If 6V is impressed across the above circuit and the voltage across the first lamp is 2V, what is the voltage across the second lamp?\nAnswer in units of V.
121840
What circuit?!? Is it a series circuit?!? Is it a parallel circuit?!? Is it a series-parallel circuit!?!
62240
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.
102791
1.Current in the 385.451 V circuit is bigger.\n\nGod is spelled B-I-G B-A-N-G\n\nPBrax@netscape.com
286098
Voltage drops for multiple lighting systems are usually handled by using 277 volt AC. \n\nThe multicore cable is something I have no working knowledge on.  Sorry better to say so than to BS you.
547393
Energy is measured in watts, so I assume you mean a voltage source of 200v.\n\n(1)\nThe voltage source of 200v applied to the light bulb (laminar) will be 200v.  Since the light bulb is designed for 130v, the supply voltage will likely blow the bulb.\n\n(2)\nHere you have two possibilities, depending on your circuit.\n(a) If you have a parallel circuit, then each light bulb will have 200v across it, and again the bulbs will likely blow because the voltage is to high.\n(b) If you have a series circuit, the voltage across the light bulbs will be subject to their resistance.  Assuming the resistance is the same, then the voltage will be the same, that is, 100v each.\nThe voltage across each light bulb will be equally proportional to it's resistance ratio.
100111
Hi! You haven't mentioned which of the resistances are in series or parallel. Too bad we can't provide images in Yahoo! Answers.\n:(\n\nDo this: Remove V2 and V3 from the circuit, and replace them by a simple wire. Then find the equivalent resistance of the circuit taking V1 as the battery source. Then find the voltage drops across each of the resistances R1, R2, R3, R4 (you can do this by finding current through them using simple Ohm's Law and then multiplying by the R value).\n\nTAKE CARE that you note the DIRECTIONS of the voltage drops. The direction of voltage drop across R is the direction of current through R. So you must see if its positive or negative.\n\nOnce you have done this, repeat the procedure by removing V1, V3 and taking only V2 in circuit. This time, again specifically note the direction (+ or -). Then do it with V3.\n\nFinally, add the drops across each R, and this is when you use the signs (directions) of the voltage drops. For instance (just an EXAMPLE. You will get diff. values), if you get VR1 with V1 battery as -6V, with V2 as 4V, and with V3 as -2V, then VR1 will be:\n\nVR1 = (-6)+(4)+(-2) volts = -4 volts\n\nYou can use the same procedure to compute IR1: find current each time with V1, V2, V3 resp.\n\nHope this helps!
76521
V = IR\nI = V/R\nR = V/I\n\nV = voltage    / V (or E)\nI = current    / A\nR = resistance / ohm\n\nOhm's Law = The voltage across a circuit is directly proportional to the current and the resistance.
140804
actualy, i think they use the same voltage, but have a transformer on the batery to up the voltage to the needed amount. are you looking for the bulb, or a batery powered fixture. you use a standard bulb, you can find it anywhere florescant bulbs are sold. i have seen those fixtures at spencer's and places like that.
379644
Some drivel here ;-)… To answer your question, you use the rating, i.e. 10kVA (note power out is ~ = to power in, there is very little power loss) and the supply voltage. US = 10kVA/110v = 90A. UK 10kVA/240 = 41A.
117333
a) E=v/l ---> E = (12/3)*1000=4000 (Volt/meter)\n\nb) v ~ 1/r ---> v2=12*3/5=7.2 Volt\n   E=(7.2/5)*1000=1440 (Volt/meter)\n\nc) w=int(f.dx)=int(Eqdx)=int(vqdx/x)=int(36qdx/x^2)\n\n    = 36q*(-1/5+1/3)=72q/15\n\nq=4*pi*e*l*v/2=2*3.14*8.854*10^(-12)*0.003*12 = 200*10^(-12)\n\n\n---> w=960*10^(-12) J
132674
When resistors are in series both have same current, i.e V/(r1+r2) = 12/(2+4) = 2 A\n\np.d across each resistor will be I*R\nSo for 2 ohm resistor p.d is 2 * 2 = 4 V\nFor 4 ohm resistor p.d is 4*2 = 8 V\n\nWhen the reistors are in series the p.d across both is the same, i.e 12 V\n\nCurrent across each resistor is given by V/R\nSo across 2 ohm resistor current = 12/2 = 6A\nAnd across 4 ohm resistor current = 12/4 = 3A\n\nResultant resistance = r1*r2/(r1+r2) = 8/6 = 4/3 ohm
641781
0.004vi = (1.110 + 0.004)vf\n\n(1/2)(1.114)vf^2 = (0.220)(1.114g)(0.300)\n\nvf^2 = 2g(0.220)(0.300)\n\nI think you should be able to do the rest...
350472
Vt is the source with R1 and R2 in series\n\nIt=Vt/(R1+R2)\n(OHMS LAW)\n\nIt=12/(120)\n---------------\nIt=0.1 A\n-------------\n\n\nVr2=Ir2 *R2\n     = It *R2\n     =0.1 * 100=10v\n\n------------------\nVr2=10 v\n------------------\n(So,Vr1=Vt(total)-Vr2=12-10=2 V)
380816
the answer is \n basic equaion man you must know this off head ...V=IR\nVOLTAGE =CURRENT(I) *RESISTANCE (R) \nSO THE ANSWER IS \nI=V/R\nTHEN ITS 24/200\n=0.12
341723
No, the voltages are different
512868
You need to wire them in parallel, with wire heavy enough for the total current and also have a battery that can deliver the current required at the voltage required without voltage drop.  You don't state the battery voltage or the lamp wattage, so I cant work it out for you!\n\n(I take it that you want EACH of the three bulbs to be as bright as one single one?)
436989
The voltage readings on the light bulbs are not what they definitely operate in, but optimim values. What is constant is their resistances, and connecting a larger voltage to a lower valued bulb will increase the current across it, and there is a possibility that this will burn the bulb.\n\nHowever, if you connect several bulbs in series to a higher voltage source, i.e. 3, 3, and 6, to a voltage source of 12 V, they should all be within the optimim values.\n\nGenerally speaking though, the bulbs in your example should still be ok (although they will shine much more brightly) unless you connect them in parrallel. (a volmeter across them would probably read about 4V and 8V if you connect them in series)