a 12.0 olm resistor, a 6.0 olm resistor, and a 8.0 olm resistor are connected in serieds to a 20 volt battery?
397971
The total resistance or effective resistance of resistors that are connected in "serieds" (series) is the sum of the values of all the resistors, in this case it is 26.0 ohms...
533465
V=IR so R=V/I so R=120/2.5 = 48 ohms
617296
The general formula for finding the equivalent resistance when the resistors are connected in parallel is:\n1/R=1/R1 + 1/R2 + 1/R3 +.........+ 1/Rn.\nWhere, R=equivalent resistance\n       R1=First resistance\n       R2=Second resistance\n       .............\n       Rn=Last resistance\n    Therefore answer for your question is:\n1/R=1/3+1/3+1/3+1/3+1/3+1/3\n1/R=6/3\n1/R=2\nTherefore, R=1/2=0.5 ohms\nThe formula for finding the equivalent resistance when the resistors are connected in series is:\nR=R1+ R2 + R3 +........+Rn\nWhere the quantities are as above.
315095
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
275873
The Heath AC voltmeter is illustrated in the link.  Click on the diagram to enlarge it.
972106
Cheap and simple solution:\nUse a good 1.5Volt battery, connect a 2.7KOhm 1% fixed resistor, and a linear 100 Ohm potentiometer in series across the battery. \nBetween the center tap and the "bottom" end of the potentiometer, you will get 0 .... 53 milli Volt. Use a knob and scale, and a milli Volt Meter to calibrate it.
899661
This will take a few steps.\n\nStep 1.  Find the equivalent resisitance of the 12 and 4 ohm\nresitor.  Reqv = R1(R2)/(R1+R2) = 12x4/16 = 3\n\nSo you have a 3 ohm and 7 ohm resistor in series.  This gives\na circuit current of V/Rtotal = 20/10 = 2amp\n\n\nNow the voltage drop across the 7ohm resitor is 7x2 = 14 volts\nand the voltage drop across the equival 3ohm resitor is 3x2 = 6 volts.\n\nThis 6 volts is across the 12 ohm resitor, so the current\nthrougth it is V/R = 6/12 = .5
32772
V=IxR where V = volts, I = Amps and R = the resistance in the circuit.  This is called Ohm's Law.\n\nManipulate the equation to get I = V/R
467003
You'll need, as the name may suggest, a voltmeter.  Also applicable is a multimeter - which you'd then set to voltmeter mode.\n\nAt which point you'd just put one contact on one one end of the battery, and one on the other, and see what the equipment reads.\n\nMy money's on 1.5V :)
227248
between 120000 and 20000 volts= 70000 volts adv.\n15500 amps
994340
V(voltage)=I(current)*R(resistance)\n\nVoltage measured in volts\nCurrent measured in amps\nResistance measured in ohms.
616731
The equivalent resistance of the parallel resistances R3, R4, R5 is given by\n\n1/Re = 1/5 + 1/10 + 1/ 15. =  55/150\nRe     = 2.73 ohm.\nThe equivalent resistance of the series resistances =R1+R2+Re =32.73 ohm\nTotal resistance =32.73 ohm.\n\nCurrent total = voltage /total resistance = 10 /32.73 ohm = 0.305 A.\n\nVolts of R1 = current x resistance = 0.305 x 10 = 3.05 V.\nVolts of R2 = current x resistance = 0.305 x 20 = 6.11 V.\nVolts of Re   = current x resistance = 0.305 x 2.73 = 0.83265 V.\n\nVolts of R3= Volts of R4= Volts of R5= Volts of Re =0.83265 V.\n\nCurrent through R1 = Current through R2 = Current through Re =  0.305 A\n\nCurrent through R3 = volts of R3/ R3 = 0.83265/5 =0.16653A\nCurrent through R4 = volts of R4/ R4 = 0.83265/10 =0.083265A\nCurrent through R5  = volts of R5/ R5 = 0.83265/15 =0.05551A
847813
This process works relying that V=IR, wich is Voltage= Amps times Ohms. Ohms are a measurement that measures resitance, for example, a lightbulb has a certain Ohm rating, and amps would be how much the batteries produced. So, by adding more batteries to the circut, depending on the circuit, you could increase the voltage. For example, let's say that you have 5 lightbulbs, each with 1 Ohm, and you have 2 batteries, producing 3 amps each. Here, we simply multiply (1*5)(2*3), which equals 30. Therefore, V=30 in this circuit. However, this would only work in a parallel circuit, which means that the batteries have to be parallel to each other and the lightbulbs. Sorry for the long response, I hope it answered your question!
380082
the way the resistors and  or other components are set in respect to the source or battery.
572940
A voltmeter consists of a galvanometer(to detect current) and a high resistance(usually thousands) connected in series. It measures the voltage across two points. Theorectically a voltmeter should have infinite resistance.
313707
Assuming you are using US household electricity (120VAC)\nP=V^2/R   R=V^2/P = (120)^2/1875= 7.68 ohms which is really small.
752024
When you copy and paste these, why don't you convert the "W"s into "ohms"?\n\nHey!  The guy who did your homework for you ("The Great One") gave three WRONG answers (as of 1930 MST 11-14.  He'll probably redo it after he reads this)...heh, heh, heh...