Used in electric circuit calculations
132812
If you change voltage say from 100 Volts to 200 Volts (perhaps by changing a switch position), the nearly instantaneous change will be a step change (like the steps on a stairs).  If you change the voltage from 100 to 200 Volts uniformly over a period of time (perhaps 10 seconds) for example by rotating a rheostat, the voltage will increase more like the ramp provided for wheelchairs than steps and will be a ramp change.  The terms are often used to describe the graphs or stripchart recorded traces of electrical signals relative to time.\n\nIn electric circuits, selecting various resistors with a switch may produce step changes and charging or discharging capacitors may produce ramp changes.
253428
I'd say that you already know.  If my memory serves (at 6 AM) it is an equation where you can determine the unknown if you have two values.  The equation can be expresssed multiple ways.  Ohms are given in impedance or interference/resistance, and I think the other values are volts(or watts) and amperage (amps).  Whether it's home, commercial or electronics, the law applies to the flow  of electron in any circuit.
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
205475
If it's for a simple hobbyist application they need to know only the resistance value and its power rating.\n\nA full specification would also include:\n\ntolerance,  thermal coefficient,  maximum voltage.
268467
This is a good time to learn dimensional analysis.  It will come in handy for all of your physics courses.\n\nFirst, µ x M cancels one another leaving ohms x Farads\n\nOhm = volt/amp\n\ncharge = capacitance x voltage  so capacitance = coluomb/voltage\n\nOhm x Farad = volt/amp X coulomb/volt\n\nso the volt cancels leaving \n\nOhm x Farad = coulomb/amp\n\nAmp = coulomb/sec\n\nOhm x Farad = coulomb/(coulomb/sec)\n\n= (coulomb/1) x (sec/coulomb) = sec\n\nBy the way, Tau = RC is the timeconstant of an RC type circuit.. you remember for a cap charging it is I = V/R[exp (-t/RC)  RC is in units of time = sec\n\nI used to have to do this stuff all the time in my undergrad physics courses.
91656
How about this one?\n\n"This calculation module allows you to quickly calculate fault currents for a complete radial system. It can simultaneously handle a main distribution transformer, service lateral, second transformer, up to 40 main-feeders, with 8 main-feeders having 40 sub-feeders for a total of 360 nodal points. \n\nAccommodating both single-phase and three-phase circuits, the module will calculate from either infinite or known fault current values at either the transformer or the main-bus. \n\nThe module accommodates multiple conduit types, bus types, multiple conductors and motor contribution. The program is based on the popular LICE (point-to-point) calculation method. \n\nOur unique Help file (of which we are quite proud) includes complete practical guidance on Fault Calculations, and even includes a theoretical derivation of the LICE method. "\n\nhttp://www.edreference.com/fault.asp
544792
C=80uF\nV=Voc1.5 volts\nI=Isc=5.0 amps (assuming short circuit current)\n\nSo thevnin's theorem implies the \nRth = Voc/Isc = 1.5/5 = 0.3ohms\n\nSo the circuit is an RL circuit,\n\nTime constant T = RC = 0.3*80uF = 24us\n\nNow you can use that formula for charging
456856
Count the lines for the voltage amplitude: 3 lines * 0.2 volts = 0.6 volts peak. In peak-to-peak form, 1.2 V.\n\nThen find the period: 5 lines wide * 0.5ms = 2.5ms period.\nThen frequency is the inverse (1/x) of the period.
234415
The link shows how to make calculations using an abacus.  Good luck.
601076
Perhaps the link will help.  At a minimum, you risk voiding your warrant if you use other that the voltage and frequency required on the unit name plates.  At worse you risk injury and a possible fire if you really don't know what you are doing.  Good luck.
107771
For series resistances and inductances, and parallel capacitances, simply sum them.\nFor parallel resistances and inductances, and series capacitances, first sum the reciprocals of each one, then take the reciprocal of the sum.\nHow you do this depends on how your calculator expects you to enter numbers and operators. You should consult the calculator manual if in doubt. If you don't have the manual, set up a test case whose answer you know. For instance, assume 2 resistors of 5 and 10 ohms. In parallel they equal 1/(1/5 + 1/10), or 1/0.3, or 3.333... ohms. Try various ways to enter the numbers and operators to get the right answer.
310680
Your circuit has 6 resitance R1, R2, R3, R4, R5, R6 and correspond currents are: I1, I2, I3, I4, I5, I6 and voltage are: U1, U2, U3, U4, U5, U6.\nYou can make many kind of circuit(direct, parallel or both) depend on total voltage as well as resistance.
615533
You can get an idea on counting the discs on the insulator string and multiplying by 12kV/disc.\nAgain it is only a general rule of thumb and is not accurate.\n\nSource 50 years as an Electrical Engineer
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.
548246
I happen to care quite a bit about imaginary numbers as a person who studies electrical engineering.  In electrical engineering, we use the imaginary number (denoted i or j) to help represent sinusoidal waveforms, work with linear systems in the frequency domain, solve sinusoidally-excited RLC circuits etc.  It shows up quite a bit in my chosen area of study.  Even something as mundane as the voltage sine wave at a common household outlet can be represented by a phasor 120*e^(i*theta) where theta is the phase shift of the sinusoid.  Also, I hear physicists use imaginary numbers in some applications.\n\nAnyway, one of the interesting properties of the imaginary number is that when you square it you have -1 (since i = sqrt(-1)).\n\nSo:\ni^300 = (i^2)(i^2)(i^2)...(i^2)\ni^300 = (-1)(-1)(-1)...(-1)\n\nThe above set of (i^2) factors will repeat 150 times (the exponents add).  Since (-1)^150 = 1 we have our answer (check with your calculator or realize that an even amount of -1 factors will always give you 1).\n\ni^300 = 1
344253
Hello, I would be glad to help you with this:\n\n a=(v-c)/t\nat=v-c\n-c=at-v\nc=v-at \n\n I hope this helps and good luck with the rest of your course work ,Please Please remember me for best answer
668425
In a circuit involving a number of sources, to find the effect at point, we use this superposition. Take circuits with each single source only. Find the effect due to the single source. The net effect is the sum of the effects due to individual sources.\nSuppose you have a circuit with sources v1, v2, i1,i2,..... Now take circuits with only one source. If there are n sources in a circuit, then u get n circuits with only one source. Now, say you want to find voltage across resister r. You find the voltage due to source1, voltage due to source2, voltage due to source3,.......