Voltage/ Voltmeter: Why 1.50 V?

[husky88]

Homework Statement

The problem is actually longer, but I will make it shorter, just to ask my question.

A flashlight battery has a voltmeter connected across its terminals. The voltmeter reads 1.50 V. (Assume the voltmeter is perfect, with an infinite resistance, drawing no current.) The voltmeter is removed and the battery is connected to a small light bulb.
It turns out the potential drop across the bulb is only 1.42 V, because of the internal resistance of the battery.

Homework Equations

The Attempt at a Solution

My question is why didn't the internal resistance of the battery record on the first reading of the voltage also? This isn't a question in the problem, I just don't understand why. There was a closed circuit in both cases.

Any ideas greatly appreciated.

 

[Pythagorean]

Battery resistance is beyond my reckoning, because it has to do with the chemical makeup of the battery as it runs out of charge.

The equation modifies the voltage by

v = v0 - I*R0

where R0 is the internal resistance of the battery and I can be expressed as V/R where R is the load, so now we have:

v = v0 - (V/R)*R0

In the first case, the R is inifnite (it says so in your translation of the problem, the voltmeter is the load, with resistance R), so the quantity V/R makes the second term 0

In the second case, R is the load created by the light, so the V/R term doesn't go to 0, since R is not infinite.

 

[husky88]

Oh, ok. That makes sense.
Thank you.

 

[Mentz114]

My question is why didn't the internal resistance of the battery record on the first reading of the voltage also? This isn't a question in the problem, I just don't understand why. There was a closed circuit in both cases.

Because the voltmeter drew no current, but the lightbulb did. In other words the voltmeter put no load on the battery, so no current flowed.

v = v0 - I*R0

if I=0 then R0 is irrelevant and v=v0.

 

[husky88]

And now I can understand it in practical terms too.
Thank you, again.