Voltmeter readings across a closed/open switch in circuits

[Krushnaraj Pandya]

Homework Statement

The following figure shows a circuit containing a bulb, switch and battery. Four voltmeters are connected in it to measure potential difference. Which voltmeters give non-zero readings and which ones read terminal potential difference when the switch is 1) open and 2) closed ?

Homework Equations

V=IR

The Attempt at a Solution

This question is raising a lot of confusion in my mind, When the switch is open, there's no current- how can there be a potential difference? I know potential difference causes current, much as in water from a filled container goes to unfilled container- if there's no water flowing; they're at the same level and so potential differences everywhere should be zero when switch is open. When the switch is closed, there'll be potential changes across the bulb and the battery as per KVL, and they'll be equal in magnitude=terminal potential differences as per KVL. Again, I have no idea what terminal potential difference even means when the switch is open. I'd really appreciate some help. Pardon the picture quality, I only have a webcam, I tried to make the simple figure as clear as possible- thank you.

 

[CWatters]

This question is raising a lot of confusion in my mind, When the switch is open, there's no current- how can there be a potential difference? I know potential difference causes current, much as in water from a filled container goes to unfilled container- if there's no water flowing; they're at the same level and so potential differences everywhere should be zero when switch is open.

There are 4 voltmeters. Which meters are you talking about? All of them?

One of them is connected to the battery. Do you think that one reads zero when no current is flowing?

When the switch is closed, there'll be potential changes across the bulb and the battery as per KVL, and they'll be equal in magnitude=terminal potential differences as per KVL.

Correct.

Again, I have no idea what terminal potential difference even means when the switch is open. I'd really appreciate some help. Pardon the picture quality, I only have a webcam, I tried to make the simple figure as clear as possible- thank you.

I believe "terminal potential difference" just means the steady voltage once the meters settle down.

 

[Krushnaraj Pandya]

There are 4 voltmeters. Which meters are you talking about? All of them?

One of them is connected to the battery. Do you think that one reads zero when no current is flowing?

Yes, all of them. I knew that one reads the emf but I just realized that maybe it's because the voltmeter provided a complete path for the battery- but voltmeters aren't supposed to take up any current ideally- am I wrong somewhere?

 

[CWatters]

I think you missunderstand...

Regarding the battery.. No it's not because of the meter. A source like a battery or PC power supply has a potential difference between its pins even with no current flowing or nothing connected to it at all. This is called the open circuit voltage and is equal to the EMF of the battery.

Capacitors can also have a potential difference between the pins with no current flowing. Q=CV or V=Q/C where Q is the charge stored in the capacitor.

Resistors only have a potential difference between their pins when there is a current flowing eg as per Ohms law.

 

[CWatters]

Sorry I deleted a reply I posted here as I think it might confuse you.

 

[Krushnaraj Pandya]

Perhaps remember that an open switch has a very very high (near infinite) resistance. So what happens to the current in the circuit if you replace the switch with a resistor with a value approaching infinity?

Ohh, that's a nice way to think about it. But the switch having infinite resistance means current tends to zero which should mean the voltmeter across the open switch should read zero as well...why isn't that the case?

 

[Krushnaraj Pandya]

Sorry I deleted a reply I posted here as I think it might confuse you.

Alright, but I still don't understand how there's a non zero reading across an open switch

 

[CWatters]

Start at the battery which has a voltage Vbat and work around to the switch. Are there any voltage drops in the circuit? If none then the voltage at the switch is the same as the battery.

More formally you can apply KVL around the circuit.

 

[Krushnaraj Pandya]

Start at the battery which has a voltage Vbat and work around to the switch. Are there any voltage drops in the circuit? If none then the voltage at the switch is the same as the battery.

More formally you can apply KVL around the circuit.

Seems reasonable, I think all my questions have been resolved, Thank you very much for your help :D

 

[CWatters]

Ohh, that's a nice way to think about it. But the switch having infinite resistance means current tends to zero which should mean the voltmeter across the open switch should read zero as well...why isn't that the case?

That's why I deleted it.

You have to consider the rest of the circuit as well. Try replacing the switch with a variable resistor...

Set it to zero and the voltage across it will be zero.

Increase it until it equals the same resistance as the bulb and the voltage across it will be half the battery voltage. Eg the battery voltage is divided or shared between the bulb and this variable resistor.

Keep increasing it and the voltage keeps increasing until at infinity its the same as the battery voltage.

You can use ohm's law to prove this or read up on the "voltage divider circuit".

The method in #8 might be easier to understand.

 

[Krushnaraj Pandya]

That's why I deleted it.

You have to consider the rest of the circuit as well. Try replacing the switch with a variable resistor...

Set it to zero and the voltage across it will be zero.

Increase it until it equals the same resistance as the bulb and the voltage across it will be half the battery voltage. Eg the battery voltage is divided or shared between the bulb and this variable resistor.

Keep increasing it and the voltage keeps increasing until at infinity its the same as the battery voltage.

You can use ohm's law to prove this or read up on the "voltage divider circuit".

The method in #8 might be easier to understand.

Oh, ok! That's a very interesting way to analyze an open switch, I understand what you meant to explain now. Thank you very much.