(rest of question) intensities of lamps a & b-to the intensity of lamp c-to current inthe circuit-to voltage drop across the 3 lamps? does power dissipated in the curcuit increase, decrease, or remain the same?
128782
in a series circuit, current is constant across every resistor.  Intensities (brightness, I assume) of A and B to C should be the same as the power across all the resistors is the same.  \n\nHowever, by the sound of your question, it sort of would make more sense if A and B are in series, but then C is connected in parallel when switch S is closed.  Sorry, these questions are hard without diagrams.  I'll watch this question and edit my answer if you clarify the question.\n\n*edit* for Sky's question...to close a circuit is to "turn it on," allowing electricity to flow.  In an open circuit, the circuit is "off," electricity cannot flow.
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.
153418
(a)\nn_o / n_i = v_o / v_i\nwhere n_i is the number of coils on the inpt side of the transformer and n_o is number of coils on the output side.  Similarly, v_i is the input voltage, v_o is the output voltage.\nYou know the ratio of the output coils to input coils is 1/30 (1 output for every 30 input), so this is the ratio of input to output voltages.\n\n(b) Power [energy] is always conserved.  If you know the input voltage and input current, you can calculate the input power.\nP = V * I\nInput power = output power (ideally), and you calculated the output voltage in step (a), just solve for output curent.\n\n(c) Energy is always conserved, thus, so is power.
462606
If I understand you question correctly, you are putting in a retrofit fluourescent bulb into a standard lamp with a 3 setting light switch.\n\nYou are probably using power on setting #1, but there is not enough power to have the ballast convert the voltage and ignite the gas in the bulb.  But it is probably using energy constantly trying to light the bulb.  \n\nIt is a different principle than an incandescent bulb where no light = no heat = no power.\n\nI am not a total expert on the retrofit bulbs, but I would say if it is humming and the base above the socket warms up, it is using power.
273597
I think you meant WATTS versus volts. You will need to inspect your power supply for that information.\n You need to check out he following website, for more info on what you will truly need 250W sounds a bit low by today's standards - http://www.jscustompcs.com/power_supply/
398472
Hmmm, curious.\n\nIf the bulb has a filament you can see, then yet, it will almost certain be dimmable.\n\nI cannot see why there would need to be a trasformer in the box, if you try and wore the lamp in series with a 400Watt rated dimmer, th e worst that is likely to happen is you kill the dimmer.\n\nTry and find a bit more about this (make/model/type) and pop up another question, we can then give better feedback
457780
The circuit has a time constant which is calculated as\nT=L/R\n\nThe circuit will have the equation\nfor current of:\n\ni(t)=(Vs/R)*e^(-t/T)\nWhere\nVs is the supply voltage\ne is Euler's constant\n\nj
248753
1.  The time constant for the circuit is Re * Ce\n\nCeffective = Ce = c1c2/(c1 + c2) since they are in series.\n\nReffective = Re = R1 + R2 since they are in series.\n\nSo Ce = 3 * 6/(3+6) = 18/9 = 2µF\n\nRe = 1 ohm + 5 ohms = 6 ohms\n\nSo the timeconstant T (tau) = RC = (2E-6F)(6 ohms) = 12E-6 sec or 12 microseconds.\n\nWith two capacitors in series, the charge on both are the same.  The voltages will be different, but the charge is the same.\n\nq =C1V1 = C2V2 \n\nAlso in your circuit using Kirchoff's law: (Vs = powersupply or battery, r = int resistance, R = 5ohms)\n\nVs = ir + V1 + iR + V2  However, we can write V2 in terms of V1:\n\nV1 = q/C1   V2 = q/C2   V2 = C1V1/C2  (substituted for the q in V2)\n\nSo the above equation becomes:\n\nVs = ir + V1 + iR + C1V1/C2\n\nVs-ir-iR = V1 + C1V1/C2\n\nVs-ir-iR = V1(1 + C1/C2)\n\nV1 = Vs-ir-iR/(1 + C1/C2)\n\ni = Vs/Re[EXP(-t/ReCe)]\n\nNow solve for V1 using i in the V1 equation.
426656
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.
422441
By touching the lamp you are changing the capacitance of the lamp and a circuit inside the lamp detects the change and turns on the light. \n\nIn order to offset sensitivity to temperature and other objects near the lamp it is designed to be triggered by a sudden change in capacitance - so when you slowly touch it, or tap it with an object with insufficient cpacitance then the circuit does not detect enough of a change to trigger.
415390
The light-bulb has a tungsten filament, which has electrical resistance.  The voltage applied across the resistance causes a current to flow.  The value of the current is: I = V/R, where I is the current in amps, V is the voltage in volts, and R is the resistance in ohms.  The current flowing through the resistance causes it to heat up--it dissipates power.  The power is: P = I*V, or P = R*I^2, where the power is in watts.  MOST of that power produces heat; in an incandescent bulb only a little goes into light.  Power, incidentally, is energy per unit time.  Watts are Joules/sec, where the joule is the SI standard unit of energy.\n\nAs for the motors, connect them in parallel.  They will each receive the same voltage, and each will draw the amount of current it needs.
500885
In a purely resistive circuit, the power factor (PF) is 1 and you *can't* adjust it.\n\nAnd, since this is the 3'rd time you've asked the question and gotten the same answer from one other person and myself, do you even know what 'Power Factor' *is*?\n\n\nDoug
649394
house is AC, lights hooked directly in- AC\nCar battery DC, lights run off battery- DC\nTransformers are AC- AC\nBattery charger run through a rectifier- DC\nElectromagnet (had to look this one up)- either
108279
this is not enough information\n# 3 if there are only two lamps in series.
436602
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?
259737
surprisingly enough, the monitor actually draws more electricity than a running computer, so if a computer tower and a monitor were the only things on your electricity bill, and you started turning off your monitor more often (such as when you fall asleep to your music...i do the same =P), you'd see a dramatic drop on your next electricity bill =)
25789
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.