Recent content by Pouyan

Thank you for the help! Now I understand that we have to take care of details. But I don't still get it... Why can't we say: We have 9 players, 4 for A and 5 for B, in percent: B = 5/9 = 56 % A = 4/9 = 44% Why can't we say B has about 11% more than A?

The solution in my book: 5/4 = 1.25. That is 25 % more. What I came up with: I thought that now we have totally 9 players. So A: 4/9 and B: 5/9. The difference is 1/9 which is about 11%! A friend told me : The difference between B & A is 5-4=1 The changing rate is (5-4)/5 = 0.2 ! So B has 20...

@Delta2 , Steve4Physics, Orodruin I'm done! Now I know that I could draw the scheme just like this: $$\frac{1}{R_{Left}}=\frac{1}{20\Omega }+\frac{1}{15\Omega }$$ $$R_{Left}=\frac{60}{7}\Omega $$ Together with the right R that is 60 Ohm: $$R_{down}= \frac{60}{7}\Omega + 60\Omega $$ Parallell...

@Delta2 I must say that my knowledge regarding circuit analysis is not sufficient and I want to read my book again ... I make examples that are inside the book, even simpler tasks, but when I get stuck and do not have access to a teacher, I turn either to the internet or if there is the...

No! I read it and do not want anyone to solve my tasks, that's my job ... But some answers make me even more hesitant

Thanks! $$R_{BC} =\frac{60}{7}\Omega $$ $$R_{BCD}=68.57\Omega$$ The total $$R=29\Omega$$

Thanks for this it. But what about C and D? are they in series with each other?

In the right part of A and D which is red; well the current to the left part of A is (12/50)A=0.24A and the same for D is (12/60)A= 0.2A The sum of these currents are 0.44A. Because A and D are parallel the R_AD = 300/11 Ohm; and 0.44 * 300/11 V = 12V. But is the current in the left part of A...

You mean here: The current through the first branch is 12/50 A = 0.24A and the current through the last branch is 12/15A which is 0.8A And the total current is I=0.24A+0.8A=1.04A ?

And you mean the same current is passing through those areas ?

Yes, I know that at intersections we have different currents but got stuck on this task, If we start from the plus pole of the voltage source I know we have a total current I. At the first junction the current will split. But at point P, will the current split again?

The true answer is : The voltage between P and Q is 1.5 V. I got stuck finding the total resistance. My question is: Is B parallel to both A and C? Is C parallel to D ? I tried many ways to find the total R but failed! My first attempt: I say the system ABD is in serie with C...

you mean that the current through B is negligible? or in other words it is so small that we can ignore it?

Thank you! But is B in serie with A,D,C or is in parallell ?

The true answer is B. But I don't understand why! I know: Kirchhoff's circuit laws : ∑In=0 If we assume that a current that goes from plus to minus, before it passes through lamp B, I know that according to Kirchhof's laws, part of the current will pass through the bottom path where there is...