A 2.0 ohm and a 4.0 ohm resistor wired in SERIES with a 12 V source\n\n\nA 2.0 ohm and a 4.0 ohm resistor wired in PARALLEL with a 12 V Source
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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
176028
here is one of the datasheets available on the internet:\nhttp://www.ee.washington.edu/stores/DataSheets/74ls/74ls73.pdf
363225
Here is a good article. To understand them, you need to get a multimeter, and electronics projects book, a breadboard (to mount circuits on), and a selection of electronic parts from Radio Shack. Use batteries initially, and then when you are more confident a power supply and rectifier to do experiments, and measure voltage, current, and resistance values around the circuit.\nhttp://en.wikipedia.org/wiki/Electronics\nHyperphysics. (Google)\nhttp://www.mathworld.wolfram.com\nhttp://www.refdesk.com\n"You can do it electronics," Needham Heights, MA\nEnjoy!
121182
Use the node voltage method. Label each node with a voltage, then apply Kirchoff's current law at as many nodes as you need to generate enough equations to solve for any unknowns you have introduced.\n\nLet the node at the bottom of the circuit be ground, and then try to figure out the voltages at the other nodes relative to ground. Clearly the voltage at the node on the top will be 14V (I'll just call this V1) but no other voltages are apparent. So call the voltage at the node between R1, R3, and R4 Vx and the voltage at the node between R2, R3, and R5 Vy.\n\nNow that we have labeled all the node voltages, you can figure out every single current in the circuit in terms of V1, Vx, and Vy (using Ohms Law, the voltage drop across a resistor is equal to the resistance times the current in the direction of the voltage drop). For example, we know the current going down through R1 is (V1-Vx)/R1, the current going right through R3 is (Vx-Vy)/R3, etc. Now we just have to solve for the unknowns Vx and Vy.\n\nKirchoff's current law (KCL) says that the sum of the currents leaving a node is equal to zero. If we apply this law to the node with voltage Vx and the node with voltage Vy, we'll get a pair of equations that we can use to solve for Vx and Vy.\n\nApplying KCL at the node with voltage Vx should give you the following equation:\n\n(Vx-V1)/R1 + (Vx-Vy)/R3 + Vx/R4 = 0\n\nThe first term is the current going up through R1 (we already figured out the current going down through R1, but we want currents that are leaving node Vx so we had to flip the sign), the second term is the current going right through R3, and the final term is the current going down through R4.\n\nI'll leave it to you to figure out the equation for node Vy. Once you have that you'll have two equations and two unknowns and you can solve for everything. Once you have Vx and Vy, you can easily figure out a numerical value for any current in the circuit.\n\nBTW, the node-voltage method is pretty powerful and can solve most linear circuits. Check out the link below, it was written by an old EE prof of mine who is very thorough.
73225
Follow the bottom 2 links that I am going to post to get a labeled schematics\n\nhttp://www.kids-online.net/learn/click/details/sblvalpci.jpg\n\nhttp://dmzweb4.europe.creative.com/wwimages/audio_int/sblive/liveplayerconnect600.gif
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?
538655
I sense that you have no clue what you're doing. Tell me I'm wrong.\n\nFor DC circuits, Ohm's law says: V = I * R, voltage equals current times resistance.\n\nYour circuit will consist of a battery (voltage source) and a load (resistance) plus the wires to connect them and the meter(s).\n\nIdeally you will test one battery at a time. That way the load will be identical (including the small resistance of the wires).\n\nBatteries typically have two ratings: voltage and capacity.\nVoltage is measured in volts and this decreases as the battery discharges. Capacity is measured in amp hours (current multiplied by time) which specifies how long the battery can deliver a current.\n\nAssuming the circuit resistance is constant (reasonable assumption for your purposes), when you connect the battery to the load, the voltage (independent variable) will cause a current (dependent variable) to flow. As the battery discharges, the voltage will decrease, and the current will correspondingly decrease. You will need a watch or timer as well. Time zero will be when you first connect the battery. Measure the battery voltage at that instant and at intervals. Depending on the load and how fast the battery discharges should determine how frequently you take measurements.\n\nmsg me when you have the data if you need help analyzing it.
