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Electricity


Niha afzal

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Look at the following correspondences.

 

When all the rates of change are uniform.

 

velocity is rate of change of distance : current is rate of change of charge

So distance ( symbol s) corresponds to charge (symbol q)

 

acceleration is rate of change of velocity : I will call rate of change of current p

 

The kinematics equation

 

s = ut + 0.5 (f t2) Where s is distance, u is initial velocity, t is time and f i acceleration.

 

This equation also applies to your uniform change of current.

 

i = initial current = 100mA

 

the current changes from 100mA to 20 mA in 8 seconds ie (20 - 100) /8 = -10 mA per second.

So p = 10 mA/s2

 

Changing our formula from

 

s = ut + 0.5 (f t2)

 

to

 

q = it + 0.5 (pt2) = 8

 

q = 100 * 8 + (-10 *64) millicoulombs

 

which I think you will find is answer A

 

Edit

 

q = 100 * 8 + 0.5(-10 *64) millicoulombs

 

which I think you will find is answer C

Edited by studiot
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Good to see you checked my arithmetic, and found the missing 0.5.

 

Students should always do this.

 

:)

 

Incidentally what are you studying

Physics or Engineering Science?

 

I chose this method because I don't know if you have any calculus, which would be the standard method.

Edited by studiot
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I did basic calculus .... im still doing my alevels ...therefore I have both physics and maths .. but basic math

This kinematics working was amazing.. who knew we could relate kinematics and elctricity like this.... this was amazing seriously yet im scared how will I figure out stuff like this. My best bet is practising a loooot! I find no words to thank you :')

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