we use it to a load for t times, the current will be wasted (the water inside the bucket will be flow out and the bucket will be empty), the voltage and current have linearity, if the current increase, the voltage will be increased too, V=I.R, the current will flow if the circuit is closed loop, if it's open loop, no current will flow. the current always flow to the lower path of the resistance.

sorry, my english is broken, hope you can understand what i said >_<

:)
 
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