anonymous
  • anonymous
!PHYSICS! Anyone willing to share their modest mental qualities you are welcomeXD I will give owl bucks
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
abb0t
  • abb0t
No.
anonymous
  • anonymous
1. Since each light bulb carries a resistance of 120 Ω, the total resistance of series circuit is calculated as follows, Rs=120Ω+120Ω+120Ω+120Ω+120Ω=600Ω The total resistance in this series circuit is therefore 6.0*10^2. On the other hand, if the light bulbs were placed in parallel manner, then the total resistance will be, 1/R=1/120+1/120+1/120+1/120+1/120 Isolating for R 1=R(1/120+1/120+1/120+1/120+1/120) 1=0.04166R R=24.000 Therefore the total resistance in parallel circuit is 24Ω 2. From the diagram, junctions of the circuit are a. H and B b. Loops of this circuit consist of the loop ABCH and BCHDFG 1. The electrical potential difference for the first diagram is calculated by reasoning with the fact that in series circuits voltages will be distributed in such ways that add up to the voltage at the battery depending on the resistance on each load. Therefore, Delta V2=60V-2.5V-1.1=56.4V Therefore the voltage at Delta V2 is 56.4V Since the voltage can be added to find the voltage at battery, 5.4V+7.5V+4.2V=17.1V Therefore the electrical potential difference at battery is 17.1V
anonymous
  • anonymous
My answers owe thanks to the following questions

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More answers

anonymous
  • anonymous
Six 120 Ω light bulbs are connected in series. What is the total resistance of this circuit? What would be the total resistance if they were connected in parallel?
anonymous
  • anonymous
that is a disappointing result @abb0t
anonymous
  • anonymous
Good try though Mr. Boo
anonymous
  • anonymous
XDXDXD
abb0t
  • abb0t
Perhaps @pooja195 can help since she seems to be so obsequious about medals.
pooja195
  • pooja195
@abb0t have nothing to say? DONT SAY IT. I havent taken physics yet so i cant help sorry.
anonymous
  • anonymous
OMG, no diagrams?
anonymous
  • anonymous
@pooja195 He was jokingXD
pooja195
  • pooja195
50 tags http://prntscr.com/84lxqy NOT a joke hes being annoying
anonymous
  • anonymous
|dw:1439458995536:dw|
anonymous
  • anonymous
What a beautiful diagram I have ever seen!
abb0t
  • abb0t
Those are in series. Yes.
anonymous
  • anonymous
That's undoubtedly the series diagram on the bus
anonymous
  • anonymous
Are you serial @abbot
anonymous
  • anonymous
now you both taking the piss!
anonymous
  • anonymous
That's a british slang
anonymous
  • anonymous
Is this too simple a problem for @abott?
anonymous
  • anonymous
There are genuine people that need help, so I will leave this problem to @abb0t
abb0t
  • abb0t
Just add them all up in series. You would add them inversely if they were parallel
anonymous
  • anonymous
For a second question for 99 smiling guy I summon you guys to kick me on this diagram
1 Attachment
anonymous
  • anonymous
Junctions, branches, and loops are my objectives. Give me the coordinates
anonymous
  • anonymous
Physicist is here!!!
anonymous
  • anonymous
@Michele_Laino
ali2x2
  • ali2x2
i have no idea ive never taken physics
Michele_Laino
  • Michele_Laino
the first answer is right!
anonymous
  • anonymous
Yay!
anonymous
  • anonymous
Second question I need to identify branches, junctions as well as loops from the diagram
anonymous
  • anonymous
but they are kind of complicated for my immature self
anonymous
  • anonymous
I forgot to include the branches but whatever
Michele_Laino
  • Michele_Laino
what we have to compute for question #2?
anonymous
  • anonymous
No we just need to name the points separately for junctions, branches, and loops
Michele_Laino
  • Michele_Laino
we can simplify your circuit like this: |dw:1439571599988:dw|
anonymous
  • anonymous
so junctions are points where more than 3 lines are intersecting ?
Michele_Laino
  • Michele_Laino
yes! at least 3 lines
anonymous
  • anonymous
So for junction it's safe to say B and H
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
The branches are extensions of junction right?
Michele_Laino
  • Michele_Laino
yes! I think so
anonymous
  • anonymous
So it's safe to say they are C and A right
Michele_Laino
  • Michele_Laino
yes A and c are branches
Michele_Laino
  • Michele_Laino
also D, F and G are branches
anonymous
  • anonymous
Great! Now loops are flowing circuits in a circular manner so they are as my answer says "Loops of this circuit consist of the loop ABCH and BCHDFG" or do I use shorter form of saying that
anonymous
  • anonymous
May I ask why?
Michele_Laino
  • Michele_Laino
it is correct, we have 2 loops ABCH and BCHGFD
anonymous
  • anonymous
Ok. That's nice. What about the question 3 with the diagram where I need to look for the missing Voltage?
1 Attachment
Michele_Laino
  • Michele_Laino
by definition, we say that a loop is a continuous path of conductors
anonymous
  • anonymous
The electrical potential difference for the first diagram is calculated by reasoning with the fact that in series circuits voltages will be distributed in such ways that add up to the voltage at the battery depending on the resistance on each load. Therefore, Delta V2=60V-2.5V-1.1=56.4V Therefore the voltage at Delta V2 is 56.4V Since the voltage can be added to find the voltage at battery, 5.4V+7.5V+4.2V=17.1V Therefore the electrical potential difference at battery is 17.1V
anonymous
  • anonymous
That's nice to know
Michele_Laino
  • Michele_Laino
first circuit, we have: 2.5 +1.1 +V= 60, where V is the missing voltage
Michele_Laino
  • Michele_Laino
so you are right!
anonymous
  • anonymous
Yay!
anonymous
  • anonymous
I assume the second problem is very identical to the first as well?
Michele_Laino
  • Michele_Laino
second circuit: we can write this: 4.2 +5.4 +7.5 = Vt
Michele_Laino
  • Michele_Laino
so you are right again!
anonymous
  • anonymous
Nice! The rules of sig figs don't apply to this particular addition problem does it?
Michele_Laino
  • Michele_Laino
more explanation: in both circuits we have applied this subsequent law, which is valid for an electrostatic field: \[\Large {\text{rot }}{\mathbf{E}} = {\mathbf{0}}\]
anonymous
  • anonymous
When all the electrical potentials are summed it results in zero
Michele_Laino
  • Michele_Laino
in other words the integral of the electric field along a circuit, or a loop, has to be equal to zero
Michele_Laino
  • Michele_Laino
when you are doing a computation for a single loop
anonymous
  • anonymous
Wow
Michele_Laino
  • Michele_Laino
more precisely, for a single loop, the algebraic sum of the voltage drops (along the resistance) and the voltage of batteries, has to be equal to zero that is the second law of Kirchhoff
anonymous
  • anonymous
Where did you complete your physics degree?
anonymous
  • anonymous
I am looking to do physics degree too after medicine
Michele_Laino
  • Michele_Laino
at the University of Pisa, here in Italy
anonymous
  • anonymous
Maybe I should have taken physics instead XD
Michele_Laino
  • Michele_Laino
ok! Physics is a good choice, congrats!! :)
anonymous
  • anonymous
Are you doing your thesis for PhD yet?
Michele_Laino
  • Michele_Laino
no, I have not the PhD title
anonymous
  • anonymous
I heard that with PhD in physics you can make like 200 grands working for national search institutions and stuff
anonymous
  • anonymous
so physicists are richer than doctors if they are really good
anonymous
  • anonymous
and physicists are bunch of smartest people on earth
Michele_Laino
  • Michele_Laino
at the moment I have no job, lol!
Michele_Laino
  • Michele_Laino
I think that a physicist is a man who tries to understand our universe
Michele_Laino
  • Michele_Laino
or more simply all what is around us
anonymous
  • anonymous
Yeah! Physicist is a revolutionist
Michele_Laino
  • Michele_Laino
at the moment please think about to get your degree in medicine, and after that you can think about physics, right?
anonymous
  • anonymous
Yeah I think so.
anonymous
  • anonymous
I can work as a doctor and still go to university for physics.
anonymous
  • anonymous
To be like you
Michele_Laino
  • Michele_Laino
thanks!! :)

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