Removing bulb causes a different fixture to go out

Question

I have ceiling fixtures, let's call them A and B. Normally, both fixtures are always on (lit) -- no switch controls them.

When I remove the bulb from A, fixture B goes out.

Here's how it looks inside the junction box that feeds both fixtures:

Fixture A is at the top of the box. Fixture B is the one on the bottom (I think, not 100% sure about that.)

As you can see, the white wire from Fixture A is connected to the other white wires in the junction box. That makes sense.

But, the black wire from Fixture A is jumpered onto a white wire, that's then connected to the white wire on Fixture B.

What's going on here?

I plan to put motion-detector bulbs in both fixtures, so I want them both hot all the time.

Can I just remove that white jumper wire and connect the black wire to the other black wires?

I have confirmed that there's nothing wired downstream from the Fixture A box - the wires terminate there.

Answer 1

It's difficult to tell for sure, as I'm not sure where all the wires in the box go. But it looks like the lights are run in series, rather than in parallel as they should be.

To fix it.

1. Turn off the power and verify that it's off.
2. Connect all the black wires together in one twist-on wire connector.
3. Connect all the white wire together in the other twist-on wire connector.

This image shows how electricity moves through the circuit.

Notice that the neutral wire "coming back from" light B connects to the hot wire "going to" light A. Then the neutral "coming back from" A, connects the the other circuit neutrals.

The only way that this wiring would make sense, is if B was a switch controlling A.

removing the bulb from either A or B interrupts the circuit, and causes both light to go out.