2 electrical currents merging into one.?
157293
well the voltage is dc,so +ve and -ve potential comes in picture.the potential of c will be differnce of potential at the two end.\nso if both +ve ends are connected at the c ends the potential of c =10 v.\nif +20v and 0v(-ve 10 v wire) is connected pot c=20v\nif 0 and +10v is conncted pot c=10v\ndir of current is from +ve to -ve\nhere in 1st case current varies from 20 to 10.\nthis is only in case of transmission and distribution. so u plz explain if it is related to the same and is externally connected. if not then current will be vector sum as in the 3 cases i said. u may refer to power system by j. b. gupta\nthis is engg book for 5th sem eee stud.
962570
It will flow in either direction if the voltage is above 0.7 volts
12356
Sounds like static electricty, which tends to develop in windy dry conditions.  Check it out.
576166
the electrical circuit?
134384
Fuses blow because the current flowing through them is too high.  It is current, not voltage, that usually causes heat.  Current is a measure of how much electricity is flowing through the wire.  The amount of current flowing through a circuit (I) depends on the circuit's voltage (V) and resistance (R).  I=V/R.\n\nWhen you connect two devices in series (one after the other), their resistances add.  If you have a 100-Ohm and a 200-Ohm device, the equivalent resistance is 300 Ohms.  But, if you connect the same devices in parallel (side-by-side), you have to invert their resistances, add them, and invert the result.  That is the resistance of both together.\n\nLet's say the voltage applied is 12V.  With a 100-Ohm device, the current through it will be 12/100=0.12 Amps.  If you instead used the 200-Ohm device, the current through it would be 12/200=0.06 Amps.  If you connected both devices in series, the resistance would be 300 Ohms; and the current would be 12/300=.04 Amps.  Now, connect the devices in parallel!  Invert the resistances: 100 becomes .01 and 200 becomes .005.  Add them to get .015, and finally invert the result.  This gives you 66.67 Ohms.  Now, stick that into the I=V/R equation.  The current is now 12/66.67=.18 Amps!\n\nSo, connecting the devices in series RAISES the total resistance, but connecting them in parallel LOWERS their total resistance. And lower resistance means higher current.
272565
You are electrically charged...it's like a mini-lighting storm... the touching of electrons causes a little clash.\nIt's an energy thing.\n\nHowstuffworks.com has a science section. It can explain electricity.
220810
No equations but this may help.... when an electrical signal is travelling along, for example a piece of copper wire, there will be a field generated aroung the cable, similar to that of a magnet (remember at school we were taught about magnetic fields  with a magnet piece of paper and iron filings)\nWhen a second piece of wire is brought in close proximity to the 1st piece then it will cut the field , and a signal will be induced into the second piece of wire.This is called mutual induction. If a signal is now added to the second wire then the induced signal will either add or detract from it dependant on the phase at the time.... this will happen, also within the 1st piece of wire, as a field is now being generated by a signal supplied to the second wire.\nNow, by twisting the pairs together the field generated by the signal in the wire will be changing direction along its length (positive then negative then positive and so on)thus cancelling any induced signals (could be crosstalk in a speach circuit) out
1086143
It's a law of electrical flow along a circuit.
469104
The algebraic sum of the currents entering a node is equal to zero. Where current entering is positive and exiting is negative (or vice versa). In other words, what comes in must come out.\nNote: this is only true for DC signals.
904615
You sure do need help....\nWhen you were cutting the LED off, you created a short circuit, first with the wire cutter (or whatever you were using), and then you most likely allowed the 2 lose wires to touch each other (or both touch a common metallic item).\nYour battery is now kaput !  They don't like short circuits, you know....
744305
Stray voltages/currents between grounded metal parts is caused by a difference in resistivity. Usually, it is because the ground rod has a greater resistivity than the metal parts connected to the water/drainage/gas system.\nAgain as has been said on other occasions in this forum, Ohm's Law comes into play..... Current is equal to voltage divided by resistance. So if there is a differing resistance between the two systems AND there is a low resistance path betweeen the electrical wiring and earth on an appliance, current will flow (continuously) to the frame of the appliance and to ground. So if current is flowing to ground and the appliance ground has a higher resistance to earth than the ground of the adjacent metal parts there will be a difference in potential between the two. This is often felt as a tingling sensation --- The reason why it doesn't trip the breaker is that the current flow is lower than the trip setting of the ELCB (earth leakage circuit breaker). It does not have anything to do with "Back EMF" back EMF is the voltage generated in a dc motor that limits the amount of current that can pass through the armature windings.......\n\nIf you have this problem, GET AN AUTHORIZED ELECTRICIAN TO INVESTIGATE AND CORRECT THE PROBLEM IMMEDIATELY. It certainly is a dangerous situation that can cause death.\n\nSource: Electrical Engineer (ret) with 45+ yrs experience.
620886
(m)\n\ngoto the following website for your question with neat diagrams\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html
944281
+ve charge on one, -ve on the other....static electricity
6099
I may be living in the past but I see transistors every day.  The 2-diode analogy is pretty close to the mark.  They meet at the base, pointing in for PNP and out for NPN.  Check them that way with your multimeter set on diode setting.  If the reading is inconclusive you may have to de-solder them from the circuit.
269821
Depending on the (source + cable) impedance, approx .01 ohms and figuring that this is a 3phase line then:\n\nFault Current is approx 365,000 amps in less than a second.\n\nCheck out the following; Type in Fault Current in search.
1100598
voltage = current x resistance\nE = I x R  It IS Ohm's law. An easy way to remember the formula and it's configurations is to draw a circle and divide it in half horizontally. Above the line write "E". In the lower half divide the semicircle vertically and write "I" on the left, "R" on the right. It should look something like this:\n                                                     E\n                                                   I    R\nNow just cover up the one you need. and either divide, or multiply.
2647
The PD will decrease as the current passing through it will drop due to current division