How does this Dark Meter work?

[AntiStrange]

How does this "Dark Meter" work?

In our physics lab we used photoresistors to make a simple circuit that is light activated.

It was essentially a battery hooked up to a photoresistor which was in series with an ammeter to measure the current. When the photoresistor is blocked there is low current (does the current go to zero ever? or just really close?) and when there was light a larger current would flow through.

Then the next circuit was a "dark meter" it was basically the same setup except the ammeter and photoresistor were connected in parallel. It allowed more current to go through the ammeter when there was no light on the photoresistor and when there was a bright light the current through the ammeter was very low.
This confused me for two reasons...

1) I thought it was improper to ever hook up an ammeter in parallel (even dangerous?)

2) how could the ammeter be affected by the photoresistor? They are in parallel so no matter what the resistance is through the photoresistor, the current through the ammeter would be constant right??

 

[negitron]

1) There was probably a series resistor in there as well, possibly internal to the meter movement. Or the battery voltage was selected so as to provide full-scale current when connected through the meter's winding resistance. Otherwise, no, you shouldn't connect an ammeter to a voltage source like that

2) Consider the extreme cases. With the photoresistor replaced by an open circuit (i.e., infinite resistance), you will read a large current through the ammeter. With the photoresistor replaced by a short circuit, you will read zero current through the meter.

 

[AntiStrange]

I'm sorry but I still don't think that I understand #2.
Here is my way of thinking:
let's say the resistance of the photoresistor is 20kΩ, and the ammeter has a resistor connected in series with it of 2kΩ and the battery supplies a potential difference of 9V.

The total resistance would be R = (1/20kΩ + 1/2kΩ)^-1 = about 1.82kΩ.
Therefore the total current the battery supplies is: I = 9V / 18.2kΩ = 4.95mA
Then the current that goes through the ammeter would be I = 9V / 2kΩ = 4.5mA
and the current through the photoresistor is I = 9V / 20kΩ = 0.45mA
which adds up right? 4.5mA + 0.45ma = 4.95mA = the total current

Now if we change the photoresistor to 10kΩ instead of 20kΩ and leave everything else the same and repeat the above process we find:
total resistance = 1.67kΩ
total current = 5.40mA
current through photoresistor = 0.9mA
current through ammeter = 4.5mA
4.5mA + 0.9mA = 5.40mA

So the current through the ammeter is unchanged even through you change the parallel resistance.

What am I doing wrong?

 

[negitron]

Well, what characteristic does a real-world voltage source, like a battery, have that limits its output current and what happens to the terminal voltage as a result of this characteristic as the total load current increases?

 

[vk6kro]

The ammeter with a resistor in series with it would really be more like a voltmeter.
If it had a full scale reading on 9 volts with a 2K resistor in series with it, the meter itself must have had a full scale deflection of 4.5 mA.

If you put it across a 9 volt battery, it would read the battery voltage which would not change much with a phototransistor also connected across it, even if the light level on the phototransistor was varying.

So, you would need a common resistor feeding the phototransistor and the voltmeter.

If the meter read full scale on 9 volts with 2 K in series with it, it would still read full scale if you put two 1 K resistors in series with it. Then put the phototransistor across the meter and its nearest 1K resistor.

This leads to some interesting calculations, but since this is classwork, I'd have to leave them to you.

 

[AntiStrange]

Well, what characteristic does a real-world voltage source, like a battery, have that limits its output current and what happens to the terminal voltage as a result of this characteristic as the total load current increases?

Sorry my responses take so long, busy schedule.

I guess real voltage sources have an upper limit to the current they can produce, but would that really affect the terminal voltage like you say?
So I'm guessing that the current is already being overdrawn when there is no resistance through the photoresistor, so when you introduce some resistance to it then most of the current goes through the ammeter??