Capacitance, unbalanced wheatstone.

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Suraj M
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Homework Statement


Hand-drawn circuit diagram of a capacitor network with nodes A, B, P, Q, R, S, T and U. Capacitors are 5 µF (P to Q), 10 µF (Q to R), 5 µF (Q to T, vertical), 10 µF (U to T) and 15 µF (T to S); charge and voltage labels are written on the wires.

Find capacitance across AB
So i divided the current into q1 and q2 as shown. then i assumed that q₂>q₁.

Homework Equations


##C=\frac{Q}{V}##
I don't know anything after KVL KCL and loop law.

The Attempt at a Solution


So i split the current at T into q₃ and q₂-q₃
by the diagram, i used Loop law for UPQTU and QRSTQ.
this gave me a relation between q₁,q₂ and q₃.
i got
$$ q₁= 15q₃$$
$$q₂=16q₃$$
$$q₁ = \frac{15}{16} q₂$$
using these, i used the equation..
$$ C_{AB} = \frac{q₂ +q₁}{\frac{q₁}{5} + \frac{10}{q₁+q₃}}$$
so i got $$C_{AB} = \frac{155}{23} µF$$
Is this right.? because i doubt it.
 
You work with charges instead of currents. The net charge should be zero at junctions Q and T.
 
can't we do it by this method? KCL necessary? I will try.
 
Suraj M said:
can't we do it by this method? KCL necessary? I will try.
You can, but write charge instead of current. And take into account that the net charge on the connected plates is zero, and the charges on the opposite plates of a capacitor are of opposite sign.

Show your work in detail. Otherwise I can not decide if you got the correct result or not.