21write q = cv for all of them and eliminate..
find the equivalent capacitance between 'a' and 'b' ..they are obviously connected to a battery.
** E of the battery isn't given.
please help me out with this ques friends.
my sir did this using kirchoffs loop law..i didn't understand a bit! please give a detailed explaination so that from now on i can do thse kinda
Qs myself..
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11 Answers
30
The main thing to do in any sum of this type is to do the charge distribution. Use symmetry arguments wherever possible.
Now, we have 3 unknowns and hence we need three equations.
The basic idea behind Kirchhoff's Loop Law is that when you take a loop and start from a point and come back to it, the potential difference will be zero, which is also quite obvious.
Now, lets apply it here. Take the smallest top left loop, we move clockwise and write the potential drop,
Q12C - Q3-Q1
2C - Q2C = 0 ...(i)
Now take the larger upper loop, and do the same thing and take the largest loop consisting of 2C and 3C capacitors in the end and get an equation.. You'll have three equations and three unknowns. All you'll need now is some good maths skills and you'll be done. [1]
1we'll get q1 , q2 and q3.. but then hw'll we find the equivalent capacitance...
** emf of the battery is not given
30Thats what.. You use this expression and find it out.
ξ - Q12C - Q33C = 0 [By KLL]
ξ = EMF of the battery = Q1+Q2+Q3Ceq
think why!
30And well you can't use that expression Q1+Q2+Q3 = 0 !
Because Q1+Q2+Q3 is the charge being stored by the equivalent(imaginary) capacitor. It has some finite value.
30I somehow seem to be getting 3C. Answer doesnt matter, I could have made a mistake in calculations. But if you've done it the right way, you'll get the answer.. You dont have the answeR?