404290
YOU ARE ASKING A QUESTION THAT IS ANSWERED BY SOLVING THE EQUATION (E=IR) OR OHMS LAW. IF YOU GIVE ME TWO OF THE  UNKNOWNS THE EQUATION CAN BE SOLVED FOR A NUMERIC ANSWER.   LETS ASSUME THE VOLTAGE(E) IS 100VOLTS.  THEN THE CURRENT(IR1) WOULD BE (100VOLTS/110 OHMS) E1/R1 OR 0.909 AMPS.\nTHE SITE BELOW GIVE A LENGTHY DESCRIPTION OF THE SIMPLE CONCEPT AND JUST SEARCHING "OHMS LAW" WILL YIELD DOZENS MORE. YOU WILL FIND THAT THE VOLTAGE APPLIED TO EACH RESISTOR IS THE SAME AND THE AMPERES THROUGH EACH RESISTOR IS INVERSELY PROPORTIONAL TO THE RESISTANCE.
481709
1. C\n2. C\n3. A  \n4. B\n\nnote: The resistances are measured in ohms, not kW.
63023
Impedance and voltage are two different things. Impedance is like a complex version of resistance. Voltage is the potential. \n\nVoltage V\nImpedance Z\nCurrent I\n\nV=Z*I
550600
Make sure you have a good, solid connection between your sample and your ohmmeter probes.  Use a scale on your meter that's appropriate for the resistance you are measuring.  If your resistance is 1/2 ohm, then it will read zero on a 0-20Kohm scale, which isn't useful.  It will read half-way on a 0-1ohm scale, where you can get a good reading.\n\nThe resistance of a sample depends not just on the material and temperature, but also the geometry.  The resistance of a uniform wire will be proportional to its length and inversely proportional to its cross-section area.
236637
use a potato and two different metals as electrodes. The link will help you, do your own homework, or web searching atleast! it was the second link in google for g** sake!
414919
This is a very simple circuit.\nhttp://www.rason.org/Projects/fmkeyer/fmkeyer.htm
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I have already answered part 1. in another question you've asked. I think you probably didn't use the right units (like cm needs to be convered to meters) and got the wrong answer.\n\nNow for part 2:\n\nInduced EMF, ℰ = -L(di/dt)\n\nwhere i is the current and L is the inductance\n\nWe are given that the magnitude of the induced current,\n\ndi = 47 mA\n\nand the oscillation frequency is 100 Hz. This means that dt = 1/f = 0.01 sec\n\nWe have L = 8.44 × 10-^3 henries from before\n\nSo |ℰ| = L(di/dt) = -(8.44 × 10-^3)(0.047)/(0.01) = 0.0397 volts\n\n------------------------------------------------------\n\nNow part 3 has to do with mutual inductance:\n\nVs = VpNs / Np\n\nwhere \n\nVs is the voltage across the secondary inductor, \nVp is the voltage across the primary inductor (the one connected to a power source), \nNs is the number of turns in the secondary inductor, and \nNp is the number of turns in the primary inductor. \n\nTherefore Vp = VsNp/Ns and Vs = 0.0397 volts as calculated before, so\n\nVp = (0.0397)(197)/(2920) = 0.00268 volts
366964
R=V/I\nhence as you have  V=8\nand I=0.5 then your answer is \nR= 8/0.5= 16 ohms
92958
Assuming a DC circuit: use Ohm's law, I=E/R and P=I*E or I=150 Volts /11 Ohms = 13.64 Amps and Watts = 13.64 Amps * 150 Volts * 24 hours or 49,090.9 Watthours or 49.091 kWh.
100980
http://www.ee.duke.edu/~cec/final/node21.html#SECTION004101000000000000000\n\nhttp://www.electronics-lab.com/projects/test/014/\n\nI'm not sure you'll find all you need, but they would help.