Question

I am considering a circuit with three light bulbs. When a series circuit goes to a series-parallel circuit with one light on the main loop and two on different branches, the total resistance decreases by 1/2 (because the two on the branches go from having a total resistance of 2 to 0.5, and 3/(1+0.5=2)), doubling the total current, although the current in the branches is split in half, making each branch's current the same as in a series circuit. And since the total resistance of the branched part is half of one light, it uses 1/3 of the total of voltage (because V∗(0.5/1.5)=v∗(1/3)), just like one of the three lights in series. And this would mean the lights on the branched parts of a series-parallel circuit have the same amount of V and I as they would in series. Is this correct?

In class, in this series-parallel setup, when all the bulbs were screwed in the two on the branches were not lit (though must still have had electricity flowing through, just with not enough power). This makes me think my thinking above is wrong since the power must be less in the branches of the series-parallel circuit for the lights to be not glowing.

Answer 1

In a series-parallel circuit, the total resistance decreases while the total current doubles. The current in each branch is split in half, but the voltage remains the same. The lights on the branches may not light up because of the lower power dissipation.

Step-by-step explanation:

In a series-parallel circuit, the total resistance decreases when two light bulbs are added in parallel to the main loop.

However, this does not mean that the resistance of the individual bulbs on the branches decreases.

The total current in the circuit doubles because the addition of parallel branches allows for more current to flow.

The current in each branch is split in half, but the voltage across each branch is the same as in a series circuit.

The lights on the branches may not light up because the power dissipation is lower in the branches due to the decrease in resistance.