# Homework Help: Voltage over bulb with broken filament

1. Nov 25, 2013

### coconut62

1. The problem statement, all variables and given/known data

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

2. Relevant equations

None

3. The attempt at a solution

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.

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2. Nov 25, 2013

### 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?

3. Nov 25, 2013

### 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.

4. Nov 25, 2013

### 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?

5. Nov 25, 2013

### 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?

6. Nov 25, 2013

### 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.

7. Nov 25, 2013

### 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.