Ohm's law is not applicable??
108296
A radio valve is a tube, which is what was used before there were transitors.\n\nMy feeling is that all these devices are nonlinear devices, so V=IR does not hold over a very wide range of parameters. So depending on what you choose for V and I, NONE of these obey ohm's law. However, they might obey ohm's law over some small range of values of V and I.\n\n------------------\nAddendum:\n\nIs there something I am not understanding here? Ohm's Law is a linear law V=I*R, where R is a constant not depending on the applied voltage. Any material or device not satisfying this relation is called nonohmic. Transistors are nonohmic. Diodes are nonohmic. Vacuum tubes (i.e., valves) are nonohmic. Of course, even for ohmic devices, Ohm's law only holds for a certain range of V and I, since outside this range, one witnesses nonlinear behavior. \n\nIf R depends on voltage, so R=R(V), then the relation continues to hold, but it is no longer linear.  \n\nThe references I have attached include multiple examples of diodes, transistors and tubes being described as nonohmic devices. \n\nONLY over certain  values of V and I will any of these devices approximate an ohmic device. Other than that, the V-I relationship will be nonlinear.
363388
The total resistance is 12 + 8 + 6 = 26 ohms. In series the total resistance is the sum of individual resistors. Apply ohm's law V=IR, I = 20/26 Amperes. The voltage drop across each resistor is found by multiplying the current with the resistance of each resistor.
140428
x\n\nthe answer below is just for voltage - see my graph above - it is quite accurate. here it is again, larger: X
820466
The voltage across each resistor will be 12 volts as each is effectively directly connected to the battery.\nIt is the current flow through the resistors which will be different.\nVolts/Ohms =Amps - that is\n12/7 amps or 12/5 amps now get your calculator out!\nRoyS
65822
You'll have to describe the circuit a little better \n\nAre you adding either the 3.00 ohm or 2.50 ohm (but not both) in parallel with the 4.00 and 3.75 ohm, and need to know the total resistance with either the 3.00 or 2.50 ohm resistor connected?\n\nOr are BOTH the 3.00 AND the 2.5 ohm being placed in parallel with the 4.00 and 3.75 ohm?\n\nAnd without knowing the applied voltage, your first question (current through a 3.00, 4.00, 3.75, 2.50 ohm resistor) can't be answered.
966971
ok resistivity formula is:\n\nresistivity = [resistance x length]/ cross sectional area..\n\nyou didnt give me enough info to work with\n\nhope it helped
946535
The standard for voltage drop in a circuit is 2%.  Remember that the current must go back to the source so if you are running 3 Phase you must multiply 750ft by 1.732 to get the proper length.\n  So, using ohms law -voltage drop=current (in amps) X resistance (in ohms).  #000 gage copper wire has .00073 ohms/ ft.  Using 200 amps in this formula, If you work out the math you will find that a 30 amp service is about all you can safely install at that distance.  Even at 30 amps you are running the risk of over heating and shortening the life of your equipment. \nexample:  750ft X 1.732 X .00073 ohms per ft = .9 ohms total\n\nSo:  .9 ohms X 30 amps = 27 volts dropped\n\nThus:  if your service is 240 volts your voltage drop will be a little over 10 % or 213 volts at the point of use..  Some equipment may endure this.  I would not chance it.
756481
Normally this is a "Restive" not voltage question/answer, and the proper neutral to ground = 2 Megg ohms (min.) this being incorporated in a residential application.\nHigher voltage in let us say a 3-phase utilization this # is on the low side. (assuming a.c. source. d.c. is a completely different animal!
304043
Because algebra skills are very lacking in most people. It's just to make sure that they know what the rearranged formula's look like. But you are right, those 2 laws are probably all you need to remember. P=IE is always true, but remember than Ohm's Law is only an approximation for materials that have a linear E vs R graph, such as metals at room temperature. Its not true for semiconductors and gas discharges.
70878
Ohm's Law (named after Georg Ohm, a German physicist) states that the electric current I flowing through a given conducting material (conductor means current can flow through it) is equal to the applied voltage V divided by the resistance R:  I=V/R. Ohm's Law deals with the relationship between voltage and current in an ideal conductor.  The resistance R is sometimes called a "lumped element abstraction."\n\nFor general application to alternating-current circuits where inductances and capacitances as well as resistances may be present, the law must be amended to I=V/Z, where Z is impedance.\n\nOhm's Law is a macro-level, aggregate rule as to how current, voltage, and resistance are related.  But it is not a true "law" of nature in that it really is an approximation of behavior that holds true in most circumstances.\n\nOhm's Law may not hold due to:  1) the kind of conducting material the current is flowing through (some conductors do not follow Ohm's Law)  2) how much voltage is applied to the conducting material; as voltage is increased or decreased to extremes, then the linear Ohm's Law relationship breaks down   3) how small the wavelength of a periodically time-varying voltage applied to the conductor is with respect to the conducting circuit\n\nThe deeper reasons behind why Ohm's Law does describe macro behavior in "ohmic" conducting materials requires diving into a little bit of solid state physics and understanding Maxwell's equations.
434630
dont forget the formula I= R/U\nsort them and u will get the answer\n\n Current: 625mA (milli amperes)\nResistance: 19.2 ohms
700354
Since you have no contact info, I have no other choice...\n\nIn reference to http://answers.yahoo.com/question/index?qid=20061115060801AAEHWQA&r=w&pa=FZptHWf.BGRX3OFMhzdTUCYEW59bygAIQoCa4DY8zIlrypTkPQ--&paid=answered#NbUvWju8VzmdRHfXRHQU\n\nYou suggested to the asker to wire his sub to 1 ohm when it clearly states in his question "500 watts X 1 @ 4 ohms". You do realize you told him to blow his amp up by wiring it that way.
1028768
Simply put, it's not a conductor. \n\nhttp://jchemed.chem.wisc.edu/JCEsoft/CCA/CCA1/MVHTM/R0540/S567.HTM
935337
The resistance is increasing at .1785 ohms!!!\n\nJust use the ohm's law\n\n10/56=.1785 that's the resistance\n\nthen multiply the resistance and the current=volts\n\n.1785 x 56 =9.99999volts\n\nThis is the result.\n\nHope this help!!!
248482
Ohm's Law is always obeyed.\nIf a conductor is non-linear it has a coefficient of resistance - thermal, pressure, tensile, even voltage-dependent.\nMeasure the current thru it and the voltage across it and you have the instantaneous resistance at the time of measurement.\nThe law stands, but the result may not be static.
211231
It depends on both the type of current and the circuit.  For example, AC and DC currents passing through a circuit with reactive elements (capacitors and inductors) will much diffreently from one another.... Ohm's law always applies however...
368244
V (Voltage) = I (Current) x R (Resistance)\n\nOr (although you probably don't need it):\n\nJ (current density) = sigma (conductivity) x E (electric field)\n\nI always use wikipedia to look up stuff like this, it's very helpful for all kinds of info.