I am a relative beginner, although I did do an Electronics course a few years ago, but have forgotten quite a bit of it.

At the moment, I'm studying Potential Divider circuits. To help me understand it a little better, I set up a little series circuit with 3 resistors and a lamp, on my breadboard. 6 Volts battery supply. Resistors of l5.2 ohms, 10.1 ohms and 56.8 ohms. I measured resistance of the 6 volt lamp, which appeared to be 4.5 ohms.

I was really looking forward to getting a nice neat result, that is, I was hoping that the sum of voltage drops across resistors and lamps would equal the battery voltage.(just like it says in the text books) Unfortunately though, I got the following results:
VOLTAGE DROPS ACROSS 3 RESISTORS & LAMP
R1= 0.724V
R2=0.482V
R3=2.72V
Lamp=0.890

According to my calculator, this makes a total of 4.816 Volts.
Yet the Voltage at my battery measured 5.55 Volts.
So what has happened to the other volts?
That is, there seem to be 734mV missing.

I would be grateful if someone who is a little more advanced in Electronics than I could make some suggestions! Thanks.

 

Did you measure your battery voltage while in the circuit or did you measure the battery voltage with it disconnected from the circuit?

It should be measured while in the circuit since it is likely that it will drop slightly under load.

The error is pretty high so it is not likely that the error is due to meter accuracy.

hgmjr

 

Batteries have internal resistance. If you're powering a circuit that has relatively low resistance, the internal resistance of the battery will dissipate a large amount of power.

Since your voltage across R2 was 0.482 and I=E/R, and it's a series circuit, we know that I = 47.7mA (rounded off). Your battery with no load = 5.55v, loaded = 4.816, for a difference of 0.734v. Since R=E/I, 0.734V/47.7mA = 15.39 Ohms internal resistance for the battery (rounded off).

Something else you should note is that the resistance of incandescent lamps increases as the temperature of the filament increases. When cold, you measured the resistance of the filament as being 4.5 Ohms - yet under power, you measured 0.89v across it.
Since circuit current is 47.7mA, and R=E/I, the lamp is 18.66 Ohms!

BTW, I can tell that you measured R3, R2, R1 in that sequence; the current in the circuit was dropping as the battery was getting depleted.

[eta]
See the attached simulation of your circuit. The voltages aren't exactly what you read because in the simulation, the battery doesn't discharge. "Rbatt" represents the internal resistance of the battery.

 

Another option is a relatively bad connection or wire. Measure the voltage drop across the wires.

 

mik3 has a good point, some of those alligator clips are surprisingly high in resistance. I have measured some at about 8-10 ohms and have used them as high current fuses when prototyping and testing.

Steve

 

Essentially, it means that you can have both voltage drop due to the internal resistance of the battery (especially if it is a small one) and due to the resistance of the wiring. I think that a voltage drop of 734mV is quite typical in such cases (since your components are being biased by a relatively large current).

I suggest you to measure the voltage of your battery while the circuit is connected. If the values still don't add up, probably the voltage drop is also due to the resistance of the wiring.

As StgWookie suggested, your battery might also be getting depleted. It is a matter of measuring the battery voltage again, in order to confirm or exclude that theory.

 

Wow! What a mine of useful information, thank you so much for your helpful answers everyone, that has really helped me. I am very pleased to have learned some new things about Electronics from your replies, eg,

A battery has internal resistance, and thus volts are dropped across it

The resistance of a lamp goes up when electricity is going through it

The voltage at a battery is different when it’s connected to a circuit than when it’s not connected(due if I have understood correctly, to the internal resistance of the battery)

With regard to the Maths suggested by Sgt Wookie, I wrote it all out for myself to help me thoroughly understand and absorb it. I didn’t know that you could work out Total Current in a series circuit by doing Voltage drop of a particular Resistor divided by its individual resistance, that’s very useful. I was very pleased to receive the schematic of my circuit, I’ve never seen anything like this done before!

Anyway, having studied your replies, it was still however,with some trepidation and nervousness that I approached the folding camp table in the corner of my room, upon which was set the circuit in question. But to my great pleasure, this time, when I took the measurements, it all added up!

I did indeed measure the voltage drop across the various wires, and it came to a total of 1.1 Ohms. Very interesting!

 

Glad that helped

I suggest that while you are experimenting, you use fairly large values of resistance; ie: 1k Ohms or more. This will make your battery's internal resistance far less significant, and your battery will last much longer. Besides, if you accidentally connect one, or several small-value resistors in parallel, across your battery terminals, the resistor(s) may get hot very quickly, and/or your battery may overheat and rupture forcefully.

 

I too am glad we were able to assist you in accounting for the missing voltage.

hgmjr

 

Indeed, we are here to help. And I'm glad to see that your question was properly answered.

 

I recommend you work your way through "Vol 1 -DC" on this very website, which will help you understand your experimental results. Doing both (theoretical and experimental) is a great way to really learn the material.