Q. In Lesson 2324B-8 on page 39 the book shows that connecting a wire between the circuits gives a voltage reading of 9v between points A and B. I’ve been racking my brain trying to figure out how we came up with 9v. Can you please explain?
A. Let's see how the value of 9V was calculated.
The first portion of that figure shows there are two separate circuits involved. The one on the left is a series circuit of two resistors, while the one on the right only has one resistor being used as the one that is not connected on the left hand side cannot and will not have a current flow through it due to the fact there is no complete path through it.
Now if we analyze those two circuits based upon that information, there will be a 15V drop across R2, and a 6V drop across R4. We know those facts from the basis of 20V/20Ω = 1A and 1A x 15Ω = 15V, while the other is just 6V across the only component of 60Ω.
The meter connected between points A and B does not have a potential measurement due to the fact that there is nothing to make a common connection between those two circuits. Granted, you could try to say the meter is the common element, but as it is measuring volts and has a very high resistance when in that mode, it becomes less of a possibility.
Now when we do connect the common line as shown in (b), those two voltages we calculated before are still there, but now we do have the two playing nicely together in the same sandbox.
Therefore, when we measure from point A which is at the 15V potential, and point B which is at the 6V potential, we arrive at the potential difference between them of 9V. That is how we got to the answer.
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