P.d. and current - One of them can exist without the other

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In a series circuit with three bulbs and a 3V battery, if one bulb (A) is blown, the voltmeter readings will show 3V across bulb A, 0V across bulb B, and 3V across both A and B combined. If both bulbs A and B are blown, the readings will be 0V across each bulb individually and 3V across both A and B together. The discussion emphasizes that a potential difference can exist without current flow, but current flow requires a potential difference. Understanding these principles clarifies the behavior of voltage in circuits with blown bulbs. The explanation provided helps resolve confusion regarding potential drops in non-functional components.
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p.d. and current -- One of them can exist without the other!

Hi! Here's a question that I'm currently thinking about. Drawing a simple circuit diagram may help...

3 bulbs A, B, C are connected in series with a 3V battery.

(a) If bulb A has blown, what value of pd will you obtain if you connect a voltmeter across
(i) bulb A
(ii) bulb B
(iii) bulbs A and B?

(b) If bulbs A and B have blown, what value of pd will you obtain if you connect a voltmeter
across
(i) bulb A
(ii) bulb B
(iii) bulbs A and B?

My guesses:-
(a) (i) 3V (ii) 0V (iii) 3V
(b) (i) 0V (ii) 0V (iii) 3V

I believe one will need to assign arbitrary values for the potential at the positive and negative terminals of the battery, and then work out the potentials "to the left and to the right" of each bulb. Since no current flows, there will not be any potential drop across each working bulb (i.e. bulb that has not blown) But what about the potential drop across a bulb that has blown?

Can someone help in explaining how the answers are obtained, especially for (b), parts (i) and (ii)?

Thank you for your attention and time!


"There can exist a potential difference even though there is no current flow, but for current flow to be present, a potential difference must exist"
 
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pizzasky,
concerning (a):
(i) you would get the full potential of the battery 3V since the volmeter is effectively connected to the battery while no current is flowing.
(ii) 0V since the volmeter is connected to only one terminal of the battery.
(iii) again 3V same reason as (i)
From this you should be able to answer the other questions.
 
Wow! I get it now...

Thank you for your help!
 
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