Parallel Circuits

To start off parallel circuits, we constucted a circuit using, a power supply, fuse, switch, and 2 light bulbs in parallel. We then measured the available voltage at each light bulb. which was 11.82v. There is the full supply voltage at each bulb because the full voltage goes through each branch. We then measured the voltage drop across each bulb and this turned out to be 11.74v for each bulb. This means that each bulb is using up the full supply voltage.

The next test was the current. Current flow through the bulb 1 circuit was .71A and current flow through bulb 2 circuit was .67A. These 2 values then add together to give the total current flowing through the circuit which is1.38A. Current will split evenly through branches in a parallel circuit as long as there is equal resistanc in each branch. If resistances are not equal, the most current will flow through the branch with the least resistance.

I then calculated the resistances for each branch and the bulb 1 circuit came to 16.6 ohms and bulb 2 circuit came to 17.7 ohms. I then calculated the total resistance in the circuit using the formula 1/Rt=1/R1+1/R2 and the total resistance came to 8.57 ohms. The total resistance of a parallel circuit should be less than the lowest resistance of any branch.

Total power output for the circuit was then calculated to be 16.3 watts and then power out put for each bulb was calculated. Bulb 1 had a power output of 8.4 watts and bulb 2 had a power output of 7.9 watts.

The bulbs are alot brighter in parallel because there is alot less resistance with the bulbs in parallel compared to in series. Less resistance means more current which means more wattage.

Next we constucted a circuit with 3 light bulbs. I measured the current flow and found that the current through the 2 smaller bulbs were about .68A and the current through the bigger bulb was 1.67A. This meant that there was less resistance through that bulb which meant more current flow, which meant a brighter light. This means that the current will get split up between the seperate circuits, with the most current going through the circuit with the least resistance.

I then measured the available voltage at each light bulb and had supply voltage of about 11.55v at each light bulb. I then measured the voltage drop over each light bulb and that came to about 11.40v for each bulb. This means that the supply voltage is the available voltage in each branch.

I then calculated the resistance of each bulb usiong Ohms law. Bulb 1 came out to be 16.72 ohms, bulb 2 came out to be 17.52 ohs and bulb 3 (bigger bulb) came out to be 6.916 ohms. I then calculated the total resistance resistance of the whole circuit which I calculated to be 3.82 ohms, which is less than the lowest value of the individual resistances. I then calculated the total power output of the circuit which I found to be 36.15 watts.