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
    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?
     
  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

    R=150/0.05=3000 Ω ------Answer/

    [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 >_<

    :)
     
  13. sellb

    New Member

    May 15, 2005
    3
    0
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