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Assuming that the rectifier will be followed by a filter capacitor, the p.i.v. should be at least twice the peak of the applied a.c. (The capacitor will charge to the peak of the applied a.c. On the next half cycle of the input, the peak of that cycle will be of the opposite sign to that of the stored voltage on the capacitor, so the two add - giving twice the peak.)
Write down the Ohm's law expression of "V = I * R" where "V" is the voltage drop across shunt resistor, "I" is the current flowing through shunt and "R" is the shunt resistance.Substitute value of voltage "V" and current "I" in the Ohm's law expression. For example, if voltage across shunt is 10 volts and current flowing through it is 1 ampere, then the expression is 10 = 100 * R.Divide the expression throughout by 100 to calculate the "R" value. Using a calculator, find the value of "R." From the example, the value of "R" will be 0.1 ohm, which is the value of shunt resistor.
Connect the two heaters in series. The supply voltage, what ever the value, will be split in half across the two heaters. If the voltage is dropped by half across the two heaters in series, the wattage of each heater will be dropped to one quarter of the heaters wattage rating at its full voltage rating. The new wattage's will be added together for a new total wattage for the circuit. When you find the heaters working voltage use these two formulas; This one to find the resistance of the heater at its working voltage value, R = E (squared)/Watts. Then to find the watts at the reduced voltage value, W = E (squared)/R. R being the heaters resistance in ohms from the first formula.
In general, line resistance = resistivity * line length / (line cross-sectional area). Check the unit consistency: for the right-hand side, [ohm-m] * [m] / [m^2] = [ohm]. The resistivity of the material should be given. You can measure it with a large chunk of the material using a four-point probe, simplistically speaking. You can also look up the value online.
M. R. Shaw has written: 'The detection of initial failure in high-voltage insulation' -- subject(s): Electric insulators and insulation
That entirely depends on whether the resistances are in series or in parallel with each other. Ohm's law states that I=V/R. i.e. current = voltage/resistance. If you know the current and voltage you can find the resistance. You can use algebra to rearrange the formula for R and get that R= V/I. Resistance = voltage/current.
#include<stdio.h> #include<stdlib.h> int ack(int,int); main() { int m,n; printf("Enter the value for m : "); scanf("%d",&m); printf("Enter the value for n : "); scanf("%d",&n); if(m<n) printf("The value is : %d\n",ack(m,n)); else printf("Evaluation is not possible"); } int ack(int m,int n) { int r; if(m==0) r=n+1; else if(m!=0 && n==0) r=ack(m-1,1); else r=ack(m-1,ack(m,n-1)); return r; }
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The power pack has internal resistance of its own, equal to r. The voltage across the load resistance R is therefore: VR = 12(R / R + r) Thus the P.D. (VR) varies according to the load resistance R; it is impossible to ascribe an absolute value to it.
$125-$200, depending on condition.
No. V =Voltage, I =current, and R =resistancein the simple equation: V=I*R. As well, V/I=R, and. V/R=Iso Current is voltage divided by resistance
J. R. M. Hosking has written: 'Estimation of the generalised extreme-value distribution by the method of probability weighted moments'