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In the circuit below, the Ohm's law calculations of the parts do not agree with the whole:

1. the voltage drops of the parts do not add up to the total voltage, and


2. the current does not calculate to be the same throughout the entire circuit.

Why? Actual VOM readings are in red and done with a digital VOM.

the most probable causses for the discrepancies are the following:

1. The DT830 is not known as a quality meter. Inaccuracies in the readings are one cause of the problem.


2. You are drawing almost 300 ma from a D battery. At that level the voltage of the battery will tend to drop fairly quickly. Depending on how fast you take the measurements, the readings can vary due to that factor. Note that your own readings show a 0.2 drop in voltage from no load to full load.
   As a test of your meter try measuring the voltage of each of your D cells and compare it to the total voltage.

Ohm's Law and KVL always applies but the voltage drop is nonlinear with time due to thermal effects so bulbs never match perfectly and the voltages are never equal.

Bulbs are nonlinear PTC ( positive temp coefficient ) conductors that rise in R by 10x at rated power with temperatures of say 2500'K.

So R (hot/cold) ratio ~ 10:1 for bright warm white light (hot) and room temp (cold).

This means if you put two 6V bulbs in series with a 6V battery, you will never ever get 3V each.

The bulb with even the slightest higher cold resistance heats up faster in temperature will rise in resistance faster and thus drop more voltage than the other bulb, resulting in a runaway condition where that bulb will have full power and the other about 10 %. Putting a resistor in series as you have done reduces the balance difference between R cold= est. 0.5 Ohms and R hot est= 5 Ohms x 0.3A 1.5V

Although this is the extreme case where the bulbs are most sensitive to resistance changes at half voltage, it means your measurements probably changed while you were taking the readings.

Is it repeatable? Is it stable? Measure again.

If the load voltage is 5.9V @ 0.3A and the no-load Vbat=6.1V then the 4 battery cells have an internal resistance ESR = 0.2V/0.3A= 0.67 Ohms total.

The Ohm's law calculations of the parts do not agree with the whole

Actually, there's a very plausible explanation which would mean that Ohm's law is alive and well, and does explain what you report.

(Note: I'm assuming that the bulbs have reached their steady-state "hot" resistances, so that those resistances don't change during the measurements. I'm also assuming that the battery voltage doesn't drop significantly during the measurements. It will drop, but I'm assuming that doesn't happen to a significant amount while measurements are being made.)

Hypothesis: Additional wiring resistances

Your diagram doesn't show additional resistances which must be present in your setup.

Let's look at just one area in your diagram as an example - the voltages across the two bulbs together and individually.

Here is your diagram again (so that readers don't have to keep scrolling up to the question):

Voltage drop across bulb 1 = 1.13V

Voltage drop across bulb 2 = 1.24V

Voltage drop across both bulbs and the wiring between bulbs = 2.6V

Therefore "unexpected" additional voltage drop = 2.6V - (1.13V + 1.24V) = 0.23V

Although the 10.2 Ω resistor value might not be very accurate (e.g. multimeter lead resistance is probably included in that, so the resistor's true value is likely lower) for now, let's use your approximate current value of (2.85V / 10.2 Ω =) 0.28A (to 2 decimal places).

Therefore the additional voltage drop between the two bulbs is only a resistance of (0.23V / 0.28A =) 0.82 Ω which is easy to believe, due to cable type / length / connection resistance etc.

The same reasoning explains why 2.85V (across the resistor) + 2.6V (across the bulbs) doesn't equal 5.9V - there are additional resistances (and therefore voltage drops) in the wiring / connections, which aren't included.

Summary: Go through the actual circuit hardware whose diagram you gave, measuring the voltage drops across not only the obvious parts, but also measure the voltage drop across every wire and connection of any kind.

You will find additional voltage drops which, when added to the voltage drops across the main components already given, I expect will show that the total voltage drops do all sum to the expected value.