Why Does Adding Resistors Change the Calculation in Circuit Analysis?

[kostantina]

I am trying to follow examples solved by the publisher of my book in order to understand the problem. However, I can't understand why he is solving it like this. What is confusing me, is why v1=(12+8)*1/8

why is v1 not 12*(1/8). Why is he adding the 8ohm resistor in there? Any help would be greatly appreciated.

Heres the link to the practice Problem. Its it P.P.4.2 (the second one) Thank you.


http://highered.mcgraw-hill.com/sites/dl/free/0073380571/938347/Chapt04PP_120121.pdf

 

[Simon Bridge]

You mean ex. P.P.4.2?

That relation is because V=IR ... he's added the 8Ω resister because the voltage is taken across the 12 and the 8 in series. Your calculation would give you the voltage across the 12Ω resister only. That would be 1.5V ... with the 1V drop across the 8 gives a total of 2.5V.

The question being answered seems to be, how does the supply voltage relate to the voltage across the 8Ω resister?

 

[kostantina]

Thank you. I m new to this topic so I appreciate your feedback on something that may seem simple to most of you in here.

 

[Simon Bridge]

To be fair - it was not obvious from the diagram what v1 was in relation to.
You need two points to make a voltage - not just one like they drew.

Sometimes the voltage symbol is clearly associated with a component like with V0 and Vs - then the voltage across the component is intended. When it is associated with a point on a wire - in this case, a junction - then it is usually safe to take it between that point and an obvious 0V point which, in this case, would be the position of the ground symbol.

If you have to guess like that - write it down as part of your working.
Cheers.