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? In this problem denotes the emf provided by the source, and is
the resistance of each bulb. Part A Bulbs A, B, and C in the figure
(Part A figure) 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 is unchanged.
The potential difference across B drops to zero.
The potential difference across A increases by 50%.
The potential difference across B drops by 50%. 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%

Also Read :   Label the schematic of the autoclave with the correct descriptions.

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