Electrical, calculate capacitances

[bsbs]

1. Homework Statement

d.c supply 200V
capacitors A and B connected in series
p.d across A is 120V

p.d. across A, increased to 140V when a 3μF capacitor is connected in parallel with B.

attached diagram for better understanding.

Find capacitance of A and B.

2. Homework Equations

C=Q/V
Capacitance total in parallel connection = sum of 'n' numbers of capacitors
Capacitance total in series connection for this case is = 1/Ct = 1/A + 1/(B+3)

3. The Attempt at a Solution

Need help, really don't know where to start with 2 unknown capacitors value.

 

[NascentOxygen]

With capacitors in series, charge on the plates of each capacitor is equal. In case 1, denote the charge Q₁ and write the equations relating Q₁, C_A, C_B, and the voltages.

 

[bsbs]

With capacitors in series, charge on the plates of each capacitor is equal. In case 1, denote the charge Q₁ and write the equations relating Q₁, C_A, C_B, and the voltages.

thanks, i am getting some headway now.

(Qa) = (Ca) x 120
=120μC if assume (Qa) is 1μF

120μC = (Cb) x 80
(Cb) = 1.5

thus, if (Ca)=1, (Cb)=1.5

the ratio is correct now, next hint please, on getting the correct answer.
Now, with the voltage across (Ca) increased to 140V after putting the 3μF in parallel with (Cb)

 

[NascentOxygen]

thanks, i am getting some headway now.

(Qa) = (Ca) x 120
=120μC if assume (Qa) is 1μF

Well, we don't make such assumptions! You keep C_A as C_A and allow the algebra to sort out things (hopefully, eventually it will cancel out).

Now, with the voltage across (Ca) increased to 140V after putting the 3μF in parallel with (Cb)

You repeat the same idea, except now one of the series capacitors has the value (C_B + 3*10⁻⁶)

 

[bsbs]

Well, we don't make such assumptions! You keep C_A as C_A and allow the algebra to sort out things (hopefully, eventually it will cancel out).

You repeat the same idea, except now one of the series capacitors has the value (C_B + 3*10⁻⁶)

unsolved, but thanks anyway.

 

[aralbrec]

unsolved, but thanks anyway.

It's exactly the same thing.

You found the first scenario already:

Q = Ca*120
Q = Cb*80

The charges on each capacitor are the same because they are in series.

So:

120 Ca = 80 Cb


The second scenario is the same.

Q = Ca * 140
Q = Ceq * 60

where Ceq is the equivalent capacitance for Cb in parallel with 3uF


This leads to two equations and two unknowns that you can solve for Ca and Cb.

===

You need to be careful with charge redistribution problems. Here you start with a live circuit from scenario A that charges capacitor B to 80 volts. Then an uncharged 3uF cap is added in parallel with B that causes instant charge redistribution. But since you know the voltages across all capacitors it is easy to determine the charges on them.

If voltages are unknown, you sometimes have to follow how the charge redistributes itself across the capacitors in the second connected scenario. This is done by assuming some unknown charge dQ moves from a plate of one capacitor to others and you follow how the charge moves to other capacitors by honouring series and parallel relationships. In parallel connected capacitors, the charge shares such that the voltage across the caps are equal. In series connected capacitors, the same charge magnitude moves onto the second capacitor's plates.

You will run into problems of this nature in the future.

 

[bsbs]

If voltages are unknown, you sometimes have to follow how the charge redistributes itself across the capacitors in the second connected scenario. This is done by assuming some unknown charge dQ moves from a plate of one capacitor to others and you follow how the charge moves to other capacitors by honouring series and parallel relationships. In parallel connected capacitors, the charge shares such that the voltage across the caps are equal. In series connected capacitors, the same charge magnitude moves onto the second capacitor's plates.

You will run into problems of this nature in the future.

Thanks, you are great help, finally solved, valuable piece of info for charge re-distribution

Ca=8cb/12....eqn 1
Ca=18+6Cb/14...eqn 2
eqn1=eqn2

therefore, Cb = 5.4μF

therefore, 120 Ca = 80 x 5.4

Ca = 3.6μF