DC Circuit, and Similtaneous Equations?
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
315859
you got Ohm law : Ur=R*i(where i stands for current) and what's more, for the coil we have Ul=-L*di/dt . And for the condenser we have C*Uc=q and i = dq/dt.Now you have a differential equation with only one variable i.\n note: if the circuit is parallel then U should be a constant , in this case you shuold use U instead of i.Goodluck
786148
Electronic engineering - circuit analysis:\n\nSolving a circuit for the current and voltage according to Kirchhoff's Laws and these device equations:\nResistor   : v = R * i\nCapacitor: i = C * dv/dt\nInductor   : v = L * di/dt\n\nAnything complicated you generally solve by using the Laplace transform, in which case the above equations simplify to\nfor V,I = Laplace transforms of v,i:\nResistor   : V = RI\nCapacitor: I = sCV\nInductor   : V = sL\n\nThen you solve the system of equations for V,I and apply a reverse Laplace transform.
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
97533
+12VDC, -12VDC, +5VDC, -5VDC
1046550
Remember the equation v/i=r, Keep the voltage constant over all the areas. If the resistance goes up, then the current goes down.\nIf you have zero ohms, you have a short. IF you have infinite resistance, you have a break.  This seems general, but this is a great help for most of the electronics applications you will need. \nIf you want more you will need to give us more info.
293151
What circuit?!? Is it a series circuit?!? Is it a parallel circuit?!? Is it a series-parallel circuit!?!
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
214344
Voltage across parallel circuts is always the same, so 7V. The current, however will be split (if the lamps are exactly identical, split in half between them).\n\nSeries circuits are the other way. Voltage is different, current is the same.\n\nThis is a consequence of Kirchoff's laws. Kirchoff's law says that all the current going into a node must equal the current going out of the node (conservation of energy, really). And also, the voltage drops around a loop must equal. Common sense really. There can't be any extra (or deficit) current or voltage.
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.
202486
E(sub l) = L(di/dt). \ni is the varying current through the circuit during transient period or in AC circuit.\nif it's DC, E(sub l) will be equal to 0 after some time..
964185
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.
924475
= 9u-10u + 6v = 6v - u
148738
Homework, huh?  Badly copied from the source.  Make it an answerable question, clean up the gibberish and try again.  For instance, 90 degrees phase lag at what frequency?  Sorry to be curmudgeonly, but this shouldn't be just about getting answers without understanding.  Anyway, here's a hint.  At 90 degrees, Xr=Xc, and RC=1/(2*pi*f).  For phase lag the R is in series and the C in shunt.
379771
http://www.mikeholt.com/mojonewsarchive/EC-HTML/HTML/WyeTransformerCalculations~20030916.htm
655128
Assuming that we have 2 series resistors R1 and R2. And the voltage applied on them is V.\n\nTo get voltage applied on R1 (V1)= V * R1/(R1+R2)\nand so V2= V * R2/(R1+R2)
93256
Try this website. Scroll down to "The Circuit" ---\n\nhttp://www.play-hookey.com/ac_theory/ac_rc_series.html