generation of impulse voltage?
822878
It's impossible to generate a perfect impulse voltage.\n\nIt's a mathematical thing to help engineers make the math work.
871859
Nominal voltage required for excitation of a circuit.\n\n- - - - - - -s-
504604
Time constants in electronic circuits.
800315
visit\nhttp://www.allaboutcircuits.com/vol_3/chpt_7/5.html
982631
The equations you post are the not correct.  If you close the switch on the circuit you describe, the CURRENT level is given by\ni = (V/R)e^-t*R/L.  The current starts out at zero, and ends up (after a long time) at V/R.  This equation is derived from the fundamental relation V=L*dI/dt.  If you open the switch, you must go back to this fundamental relation.  If you stop the current instantly, dI/dt = ∞, so the voltage must also go to ∞.  Of course this cannot happen.  What does happen is that the air in the switch breaks down at some voltage, and that limits the level to which the voltage can rise.  There will also be a big spark at the switch.
211994
V = Er = 3E6 x 0,30 = 9E5 volt = 900 kV
393875
The way I see it.... u calculate its impulse with the ground.... mass1*velocity1 (pre-shock)=m2*v2 (after shock) then caculate it's potential energy by newton's law of gravity using v2....(god I'm being a GEEK!)
679488
simple use two resistors, 6 and 3k ohm , in series and take the volatage across the 3 ohm resistor, it will be 3V.....\n\nuse the laws,\n\nif you have a voltage source, and two resistors p and q in series with it, \n\nthe voltage across p is given by  Vx(p/(p+q)), here u know voltage across p(Vp)= 3V, or 4.7 V or whatever....\n\nso, Vp = V x p/(p+q)....\n\nu know the value Vp, and V so substitute them......\n\n3 = 9 x p/(p+q)\n\nso p + q = 3p, \n\nso "2p = q", take two resistors ( in kilo ohms, coz u cant use just ohms, coz they are very large) such that one is double the value of other.......\n\nand connect the circuit across the 'p' resistor......\n\nwith 4.7, u get,\n\n4.7 = 9 x p/(p+q).....solve ...and choose resistors accordingly
662856
First answer said convert Joules in electron volts and then compute. You don't need to convert to eV; just divide joules by potential difference (volts) to get 5 coulombs.
604716
Need to know a need or application. AC voltage by definition is bipolar. What are you trying to do?
1114398
d: Subjected to a small voltage.
924875
50,000V\n\nOtherwise, if you use Kilo, you just use the K - eg: 50 KV
802288
the impuls (m*v) is conserved, so the impulse in one direction;\n0.150*20*sin(30deg)\nEquals the impulse in the other direction\n50*v_recoil\nthe recoil velocity is thus 0.03 m/s
353367
if f(t) = [delta](0) (unit impulse at t = 0), its Laplace transform is F(s) = 1, therefore, the inverse transform of F(s) = 1 is f(t) = [delta](0).\n\n And given that {L(a*f(t)) = a*L(f(t))}\n\nL(2*[delta](0)) = 2 * (1).\n\nThe answer is:\n  2*[delta](0)
892747
Yes. That is Ohm's law. It is a good approximation except for when large currents are involved.
788402
You can find the current flowing through a Wheatstone bridge by using Kirchoff's Loop Rules and Junction Rules.      \n\nThe sum of the Voltage Sources - Voltage Drops = 0 in a closed circuit.  \n\nThis setup can be solved using a matrix or a CAS. \n\nGood Luck
713107
range/amount\n\nrange = 120 mV (90mv - (-30mV))\namount = 2^16 = 64*1024\n\nplug-n-chug