if there's an open between points E and G in the circuit diagram above, shouldn't points D and C become a short circuit? If not, why?

our prof said that D and C is not short but i believe it's otherwise.

 

Does the open affect points E -H - G, or E - D - G - or even both paths? It makes a difference to the circuit.

On the other hand, if C & D are connected by a wire, how are they not shorted together? This may be some distinction that equates a short to an accidental or unintended connection, so a deliberate connection is not a short.

 

in the case that the open affect points E -H - G, C and D is short, right? so resistors 2, 3 and 1 will be in series. but if the open only affects points E and D, resistors 4 and 5 will be parallel and they will be in series with resistors 2, 7 and 1. is this correct?

 

Redraw the circuit with those elements left out to see what current paths are left.

 

our prof said that D and C is not short but i believe it's otherwise.

It is and it isn't; depending on your definition of a short.

1) D and C are shorted if you consider that a short means having no resistance between 2 points.

2) D and C are not shorted if you consider that the voltage source has a total resistance (load) of greater than 0 ohms.

(The same goes for points G and F, as well as E and H)

 

... either way, it doesn't affect the outcome of the problem.

 

It is and it isn't; depending on your definition of a short.

1) D and C are shorted if you consider that a short means having no resistance between 2 points.

2) D and C are not shorted if you consider that the voltage source has a total resistance (load) of greater than 0 ohms.

(The same goes for points G and F, as well as E and H)

i mean 1) for short.

here's the circuit ( there's an open between points E and H )

what's left of the circuit:

to get the total current of the circuit, using ohm's law, i have to get what the total resistance first. what i really want to know is, do i have to include R4 and R5 in the total resistance of the circuit?

 

You can redraw your original circuit by making resistors 4 and 5, (parrallel combo) into one resistor, than you have a curret path to the top of it as well as a path through it, but when you open up point E and G, you now have a dangling resistor, (the parrallel combo), which is not connected to anything at the bottom side of it.

 

In the original question the open could occur at points E and D, at points D and G, or both. Thus eliminating resistor 3, 5 or both. I would say there is a number of different ways to look at this problem.

(images shown respectively)