Simple KVL (kirchhoff's voltage law) problem

[PainterGuy]

hello everyone,

my book says:
\sum E = \sumIR

where E is emf. it says the sum of all emfs equal sum of all IR terms.

1:-- please have a look on this PDF
https://docs.google.com/viewer?a=v&...ljMzctZjQxNzNlNjZlMGMy&hl=en&authkey=CNnjye8P

in the highlighted part - the equation #3 - there is no "E". how can we apply KVL when there is no "E".

2:-- and in this PDF (the highlighted part):
https://docs.google.com/viewer?a=v&...gwOTYtOGQzN2NmMmFjNTRh&hl=en&authkey=CM62nsYK

there is no IR terms and V_cd is neither E nor IR. so how can we apply KVL here?

please help me with this KVL problem as soon as possible. i am very grateful for your helping me out. many thanks.

cheers

 

[PainterGuy]

hello,

will someone please help me with this? it will be generous of you. is what book says incorrect?

cheers

 

[PainterGuy]

please, please, help me on this.

cheers

 

[varialectio]

It's a very long time since I did anything like this but I can't see a problem.

In (1) in loop 3, if E=0, then sum(IxR)=0 which can only be achieved by having I=0. There is no voltage driving a net current round the triangle of three resistors so that seems sensible.

In (2) we know sum(IxR)=0 because it is an open circuit so sum(V)=0 round the closed loop You can think of it as calculating the battery that would have to be connected across C-D to exactly cancel out the emf provided by the two that are already there so that no current would flow when the loop is closed.

 

[PainterGuy]

hello everyone,

my book says:
\sum E = \sumIR

where E is emf. it says the sum of all emfs equal sum of all IR terms.

1:-- please have a look on this PDF
https://docs.google.com/viewer?a=v&...ljMzctZjQxNzNlNjZlMGMy&hl=en&authkey=CNnjye8P

in the highlighted part - the equation #3 - there is no "E". how can we apply KVL when there is no "E".

2:-- and in this PDF (the highlighted part):
https://docs.google.com/viewer?a=v&...gwOTYtOGQzN2NmMmFjNTRh&hl=en&authkey=CM62nsYK

there is no IR terms and V_cd is neither E nor IR. so how can we apply KVL here?

please help me with this KVL problem as soon as possible. i am very grateful for your helping me out. many thanks.

cheers

the page in the link "1" was not properly rotated. here is proper page:--
https://docs.google.com/viewer?a=v&...I4ODgtNGUwODhiZTE4MDM5&hl=en&authkey=CPzZ4vsE

It's a very long time since I did anything like this but I can't see a problem.

In (1) in loop 3, if E=0, then sum(IxR)=0 which can only be achieved by having I=0. There is no voltage driving a net current round the triangle of three resistors so that seems sensible.

In (2) we know sum(IxR)=0 because it is an open circuit so sum(V)=0 round the closed loop You can think of it as calculating the battery that would have to be connected across C-D to exactly cancel out the emf provided by the two that are already there so that no current would flow when the loop is closed.

Hi varialectio,

Many thanks for this help. I would keep on reading your reply several times and if I have any question I will ask. Much grateful this kind help.

Cheers