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Physics: Post your doubts here!

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so frst well calculate I3 i.e : 12/6 = 2 A
dn I2 i.e = 12/10 ( resistance of wire given in ques) =1.2A
nd as I1 = I2+I3 so it wud b = 1.2 + 2 = 3.2 A
 
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Do u know how to do part b of this question ?

um y dun u take a look at the marking scheme im sure ull gt it :) here :

p.d. BC: 12 – 12 × 0.4 = 7.2 (V) / p.d. AC = 4.8(V) C1
p.d. BD: 12 – 12 × 4 / 6 = 4.0 (V) / p.d. AD = 8.0(V) C1
p.d. = 3.2V
 
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um y dun u take a look at the marking scheme im sure ull gt it :) here :

p.d. BC: 12 – 12 × 0.4 = 7.2 (V) / p.d. AC = 4.8(V) C1
p.d. BD: 12 – 12 × 4 / 6 = 4.0 (V) / p.d. AD = 8.0(V) C1
p.d. = 3.2V
weeelll ... actually i didn't get it .. :(
 
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Is this correct?
Blue line is where maxima would be observed and red line is where minima would be observed.
 
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because 1m is the total length and the part of the length they are taking is 0.9 as u know R is directly proportional to length.
and then the resistance of the total length is 4 ohm they multiply by 4.
if u don't get it this way then use the ratio method: -

0.9 : x
1 : 4

cross multiply and u willl get (0.9/1) * 4
 
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