!PHYSICS! Anyone willing to share their modest mental qualities you are welcomeXD I will give owl bucks

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!PHYSICS! Anyone willing to share their modest mental qualities you are welcomeXD I will give owl bucks

Mathematics
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No.
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
My answers owe thanks to the following questions

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

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