If the total current into a set of parallel resistors is known, the current through any individual resistor can be determined by the concept of current division. In essence, the total current into the parallel combination is divided among the individual resistors.
For the special case of two parallel resistors shown in the figure below, the individual current through each resistor can be calculated according to the current divider formulae below:

One application of parallel resistances is to reduce the power dissipated by any individual resistor in a circuit. Power is the product of voltage and current, so the total power applied to the circuit above is:
\[P = i \cdot v\]
By dividing the total current among several resistors, we also divide the total power dissipation among the resistors. Thus, we can draw a larger amount of power from the source, without exceeding the power limitations of any individual resistor. This project illustrates the overall process.
Qty  Description  Typical Image  Schematic Symbol  Breadboard Image 

2  1kΩ Resistor  
1  470Ω Resistor  
1  Digital Multimeter (DMM) 
Implement the circuit shown schematically to the right. Use V+ to apply the 5V source.
Measure the voltage V_{OUT} and the current I.
Calculate the power absorbed by the resistor. Also determine the power delivered by the source. (Hint: conservation of power says that the power delivered by the source is the same as the power absorbed by the resistor.)
Implement the circuit shown schematically to the right. Use V+ to apply the 5V source.
Measure the voltage V_{OUT} and the currents I_{1} and I_{2}.
Calculate the power absorbed by each resistor. Also determine the power delivered by the source. (Hint: conservation of power says that the power delivered by the source is the same as the power absorbed by both resistors.)
Compare the power delivered by the source for the circuits in Step 1 Parts A and B.
Compare the power absorbed by the resistor in Part A to the power absorbed by the resistors in Part B.
What is the effect using two resistors in parallel to draw approximately the same power from a source?
We can also use resistors in series to reduce the power dissipated by any individual resistor in a circuit. To verify this, use 5, 100Ω resistors in series as shown in the circuit below to create approximately the same “load” on the source as in the circuits in Parts A and B above.
Measure the current I and the voltage V_{out} in the above circuit. Use these measurements to estimate the power absorbed by each resistor in the circuit. Also estimate the power delivered by the source for this circuit. (Hint: just assume that the source is delivering 5V.)
Compare the total power delivered by the source and the power absorbed by each resistor in the circuit above with the corresponding values in Part D above. Has the series combination reduced the power dissipated by each resistor, as expected?