How to find the power dissipated in circuits with only parallel connections
deflashing to do
As with series circuits, start with the analysis in episode 01. This provides values for the currents in the loops and the potential differences across each circuit element. Again, taking care to identify the values, multiply the correct values of potential difference and current to find the power:
For the left hand lamp, P1 = I × V1.
For the right hand resistor, P2 = I × V2.
For the cell, P = I × V.
Again, you need only do two of the three multiplications to be able to work out the third, by simple addition or subtraction. Once more, the conservation of energy comes to your assistance: the pathway shifting energy from the chemical store and the sum of the pathways shifting energy to the thermal stores are of equal power.
Here are the calculations, laid out:
Something for nothing revisited
Now you are in a position to revisit the puzzle that adding loops in parallel seemed to give something for nothing (SPT: Electricity and magnetism topic, episode 03). You do get something: the thermal stores are being filled more rapidly, here by the extra glowing provided by the additional lamps. But not for nothing: the chemical store is being emptied at a much greater rate.
The necessary link between the depleted and augmented stores of energy is the electrical pathway, where the power sets the energy shifted to stores as well as from stores over the selected duration. The same power, so the same energy shifted in and out. Energy is conserved: the pathway is the mechanism that links the two stores.