This may be a vintage question, I'm not sure. I am self-studying electronics using some old books and came upon the problem below from "Basic Electronics" by RCA Institutes, 1965. The book is fantastic, but some of the terminology and tech is ancient. The answer to the problem in the book is as follows:
The difference in potential between points C and D is 3 Volts. The potential at point B with respect to ground is +1 Volts; at point C 0 Volts; and at point D -3 Volts.
My read of the circuit is that there are 6 volts in the circuit, but I'm confused by the ground at point C. If C is ground and ground is 0 Volts, then it would seem that 6 volts would drop between the positive terminal and point C, meaning that 4 Volts would drop between A and B and 2 Volts between B and C. Similarly, there would be a 6V drop between the source and point C going from the negative pole to ground.
Obviously, I am mistaken, but I'm at a loss as to how to analyze the circuit and pull out the different potentials. I am hoping y'all can shed some light on how they came up with their values.
Best Answer
Ground is just our reference point, against which we measure voltages. Current is not flowing into or out of it (in this example).
Because the resistance above and below the ground point is equal, we know that current passing over these resistances will drop the same amount of voltage.
The total resistance of the circuit is represented between points A and D...this is where 6 V will be dropped.
Point C is midway through this resistance. Therefore, there is 3 V drop between A and C and another 3 V between C and D.
Using Point C as our reference for "zero volts", point D will be -3 V and point A will be +3 V. This, of course, shows our total of 6 V across the whole circuit.