In this problem e denotes the emf provided by the source. and r is the resistance of each bulb.

On this downside denotes the emf supplied by the supply, and is
the resistance of every bulb. Half A Bulbs A, B, and C within the determine
(Half A determine) are an identical and the swap is a perfect conductor.
How does closing the swap within the determine have an effect on the potential
distinction? Verify all that apply. The potential distinction throughout A is unchanged. The potential distinction throughout B drops to zero. The potential distinction throughout A will increase by 50%. The potential distinction throughout B drops by 50%.

by closing the swap the present will get a path that has least
resistance as in comparison with B and C.it has 0 resistance.so complete
present flows by way of swap and no present flows by way of B or
C
POTENTIAL DIFFERENCE ACROSS B DROPS TO ZERO.
POTENTIAL DIFFERENCE ACROSS A INCREASES BY 50%.
intially web resistance=r+r/2=3r/2
i=E/(3r/2)=2E/3r
voltage drop throughout A=2E/3r*r=2E/3
when swap is closed i=E/r
voltage drop=E
so it will increase by 50%