Anabolics
Search More Than 6,000,000 Posts
Results 1 to 12 of 12
  1. #1
    needmorestrength's Avatar
    needmorestrength is offline Anabolic Member
    Join Date
    May 2004
    Location
    Canada eh
    Posts
    7,079

    Exclamation Anyone who knows electrical formula's, please help

    Hey, under Ohm's law, subsection Power I am trying to figure out a question.. Any help would be greatly appreciated.

    "A 1/2 watt resistor has a resistance of 1000 ohms. What is hte maximum current it can handle?

  2. #2
    tonytone's Avatar
    tonytone is offline Registered User
    Join Date
    Apr 2004
    Location
    TEXAS
    Posts
    2,505
    V = I R.........sorry, that's all i remember....i hate physics

  3. #3
    Rob's Avatar
    Rob
    Rob is offline Canadian Legend
    Join Date
    Aug 2004
    Location
    ONTARIO
    Posts
    3,201
    I know the answer.

    But since you wanted to go to Queens.

    Do your own homework!

  4. #4
    symatech's Avatar
    symatech is offline Retired Moderator
    Join Date
    May 2002
    Location
    not where I want to be
    Posts
    7,546
    1/2 watt resistor? watts are J/s and resistors are measured in ohm meters. max current would be potential difference over resistance. I=V/R and R = p l/a which is the resistivity of the metal times the length over area. resistivity varies depending on what metal it is (provided it is ohmic). Is there anymore to the question? It's been awhile since I've done this stuff.

  5. #5
    symatech's Avatar
    symatech is offline Retired Moderator
    Join Date
    May 2002
    Location
    not where I want to be
    Posts
    7,546
    Ignore my last post I'm a little slow when I've been drinkin. so you know power is equal to the current squared times the resistance. essentailly P=I^2 R so I = sqrt(P/R) = sqrt(.5watt/1000 ohm) = .022 amps.

    this could be wrong I don't know like I said it's been long time.

    edit: I think i am wrong, just doesn't seem right. sorry man
    Last edited by symatech; 09-27-2005 at 09:54 PM.

  6. #6
    CRUISECONTROL's Avatar
    CRUISECONTROL is offline Post Whore Extraordinaire Cruising On Autopilot
    Join Date
    Jun 2005
    Location
    La La Land
    Posts
    5,413
    wrong one just a minute

  7. #7
    CRUISECONTROL's Avatar
    CRUISECONTROL is offline Post Whore Extraordinaire Cruising On Autopilot
    Join Date
    Jun 2005
    Location
    La La Land
    Posts
    5,413
    The voltage divider rule can be used with resistive, inductive, or capacitive circuit elements. It can also be used with AC or DC input sources. The equation for calculating the output voltage is different, however, depending on the type of circuit element. Following are the three general cases of two like elements connected in series:

    Resistive divider:
    The formula for determining the DC or AC output voltage of a resistive divider is:
    Vout = Vin*R2/(R1+R2)
    Example: In the following circuit, the output voltage would be: Vout = 9V*10K/(10K + 5K) = 6V

    Inductive divider:
    Inductive dividers can be used with AC input signals. A DC input voltage would split according to the relative resistances of the two inductors by using the resistive divider formula above. The formula for determining the AC output voltage of an inductive divider (provided the inductors are separate, i.e. not wound on the same core, and have no mutual inductance) is:
    Vout = Vin*L2/(L1+L2)
    Example: In the following circuit, the output voltage would be: Vout = 10VAC*50mH/(50mH + 100mH) = 3.33VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the inductors is not high at the frequency of operation (i.e. inductance not large enough), there will be a very large current drawn by the shunt element.


    Capacitive divider:





    Capacitive dividers can be used with AC input signals. A DC input voltage would not pass through the capacitors, so the DC case is not relevant. The formula for determining the AC output voltage of a capacitive divider is different from the resistive and inductive dividers, because the series element, C1 is in the numerator instead of the shunt element, as shown below:

    Vout = Vin*C1/(C1+C2)
    Example: In the following circuit, the output voltage would be: Vout = 10VAC*0.022uF/(0.022uF + 0.01uF) = 6.875VAC. Note that the output voltage is not dependent on the input frequency. However, if the reactance of the capacitors is not large at the frequency of interest (i.e. capacitance not large enough), the output current capability will be very low.





    --------------------------------------------------------------------------------

    Copyright © 1999, Randall Aiken. May not be reproduced in any form without written approval from Aiken Amplification.

    Revised 01/01/00

  8. #8
    CRUISECONTROL's Avatar
    CRUISECONTROL is offline Post Whore Extraordinaire Cruising On Autopilot
    Join Date
    Jun 2005
    Location
    La La Land
    Posts
    5,413
    Search: Lycos Angelfire Dating Search
    Share This Page Report Abuse Build a Site Browse Sites
    « Previous | Top 100 | Next »


    Electrical & Electronics, Ohm's Law, Formulas & Equations



    --------------------------------------------------------------------------------

    OHM’S LAW CALCULATOR
    VOLTS=Voltage, AMPS=Current, OHMS=Resistance, WATTS=Power
    Give me any TWO numeric values and I'll give you all FOUR. Press the Ohm's Law button after you have made your entries: VOLTS AMPS OHMS WATTS




    keywords = Ohms, Ohm’s Law, Volts, Amps, Current, Watts, Power, Calculator, Electricity, Electronics, Electrical, Equations, Formulas, Pi, Math, Henry, Bacon




    FORMULAS, EQUATIONS & LAWS


    Symbolic:

