How do I solve a resistor network in SPICE to get the equivalent resistance?
821458
Use Ohm's Law.\n\nMake the network, put 10 volts across it, find how much current is being drawn across the 10 volts.\n\nE = iR so R = E/i, where E is 10 volts, and i is the current, amps, then R is the resistance in Ohms.
935337
The resistance is increasing at .1785 ohms!!!\n\nJust use the ohm's law\n\n10/56=.1785 that's the resistance\n\nthen multiply the resistance and the current=volts\n\n.1785 x 56 =9.99999volts\n\nThis is the result.\n\nHope this help!!!
900061
a) combined resistance in series - add them up \n    That is, 8ohm + 4 ohm = 12ohm\n\nb)  V=IR so, I = V/R = 6/12 = 0.5A\n\nc) Ammeter in series reads the same value throughout.  So, 0.5A\n\nSorry but I can't understand your circuit diagram!! Try explainig it!
31718
for an ohmic material:\n\nP=VI\nV=IR\nso\nP=I^2 * R\n\ncheers!
15184
Using Ohm's Law, V=IR, we have 10=I(5).  Solving for I, the current is 2 Amps.
368244
V (Voltage) = I (Current) x R (Resistance)\n\nOr (although you probably don't need it):\n\nJ (current density) = sigma (conductivity) x E (electric field)\n\nI always use wikipedia to look up stuff like this, it's very helpful for all kinds of info.
815065
I tried to import a spreadheet for you but I coouldn't make it come through, if you email me I'll send it to you, like1m@yahoo.com  \n\nYou can fill in the two variables, just write it in excel and you will be able to calculate your values for voltage and current dependent on the resistance for any given values.  All answers here are correct as it is the basic premise of \nOhms law, but you asked for a chart.  You do need to know the resistance though.  The interrelation is such that for at a given voltage with infinate resistance there will be no current (flow), for that same voltage with minimum resistance there will be maximum current (flow) this would be referred to as a dead short and is comparable basically to welding. Either extreme is not normally practical unless you are turning the power "off" in one case or welding in the other (dead short).  Dependent on the requirement usually the interrelation is desired to be somewhere in the middle.  There is also the element of reactance which is slightly more dynamic that static resistance. but for now this is what your auto calc chart should look like....  \n\nInput ohms & volts to autocalc Amps\n\nInput OhmsInput VoltsI=V/R\n      200                     24                 0.12\n\nInput ohms & Amps to autocalc Volts\n\nInput OhmsInput AmpsV=I*R\n    200                     0.12                   24
1165503
For a steady state DC circiut P=EI \nwhere\nP=power in watts\nE= Potential in volts\nI=Current in amperes\n\nfrom Ohm's I=E/R\nwhere R = resistance in ohms\n\nsubstituting and solving for E we obtain\nE = SQRT (PR).\n\nSo what is the resistance of your circuit.
87871
let's consider a network like this:\n\n                           \n-node1(v1)----(+)V(-)----node 2 (v2)- \n                         \n\n\nnow this region is a supernode as voltage sourve(V) connects\nnode1 and node2 with no other element bet. node 1 and 2.\nAlso lets say that v1 and v2 are potentials of node 1 & 2 resp.\nSO how do u solve the equation? its simple.follow these 2 steps:\n\n1)write v1 = v + v2               \n2)then express v1 and v2 using KCL as seperate eqns.\nWhat is KCL?equate all currents thro the node as zero.\n\nU will get 3 equations.eqn 1 in step (1).\nand eqn 2 and 3 for KCL at node 1 and 2 respectively.\n Solve them using linear method or use Kramer's rule.(use determinants)
238179
Resistance=Voltage/Amperes
966971
ok resistivity formula is:\n\nresistivity = [resistance x length]/ cross sectional area..\n\nyou didnt give me enough info to work with\n\nhope it helped
970983
Rt - Total resistance.\n R1 - Resistor 1.\n R2 - Resistor 2.\n ì1 - Current 1.\n ì2 - Current 2.\n ì3 - Current 3.\n║ - Parallel.\n\nFrom the circuit given:\nRt = R1║R2 \nRt = (R1*R2 )/(R1+R2)\nRt = (3k*5k )/(3k+5k)\nRt = (15k )/(8k)\nRt = 1∙875kΩ\n\n ì1 = V/R1\n ì1 = 10V / 1∙875kΩ\n ì1 = 0∙0053333....\n ì1 = 5∙3333....mA.\n\n ì2 =  ì1[R1/(R1 + R2)]\n ì2 =  5∙3333....mA[3k/(3k + 5k)]\n ì2 =  0∙002 \n ì2 =  2∙0 mA.\n\n ì3 =  ì1[R2/(R1 + R2)]\n ì3 =  5∙3333....mA[5k/(3k + 5k)]\n ì3 =  0∙0033333...\n ì3 =  3∙3333...mA
898940
You can find printable schematics on the page I've linked below.\n\nGood luck.
94961
Well, not everything obeys Ohm's Law (for example, diodes), so it's not really a law.  It is more of a definition of R (resistance) in the sense that IF a system obeys E = I/R then that system has a well-defined resistance, R.
616704
pie   p power = i current x e voltage\n      e = i x r resistance
291589
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.
1130014
V=IR\n\nWire 1:\n50 = 2 * R1 => R1 is 25 ohms\nCurrent = 2 amps\npower = V*I = 2*50 = 100 watts\n\nWire 2: \n100= 2* R2 => R2 = 50 ohms\nCurrent=2amps\npower = V*I = 100*2 = 200 watts\n\nare you looking for ratios when you say "compare" ?\nif yes: ratio of resistances is  1:2\nratio of currents is 1:1\nand ratio of power is 1:2