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Beginner Question--Current Direction/Component Polarity in Circuits

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paperking:
Hi,

So I am reading through my circuit book to learn electronics and I am getting confused about current flows and how the negative and positive signs on elements in a circuit are determined.

What has really confused me is these two example problems in the book that I have attached. One shows a resistor that has its signs reversed and the other I dont understand the algebra they are using to get the answers they are getting.

Can someone clearly explain to me how to determine current direction and stuff to do these problems and for future problems. Thanks.

rstofer:
In the first example, the polarity of V2 is deliberately reversed.  We know the current is circulating clockwise yet they assume a different direction.  Why?  To point out that your assumption doesn't matter.  It just results in a negative voltage.  This is an important concept.  Sometimes you won't really know which direction the current is flowing but Kirchhoff is there to help you out.  Just make an assumption!  If you are wrong, the result will be of an opposite polarity.

In 2.21a, I would start in the lower left corner and run clockwise so +20 - 2i - 3i = 0  or 20 = 5i or i = 4A.  Then V1 = 2Ohm * 4A = 8V and V2 = 3Ohm * 4A = 12V and 8V + 12V = 20V, the original battery voltage.  It all works out!

In figure 2.22, the -8V battery has polarity markings of a voltage gain in the clockwise direction (negative to positive) but the value is negative.  Again, they're trying to point out that your assumption doesn't matter.  You assume a gain of -8V and you wind up with a total battery voltage of 24V and a total current of 4A.

Kirill V.:
Current "direction" in Kirchhoff's laws is a purely conditional concept and you have the right to choose your own direction of current for analysis.If as a result you get a negative value for some current it will mean that "your" direction is not the same as the actual one

ArthurDent:
Someone once told me: “things that are in series are in series with each other”. So in a series circuit like figure 2.22 with a number of components, if you redraw (or just visualize) like components grouped it will be easier to reduce the circuit. Below is what figure 2.22 looks like with some rearranging.

Number 2 on my drawing is what they would lead you to believe is correct but it isn’t. Number 3 is closer but the resistors are still marked wrong. The way I see it is you always measure a voltage drop across any resistor, never gain.  So if the voltage you measure across any resistor with your DMM is positive, swap your positive and negative probes to measure the voltage drop. As far as the battery polarity symbols shown in the original drawing, if you own a piece of battery powered electronic equipment, try ignoring the + and – symbols when you are inserting the batteries and tell me how well that piece of equipment works.

While I understand what they are trying to show with these exercise problems, I look on it like when they were pushing ‘New Math’ which I also just totally ignored.


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