two resistors are connected in series with a 15 volt power supply.  which is FALSE?
813641
using KCL (current law), A is corrent.\nusing KVL (voltage law), B is correct\nC doesn't make any sense, and you can support that with the definition of the KCL.
736596
It depends on the current.  You want to drop 6 V across the resistor to get the voltage to the load to 6 V.  The voltage drop across the resistor is V = R*I.  The resistor value you need is R = V/I.  Once you know the resistance, you need to determine its power rating.  The power dissipated in the resistor is P = V*I, where V is the voltage across it and I is the current through it.  Or, you can use P = (I^2)*R.  You should choose a power rating at least 1.5 times the power you've calculated so it doesn't run too hot.
679488
simple use two resistors, 6 and 3k ohm , in series and take the volatage across the 3 ohm resistor, it will be 3V.....\n\nuse the laws,\n\nif you have a voltage source, and two resistors p and q in series with it, \n\nthe voltage across p is given by  Vx(p/(p+q)), here u know voltage across p(Vp)= 3V, or 4.7 V or whatever....\n\nso, Vp = V x p/(p+q)....\n\nu know the value Vp, and V so substitute them......\n\n3 = 9 x p/(p+q)\n\nso p + q = 3p, \n\nso "2p = q", take two resistors ( in kilo ohms, coz u cant use just ohms, coz they are very large) such that one is double the value of other.......\n\nand connect the circuit across the 'p' resistor......\n\nwith 4.7, u get,\n\n4.7 = 9 x p/(p+q).....solve ...and choose resistors accordingly
980170
You aren't clear about whether the switch is on in both measurements.  If it is, then the voltage will be the same.  If not, then the voltage will be different because of the voltage drop across the internal resistance of the battery.
827050
Yupp... but unfortunately is a hardcopy on my desk. Perhaps you may straight to tell us what\n\n1. You are designing from HT to Transformer 11kV / 415V\n2. You are designinig from Transformer to LV ??\n3. What is the kVA of the transfomer?
207430
I suspect the protection devices are sharing the load or branch circuit which they are connected.\n\nThere is probably a ground fault from one of the devices or loads connected to the circuit. Disconnect each load to verify that the circuit is ok, then proceed by adding back in the loads. \n\nAs you check out each load, verify that there is no obvious wiring short to ground. If you can, use a Ohm meter to verify the load is not shorted before you connect the load back into the circuit.
597402
let the resistances be r1 and r2.\nwe know the equation for power'p' is p=V*I\nV-Voltage applied,\nI-current in the circuit,\nby ohms law,\nV=IR\nsubstitute the value of V in power,\nwe get P=I^2*R  \nif the resistances  are in series then the value of R is,\nR=r1+r2.\nso the power dissipated when the resistance in series is,\np=I^2*(r1+r2),\nsimillarly power dissipated when the resistances are in parallel is\np=I^2*((r1*r2)/(r1+r2))
261103
10 amps if the voltage difference is 120 volt as usual
821020
Each should be marked with a Vector Group Reference.\nThis gives phase angle and lead or lag angles. \nThese should be matched before connecting them.\nAfter connection the load should be introduced to the board in synchronism with the supply.
179861
If the breaker is a two pole breaker which it should be it  is 240 volt, the voltage between the two red  should be 240 one of the blks are a ground not marked (green) and the other a neutral (white) not marked
323912
acrtoss the power supply
698161
No, the voltages are different
143811
Aboslutely, provided the fans are for 12V service.  Leave the secondary center tap unconnected, and put each end of the secondary to the AC input of the rectifier.  The + goes to the red lead of the fan; the - goes to the black lead (traditionally speaking of course.) Now the supply will be very ripply, with a 120 Hz waveform which could be more than a volt.  If you put a 470uF 35V capacitor across the plus and minus of the rectifier, you will smooth out the ripple, but increase the voltage above 12 volts.  Make sure the fans can take the increased voltage.  Transformers are rated at the RMS output at the 3 amps in your case.  The peak voltage will be around 1.4*12.6, but in practice I have never seen it go above 15 V with a load.  Therefore at 1 amp, if you put a cap on the supply, it may produce excessive voltage.  Check it with a voltmeter-- you may luck out.\n\nI have been in a similar situation.  Get a 5 watt resistor at maybe 1 to 3 ohms.  Insert it in series between the capactitor positive and the fans wired in parallel.  When running, use a voltmeter on the fan circuit.  Increase the resistor till you get 12V or thereabouts.  The resistor will get hot, so leave the leads long and connect them to a terminal strip or block if possible, and mount it in the airstream.\n\nCheck out the LM7812, however, at 1 Amp.  Google it, or check out jameco.com for the part.  Put a 470uF at 35 V at each end of the regulator to keep things very smooth.  Your fan will run perfectly with the voltage regulator.  Be sure to mount the tab on a heatsink to keep it cool since you are running it at full capacity.
391913
Do you mean power supply or cable?\n\nI need more info
1079800
in one case you have half an ohm, in the other you have over 500 ohms.  It seems you may be reading one of the meters incorrectly.  One of the meters could be set to auto, while the other is not.
632448
Yes. Definitely.\n\nA/C adaptor is an Alternating Current.\nPower supplies are custom units of power management, it might be D/C (Direct Current), so it's really important to pay attention to them, or you could fry your electronic.
972205
If the resistors are in parallel, then the voltage is the same across all of them - that's pretty much the definition of parallel.  If they are the only circuit elements, then that voltage will be the voltage of the circuit.  Since you didn't tell us what that voltage is, there's no way for us to figure the rest of the problem.  If you know the voltage, though, then Ohm's Law will give you the currents.