Voltage Polarity Example

The bottom line (points 7,8,9,10) is the common point - all points on that line are at the same voltage - so you should call that line "common", or "ground", and calculate all voltages relative to that line. You calculated the voltage at 4 relative to 5, rather than relative to 9, and mixed up signs somewhere.

If you calculate the voltages at 3 and 4 relative to 7,8,9,10, it should work out correctly.

Point 3 is 20 volts positive from the common point (7,8,9,10), and point 4 is 12 volts negative from the common point, so there is 32 volts between them, with 3 being more positive. So, if you declare 3 as "zero volts), 4 is at -32. But if you declare 4 as "zero volts", 3 is at +32.

Well, you could connect any point in one loop to any point in the other - but you would get a different problem for each pair of points you connect.

Before you can measure the voltage between two points, there must be some path for current to flow between those points. In this case, the author chose to make the path between points 8 and 9.

If you remove the connection between 8 and 9, you won't measure any voltage between 3 and 4, or between 8 and 5, or in general between any point on the left loop, and any point on the right loop. You would, however, still be able to measure voltage differences within either loop.

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Peter Bennett, VE7CEI  
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Just try visualizing the circuit a little differently, and it might become clear (view in fixed font or M$ Notepad):

1.----------.2 | +| | .-. | | | 15V | | | | '-' +| -| 35V --- 3o---------------. - +| | | .-. | | | | 20V | | | | | | '-' | | -| / \\ o----------o ( V ) |7/8/9/10 +| \\_/ | .-. | | | | 12V | | | | | +| '-' | 25V --- -| | - 4o---------------' | +| | .-. | | | 13V | | | | '-' | -| 6'----------'5 created by Andy=B4s ASCII-Circuit v1.24.140803 Beta

This is exactly the same circuit, but by adjusting your perspective a little, it's apparent that there's 32V between the two points of the voltmeter. "Adjusting your perspective" is a lot easier with a just little practice.

You applied the math incorrectly. In fact, if you visualize node

7/8/9/10 as 0V, then node 1/2 is at +35V, and node 5/6 is -25V. That would make node 3 at +20V, and node 4 at -12V relative to the artificial GND at node 7/8/9/10. The difference is 32V.

This also works no matter what node you call 0V/GND. I'll leave that as an instructional exercise for you. And since you've evoked the spirit of J.S. Mill, you might also want to ponder the philosophical implications of this business of arbitrarily putting 0V/GND at any node in the circuit, and everything still works the same. You will greatly profit from this, which would make the neoliberal spirit of Mr. Mill happy.

Good question, and good luck in your studies. Chris

You mean (35V-15V) - (25V-13V) except that the polarity of the second term is reversed too.

^^^

So it's (35V-15V) - ( - ( 25V-13V))

or 20V + 12V = 32V.

And the way the meter's drawn it'll actually read -32V since the postive terminal is connected to the negative voltage.

If 8 and 9 aren't connected there is no common current path and there would be effectively no indication. It helps a bit if you think of the meter needing to draw some current to get this idea straight.

Graham

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Have a look at http://homepage.ntlworld.com/g.knott/elect110.htm

The voltage source on the right is "upside down."

points 7,8,9,and 10 are connected by conductors of negligable resistance. (that's what the straight black lines represent)

for that reason they have the same potential (voltage)

point 3 is 20V more positive than point 8 , point 9 has the same potential as 9 and point 9 is 12V more positive than point 4

thus point 3 is 32V more positive than point 4

if either voltage source was reversed the voltages on the resistors in that half of the circuit would reverse and the 2V you predicted would be present between points 3 and 4.

Bye. Jasen