    E =VOLTS ~or~ (V = VOLTS)
    P =WATTS ~or~ (W = WATTS)
    R = OHMS ~or~ (R = RESISTANCE)
    I =AMPERES ~or~ (A = AMPERES)
    HP = HORSEPOWER
    PF = POWER FACTOR
    kW = KILOWATTS
    kWh = KILOWATT HOUR
    VA = VOLT-AMPERES
    kVA = KILOVOLT-AMPERES
    C = CAPACITANCE
    EFF = EFFICIENCY (expressed as a decimal)


    DIRECT CURRENT

    AMPS= WATTS÷VOLTS I = P ÷ E A = W ÷ V
    WATTS= VOLTS x AMPS P = E x I W = V x A
    VOLTS= WATTS ÷ AMPS E = P ÷ I V = W ÷ A
    HORSEPOWER= (V x A x EFF)÷746
    EFFICIENCY= (746 x HP)÷(V x A)
    AC SINGLE PHASE ~ 1ø

    AMPS= WATTS÷(VOLTS x PF) I=P÷(E x PF) A=W÷(V x PF)
    WATTS= VOLTS x AMPS x PF P=E x I x PF W=V x A x PF
    VOLTS= WATTS÷AMPS E=P÷I V=W÷A
    VOLT-AMPS= VOLTS x AMPS VA=E x I VA=V x A
    HORSEPOWER= (V x A x EFF x PF)÷746
    POWERFACTOR= INPUT WATTS÷(V x A)
    EFFICIENCY= (746 x HP)÷(V x A x PF)
    AC THREE PHASE ~ 3ø

    AMPS= WATTS÷(1.732 x VOLTS x PF) I = P÷(1.732 x E x PF)
    WATTS= 1.732 x VOLTS x AMPS x PF P = 1.732 x E x I x PF
    VOLTS= WATTS÷AMPS E=P÷I
    VOLT-AMPS= 1.732 x VOLTS x AMPS VA=1.732 x E x I
    HORSEPOWER= (1.732 x V x A x EFF x PF)÷746
    POWERFACTOR= INPUT WATTS÷(1.732 x V x A)
    EFFICIENCY= (746 x HP)÷(1.732 x V x A x PF)

    Main Page & Introduction
    Deca -Scientific Calculator, Conversion Charts, Industrial Math
    Definitions & Terminology
    Awards This Website Has Received


    --------------------------------------------------------------------------------



    --------------------------------------------------------------------------------

    Please sign my Guestbook. Thank you!
    Sign My Guestbook View My Guestbook




    --------------------------------------------------------------------------------

    --------------------------------------------------------------------------------



    --------------------------------------------------------------------------------

    Every effort is taken in order to maintain accuracy in the information contained within this web site. However, the author of this site shall not be held liable for the use of the information or of its consequences. Although all of the information is believed to be accurate you are encouraged to consult other sources for additional information or clarification. If you are a student your instructor may have very specific terminology that is required to be used for your particular field of study and therefore you should consult your textbooks and study notes for that information. It is my hope that this site will be both educational and beneficial to those of you who use it. I welcome your comments and suggestions in regards to this site. Thanks for visiting!

    Henry J. Bacon

    --------------------------------------------------------------------------------


    Email: voltampkva@aol.com

  9. #9
    Tock's Avatar
    Tock is offline Anabolic Member
    Join Date
    May 2002
    Location
    Fort Worth
    Posts
    7,103
    Quote Originally Posted by needmorestrength
    Hey, under Ohm's law, subsection Power I am trying to figure out a question.. Any help would be greatly appreciated.

    "A 1/2 watt resistor has a resistance of 1000 ohms. What is hte maximum current it can handle?

    P=IE (Power, in Watts, = current times the voltage)
    E=IR (Voltage = current time the resistance)

    So, substitute (IR) for the E in P=IE, and you get

    P=I(IR)

    If P=.5 watts
    If R=1000

    then .5=I(I*1000)

    divide both sides by 1000, and you get

    .5
    ---- = I squared
    1000

    or,

    1
    ---- = I squared
    2000

    or

    .0005 = I squared

    take the square root of both sides, and you get

    .022 = I

    So, the current would be .022 amps, or 22 milliamps.


    What do I win? It's been 25 years since I screwed with this stuff . . .

    If you have .022 amps flowing through a 1000 ohm resistor, what would the voltage be across it?

    Well, E=IR,
    so you'd multiply .022 by 1000, and you'd get 22 volts.

    -Tock

  10. #10
    symatech's Avatar
    symatech is offline Retired Moderator
    Join Date
    May 2002
    Location
    not where I want to be
    Posts
    7,546
    hey that's what I gots!! except he's not looking for voltage he's looking for current. .022 amps. I hope you're right b/c that would mean I was right and because of whiskey and beer I like being right.

    *sigh* i think it's time for bed, im almost drunk

  11. #11
    symatech's Avatar
    symatech is offline Retired Moderator
    Join Date
    May 2002
    Location
    not where I want to be
    Posts
    7,546
    no no NO!!! scratch that!.. I am drunk

  12. #12
    almostgone's Avatar
    almostgone is online now AR-Elite Hall of Famer
    Join Date
    Jun 2004
    Location
    the lower carolina
    Posts
    14,217
    ....As Tock and Sym stated P=I(IR)..
    so, I=√P/R

    Current = √.5w/1kΩ =.o223A= 22.3mA
    There are 3 loves in my life: my wife, my English mastiffs, and my weightlifting....Man, my wife gets really pissed when I get the 3 confused...
    A minimum of 100 posts and 45 days membership required for source checks. Source checks are performed at my discretion.

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •