Question about an electronic homework (find the current)

[TetraSky]

Hi!

Got a question for anyone who can help on this. 

So my electronic teacher just put our a new homework... and it's literally something we've never seen so far in class (I would know, I recorded every single online classes so far and scrubbed through each recordings... NOTHING. Same with the documents received over the course of the past 6 weeks)

 

 

Knowing that the circuit is 12V, and the total resistance is 27ohm, we're supposed to calculate the current in the circuit... Which normally would be as easy as 12V/27ohm.

But there's these weird looking shower head crap there and that's throwing me off. I believe those are the "chassis ground", even if there's one prong too many?

I know those are supposed to represent 0V...

But from my understanding, does that mean R3 just... Disappear because the power doesn't reach it and dissipate into the chassis ground before reaching it between R2 and R3 ?

Which means to find the current, I only use R1 and R2 ? I = V/R ... So 12V/16ohm = 0.75A for the circuit. Am I correct in my assumption here or am I completely off my rocker ? I literally can't find anything about this sort of dumb problems on Google.

[mahyar]

yeah 12v/16ohm=0.75 amps 

[xentropa]

The ground on the negative terminal of the battery and at the end of R2 causes the R3 to be bypassed.

[Moonzy]

u can draw the common ground together with wires, then it would make more sense

 

but it's 12/(9+7)

[TetraSky]

I see, so it really was like I thought. Good to know. Thanks for clarifying this !

 

I really don't like having to make assumption about stuff without knowing for sure it's right (even if technically, that's good and all to develop "critical thinking"...). The fact we had literally never seen this in classes or in any documentations, really fucked with my head.

 

Will update in a week or two when I get the noted copy back and if for some reason, it wasn't that.