Question

In this problem E denotes the emf provided by the source, and R is the resistance of each bulb.

One more bulb is added to the circuit and the location of the switch is changed. The new circuit is shown in the figure. Bulbs A, B, C, and D are identical and the switch is an ideal conductor. How does closing the switch in the figure affect the potential difference?

Check all that apply.

The potential difference across A increases. The potential difference across B doubles. The potential difference across B drops to zero. The potential difference across D is unchanged.

Answer 1

When you close the switch, you short out bulb C. When this happens, the total resistance of the circuit will go down (since C was in series with B). This will cause the total amount of current in the circuit to increase. Since A is getting the whole current (it is in series with the battery) its potential difference will increase. The last two answers are silly, but the second answer is interesting. The total resistance of the parallel branches is going to go down, so the potential across D is going to go down. If the bulbs are equal resistance, then A was getting half of the potential difference across D, but now its getting all of it. BUT that potential difference went down. So, the potential across B will not quite double if the switch is closed, so that one can't be right either. I'd say:

The potential difference across A increases.