@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...
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 ?
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...
The true answer is B. But I dont 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 no...