Hello, I have this circuit for which I have solved the current for the two loops indicated.
I am now trying to understand how Kirchoff's voltage law applies to the right hand loop.

eg. It starts off with 15V, then drops by 9V (9 ohm * 1A) (due to the parallel and series resistor combination represented by my diagram B). That brings it to 6V.
But I am now struggling to understand how the remaining 6V is dropped.
I have attempted to do the remaining drops on the lower right hand side of the diagram.
I think it is like this:
6v dropped by 1A * 2 ohm = 2V (6-2=4), and then (from loop 1) 2A * 2 Ohm = 4V (4-4=0 ?

Can anyone please confirm if I am correct here.

Thanks kindly for any help.

 

Hello, I have this circuit for which I have solved the current for the two loops indicated.
I am now trying to understand how Kirchoff's voltage law applies to the right hand loop.

eg. It starts off with 15V, then drops by 9V (9 ohm * 1A) (due to the parallel and series resistor combination represented by my diagram B). That brings it to 6V.
But I am now struggling to understand how the remaining 6V is dropped.
I have attempted to do the remaining drops on the lower right hand side of the diagram.
I think it is like this:
6v dropped by 1A * 2 ohm = 2V (6-2=4), and then (from loop 1) 2A * 2 Ohm = 4V (4-4=0 ?

Can anyone please confirm if I am correct here.

Thanks kindly for any help.

Do a KCL at the top node, the current through the branch with the 15V source will be 3amps.

3A * 2Ω = 6V, is that the 6V you are looking for?