- All of them
A,M,P,K,I,N,G
#1. N/A
10 char
- This is funny,
A will turn on with the switch
I, N, and G will light continuously
Not sure that was what the author thought he was doing, lol.
- Bummer! Sorry no share. I’m an Ogre…
#1
None as the picture is
If the switch is closed then only A will light.
M & P are not connected to negative so no light
K is only connected to negative so no light
I N & G are connected to a set of batteries with only the positive terminal so no light.
#2
Shared on twatter.
Hey @jhmiller, Look at the diagram again. This is where I think the author did something he or she didn’t intend to. I agree with you in that I believe the author wanted to show no connection from the negative side of the battery set but the connection is made to the negative side of the previous battery in the row via the connecting wire between the two batteries. The last or first battery in the row (perspective) does nothing.
I have no use for these batteries, so I’m not going to participate…
The question is, which bulbs can be lighted up?
So technically speaking, without changing anything in this diagram only A can be.
I, N, G are constantly lighted due to the lack of a switch. You can’t light what’s already lighted, right? 
You could disconnect the lower batteries and reconnect them but that way you’re changing the diagram.
Look again my friend! 
The artwork is deceptive…
agreed that 6 of the 7 in bottom row are connected in series but how they are wired is blocking every bulb from getting both poles connected to the source, right??lol, idk, whatever
Yep I see it now, I, N, and G will light continuously.
Speaking strictly to the 6 connected cells in series (at the bottom), and without regard to the rest of the circuit, the important thing to note is that the subsection there is in effect, it’s own compete circuit.
The 3 “bulbs” (I, N, and G) comprise what’s referred to as a series-parallel circuit. With I being the series component, and N and G comprising the parallel portion.
The orientation of the “external” connection points of the batteries has no effect (because a traditional bulb has no polarity, though LED’s do), the importance lies in the fact that there’s still a difference in potential, which creates the circuit (in simplest terms, one end is still going to be more positive than the other end). I would burn brightest, with N and G splitting the other portion of the available power.
Once the switch is thrown, then polarity and voltages become considerably more important.
#1
I’m not so sure, but I think the answer is “A”