Voltage Drop?

Question

I have a series circuit ( 9VDC + 2 resistors [ 100 ohm + 220 ohm])

1. When testing the battery is it common practice to test its voltage under full load only?
I ask because when i tested the battery I was getting a reading of 9.61 so then I decided to test it with the circuit powered up and it was closer to what it should be at 9.02 volts.

2. My voltage drops do not add up to my Source Voltage ( under load), they differ by .41 volts. I am thinking this should be acceptable because the formulas should be used as a guideline and not a strict rule.
I am thinking anything under .8 would be good.
I am also thinking, the formulas are only if all things were perfect what the results would be, but since nothing is perfect there will (most of the time) be a difference.

How much of a difference should I allow before thinking there is a problem?

Should i even bother calculating the formula or just take the actual readings and go by them - Real World Standards?.

Should be .....................Actual
9VDC ................9.61(no load) 9.02(load)
220 + 100 ohm......98 + 216 ohms
(adds up to 314 ohms)
Total Circuit Resistance = 315 ohms
.028 A ...............................026 A
Drop(98ohm) = 2.70 Volts
Drop(216ohm) = 5.91 Volts
Drop(total) = 8.61 ( less .41 than 9.02V)

Should be .....................Actual
9VDC ................9.61(no load) 9.02(load)
220 + 100 ohm......98 + 216 ohms
(adds up to 314 ohms)
Total Circuit Resistance = 315 ohms
.028 A ................................. ..026A
Drop(98ohm) = 2.70 Volts
Drop(216ohm) = 5.91 Volts
Drop(total) = 8.61 ( less .41 than 9.02V)

Resistors are 1/4Watt 5% tollerance

Answer 1

Normally, you test a battery without it being under load. A new "9 V" battery should be about 9.6 V, and still has usable power until its open circuit voltage drops to about 5V.

Under load, a battery will drop in voltage, but the exact amount of the drop is determined by the type of battery and the resistance of the load. 320 ohms is not very much resistance; 28 mA of current will flow through the resistors, so I'm not surprised of the voltage drop. As the batteries wear out, the drop will be greater; it will also be greater if you reduce the resistance (a zero ohm dead short will reduce the resistance to near zero as the wire gets very hot, and the battery quickly dies!).

Your voltage drops should exactly add up to your total measured voltage. It could be you have a cheap (low impedance) volt meter that is shorting out your battery and causing measurement errors. But in the case you describe, you should measure 6.20 V across the 220 ohm resistor and 2.82 V across the 100 ohm resistor. V=I*R. Also consider that your resistors are likely not that accurate, so the exact voltages will be different, but they MUST add up to the total under-load volatage!

Answer 2

1. Batteries have internal resistance which will lower their effective voltage when producing current. Nevertheless we usually measure them with no load.

2. This much of an error (0.41V) is worrisome. Instinctively I would assume it is due to you taking incorrect measurements. Otherwise your tester might be very low quality. Remember that there are resistance at every connecting points. Also the resistors have a lot of imprecision, so don't trust those values of 100 ohms or 220 ohms. Measure them yourself with a good tester (Fluke).

Hope that helps eh!

Answer 3

You have to take a 2 things into account.

1) How accurate is the meter you are using? When was it last calibrated?

2) You mention the resistor values, are these precision resistors? Non-precision resistors have a nominal value which will vary up to 20%.

Answer 4

1.You can check you bettery in open circuit.
If the reading value is greater than the say value, then ussually it's new battery. (9.61 v in your case)
If the reading value is smaller than the say value, then the bettery is old.
2. voltage drops do not add up to the souce voltage due to the fact that there is internal resistant inside the battery.

Answer 5

The battery has an internal resistance. You should not expect the individual drops to add up to the open battery voltage.

Your meter will have some error. If you are careful you should see <2% error.

Answer 6

You either have a REALLY BAD meter, or a REALLY BAD meter. Either it has errors or it's directly affecting the measurements you're trying to make.

Your assertion about equations being guidelines is, of course, absurd. Everything should add together to the accuracy of the meter used, unless you have a REALLY BAD meter.

Answer 7

You are putting too much thought into this. You won't get readings that line up to the exact 1000th of a volt.