matter what parameters are given in the problem, move them to the SI system.
If the condition given load resistance and released her power, guided by the following considerations: R = U / I, where R - resistance, ohms, U - voltage, V, I - current, A.P = UI, where P - power, W, U - drop voltage , B, I - current, A.Otsyuda follows that P = I ^ 2 * R, ie, I ^ 2 = P/ R, or I = sqrt (P / R).Consequently, U = R (sqrt (P / R)) or, after simplification of the expression, U = sqrt (P) * sqrt (R), where U - desired drop voltage load, B, R - resistance,Ohm, P - power watts.
much simpler case occurs if found voltage drop is necessary, knowing the power and strength of the current.Convert expression is not required, so just use the following formula
: U = P / I, where U - required drop voltage , B, P - power dissipated in the load, W, I - the current passing through the load, and.
If you know the load impedance and passing through it current, drop voltage her as calculated in one action: U = IR, where U - required drop voltage , in, I - current passing through the load, A, R - load impedance, ohms.
addition to the above the most frequently encountered problems, there are also other books in which you need to know drop voltage to a length of a homogeneous rod made of a material having a high resistance.To do this, first calculate drop voltage the entire length of the bar (if it is not given in the problem initially).Then, subtract one from the other horizontal coordinates of points, drop tension between them to be determined.
voltage over the entire length of the rod to divide its length and then multiply by the calculated length of the segment you, and you get a drop voltage between points.Such dividers are encountered in the apparatus with a transformerless power supply and used as network switches voltage - in this case, the simple design sacrificed efficiency and safety.
completing calculations, if necessary, turn the result into convenient for presentation units: volts, millivolts, kilovolt, etc.