Hi sir , I have two questions ,one is that the answers given about the circuit are correct or not in the images corr or not
when i am solving ,dont getting these answers ,and i have read in comments somewhere something like that “in this book there are mistakes in answers or solutions”(i dont exactly remember the words) and the book is fundamental of electric circuit 5th edition charles k alexander and mathew n.o sadiku., another problem that i solve from this book and found same issue that my answer is 18.6 and book's answer is 19 and it was practice problem 2.10 of that book,and its difficult for me to find my solved answer of corr or not,from my register,but that was not the same as given in book. .
my second question is about the image 5 o 6 ,and the question is that when they combine the last resistor of 5 ohm, with 1 ohm in its series,why they place the equivalent 6 ohm in the place of 5 ohm and why not to the place of 1 ohm,?is there any general rule?or just because there will a parallel combination?and why they did’nt solve in the way to add 4 2 2 and 8 ohm in series and make parallel with six,and in this case,answer would be different ,actual answer is 14.4 and in this case,the answer will be 4.3. .
For homework problems we like to give advice and counsel but not the work of the solution.
So, I'll answer your second question about reducing the resistor network to a single equivalent value.
ONE: You cannot combine the circuits in a way that changes the nature of the circuit.
TWO: Series resistors can be added together to give and equiv. of that sum.
THREE: Parallel resistors can be replaced with an equivalent R of (R1 X R2)/(R1 + R2).
Let's look at it. First add the series components: 5 +1 = 6 ohms. That's as far as we can go for now. Then look at the parallel components: 6||3 = (6X3)/9 = 2. Now we have another series that can be added: 3 +2 = 5. Now we have the 5 in parallel with the 2 + 2= 4. That 4 is in parallel with the 6, or 6x4/10 = 2.4. Now you have a series of the 4 + 2.4 + 8 = 14.4.
Your solution changed the circuit.