# ohms law help

Discussion in 'General Electronics Chat' started by beckyj43, May 10, 2005.

1. ### beckyj43 Thread Starter New Member

May 10, 2005
1
0
Assume a device draws 50 mA of current and a 150 vdc across it . what is the resistance?

2. ### Erin G. Senior Member

Mar 3, 2005
167
1

Resistance = Voltage / Current

R = E/I

R = 150 / .05

R = ???

3. ### Brandon Senior Member

Dec 14, 2004
306
0
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.

4. ### n9xv Senior Member

Jan 18, 2005
329
1
Resistance = Voltage / Current

Voltage = Current X Resistance

Current = Voltage / Resistance

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

5. ### legac Well-Known Member

May 4, 2005
54
0
Hi
I think the term INTERNAL RESISTANCE of the DEVICE is more precise. For a circuit we use the term IMPEDANCE.
Cheers
Legac

6. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
No, the term *resistance* of the device is correct.

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

7. ### Brandon Senior Member

Dec 14, 2004
306
0

Heh, got to get in on this.

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

8. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
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

9. ### susi Active Member

Jun 4, 2004
31
0
^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

[I hope I solved it right..lol..check the conversion of mA into Amp]

10. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
5mA is 0.005A

11. ### susi Active Member

Jun 4, 2004
31
0
^lol yeh sorry its milli amp :huh:

12. ### zhi_yi New Member

May 15, 2005
9
0
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 >_<

May 15, 2005
3
0