Voltage over bulb with broken filament

[coconut62]

Homework Statement

A main circuit contains six similar bulbs connected in series. One of the bulbs has a broken filament. Voltmeters X and Y of infinite resistance are placed in the circuit as shown.

What are the voltmeter readings?

A) X=0 Y=0
B) X=0 Y=240
C) X=40 Y=40
D) X=240 Y=0

Homework Equations

None

The Attempt at a Solution

Answer: D.

This is my reasoning:
A broken filament has infinite resistance because no current can flow. So the it would "draw" all the voltage from the source.

Am I correct?

Also, I would like to ask, why can't the first bulb light up? Since the source is continuously supplying a current, and the current should pass through the first bulb first before passing the second bulb.

Thank you.

 

[voko]

I cannot see from the picture which bulb has a broken filament. What can be said safely is that one voltmeter is connected to a good bulb - what voltage will it read? If another is connected to the broken bulb, what is its reading?

 

[coconut62]

Sorry, the second bulb has the broken filament.

The voltmeter across the broken bulb would read 240V and that across the good bulb would read 0V.

 

[voko]

All correct, then.

Re why the first bulb would not light up: that needs current. Current needs an uninterrupted loop.

Or put another way: if the current flows through an infinite resistance, how large is the current?

 

[coconut62]

During the moment the switch is turned on, is there no current flowing through the first bulb?
Charge did flow through the first bulb, only that they're blocked at the second bulb, no?

 

[voko]

I do not see any switch in the circuit.

Regardless, any break in a circuit is essentially a capacitor. As soon as voltage is applied to a circuit with a capacitor, there is some current that charges the capacitor. But as soon as the capacitor accumulates enough charge so that its voltage becomes equal in magnitude to the source voltage, current stops. This happens so quickly, the bulb does not have enough time to heat up. But a sensitive voltmeter and a trained eye could see that very briefly the bulb was energized.

 

[Basic_Physics]

It is a series circuit. A break anywhere in the circuit will result in no charge being able to flow through the circuit. So none of the components in the circuit will operate. With a capacitor you do get a momentary flow of electric charge though.