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Physics: Post your doubts here!

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using kirchoff 1st law conservation of charge we can say the current coming out of the cell is 1.2+0.4=1.6A
no we find the potential difference of the 80 ohm resistor. v=ir so v=96 since 1.2*80=96v
so terminal potential is also 96v because parallel components have the same potential difference.
using the formula
terminal p.d=EMF*current(internal resistance) or v=E-IR
we get 96=120-1.6r
so r is equal to 15 so the answr is A
 
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using kirchoff 1st law conservation of charge we can say the current coming out of the cell is 1.2+0.4=1.6A
no we find the potential difference of the 80 ohm resistor. v=ir so v=96 since 1.2*80=96v
so terminal potential is also 96v because parallel components have the same potential difference.
using the formula
terminal p.d=EMF*current(internal resistance) or v=E-IR
we get 96=120-1.6r
so r is equal to 15 so the answr is A
thnx
 
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The Electric Field is upwards hence the Potential of the upper plate should be lower than the lower plate.
EFS= V/d
60000= - x - 80/4x10^-3
x= -320 (A)
I'm getting +320 instead of -320
x-80/0.004=60000 then x is 320
is it -320+-80 or -320--80?
bcos x-80=240
 
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E = (F/A) /( extension /length)
--> Stress : F/A
F = W = mg ( g is constant)
m = density * volume (density is constant as same material )
V = l * l* l ( The Load is a cube )
[The model is one tenth full-size in all linear dimensions.]
V = 1/10*1/10*1/10
V = 1/1000
When V decreases mass decreases and hence Force decreases
F = 1/1000
Area = A = l*l
= 1/10 * 1/10
= 1/100
So Stress = F/A = (1/1000)/(1/100) ---->1/10

----> Strain : e/l
Suppose extension = 1
length = 1/10
Strain = 10

E = stress/ strain = 0.1 / (10)
=> 1/100

Ration of original to model = 1/(1/100)
= 10^2 ANSWER = C
 
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op
I'm getting +320 instead of -320
x-80/0.004=60000 then x is 320
is it -320+-80 or -320--80?
bcos x-80=240
Opps I wrote it the opposite way around -80--x / 4x10^-3 = 60 ,000
We always subtract the smaller potential from the bigger -80 > -320 .
My apologies.
 
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View attachment 62389
O/N 2015 (13)
help out
This is the hardest question!!! I spent a 5 hours trying to solve this!

Physics  Post your doubts here    Page 829   XtremePapers Community.png

First we calculate the Total resistance :
R2 = (R3 + R1) = [1+ 1/2]^-1 = 2/3
R5 + (R4 + 2/3 ) =[ 1 + 1 / (2/3 + 1)]^-1 = 0.625
0.625 + R6 = 1.675 ohm
Total R = 1.675 ohm.

Now we will use V= IR to calculate potential and Current across the resistors and loops.

Current R1 + R3 = 1 A
P.D R1 + R3 = 2 V (( V = 1*(1+10)) = 2V)

(R1 R3) are in parallel with R2 so P.D R2 = 2V

R2 Current = 2/1 = 2 A

Kirchoffs 1st law ; Current In = Current out

Looking at first junction from the left we have 2 A + 1A = 3A ( 2 A--> R2 and 1A--->R3+R1)

3A is leaving hence Current in R 4 = 3A
P.D across R4 = V= IR
V = 3*1 = 3V

Now lwt R1+R2+R3 = Resistor X

Resistor X is in series with R4( P.D = 3V)

Total potential = 2V + 3V = 5V across the whole loop.

So 5V is the potential across R5 as it is in parallel with (R4 + X)
Current across R5 = 5/1 = 5A

Second junction from left :
I in = I out
5 A + 3 A = 8 A

8 A iss the current across the resistor R 6
P.d across R6 = 8/1 = 8 V

E.M.F of the supply = Total potantial across all the loops = 3 V + 2 V + 8 = 13 V

OR

V = I R
V = total R * 8
= 1.635 * 8
= 13 V


AND YOUR ANSWER IS D!! !!! WOW I CAN'T BELIEVE I DID IT!!

You can do it by the Emf thing too for that you dont need to calculate the total resistance but I guess its harder to understand that.

I'd recommend you note all of this down on a note book view it step by step and solve it again for yourself because damn it was hard.
 
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