If the calculations have not be implemented for this task from a textbook, and in the parameters of a real experiment, to measure voltage , connect the voltmeter in parallel with the load to measure current - ammeter in series with the load for measuring resistance - the ohmmeter in parallel, the load is de-energized, and to measure emitted power, put the load inside the calorimeter. In all cases, use caution. In this case, it is assumed that the measurement voltage at the load for one reason or another impossible, therefore we have to measure other parameters (the combination of resistance and current, the combination of resistance and capacity or a combination of current and power), then use in calculations.
Be sure to translate all quantities to SI before performing calculation. It is much easier, than then to be transferred to the system result.
In case you do not know the current through the load and the resistance, to calculate the drop of voltage it can use Ohm's law: U=RI where U is the desired drop in voltage on load (In); R - load resistance (Ohms); I is the electric current passing through the load.
If you know the load resistance, and allocated to her power, output the formula for calculating voltage on it in the following way: P=UI, U=RI. Therefore, I=U/R, P=(U^2)/R. this implies that U^2=PR, U=sqrt(PR) where U is the desired drop in voltage on load (V); P is the power produced at the load (W); R - load resistance (Ohms).
If you know the current through the load and allocated to it the power use when calculating the falling voltage at the load by the following considerations: P=UI. Therefore, U=P/I, where U is the desired drop in voltage on load (V); P is the power produced at the load (W); I is the electric current passing through the load.
If you have multiple series-connected loads, and the known ratio of their resistances or allocated to capacity take note of the fact that the current through each of them the same and equal to the force of the current in the entire circuit.