# Homework Help: Calculate capacitances of 2 unknown

1. Nov 7, 2012

### bsbs

1. The problem statement, all variables and given/known data

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.

Find capacitance of A and B.

2. Relevant equations

The only hint is the voltage drop (20V) across B when 3μF capacitor is connected in parallel with capacitor B. I have problem coming up with logical equations to solve this, can anyone help?

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

#### Attached Files:

• ###### q15.jpg
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Last edited: Nov 7, 2012
2. Nov 7, 2012

### Staff: Mentor

How do do you determine the potential drops across capacitors in series?

Call the capacities A and B. If 200V is applied across the series connection of A and B, what would be the expression for the potential across B?

3. Nov 7, 2012

### bsbs

Blur on forming a maths equation to solve this problem.

Last edited: Nov 7, 2012
4. Nov 7, 2012

### Staff: Mentor

That just restates the given conditions. Suppose you didn't know what the potentials were but were given the capacitor values. How would you calculate the potential across capacitor B?

5. Nov 7, 2012

### bsbs

ok, if supply is 200V, A is 3μF and B is 5μF connected in series.

total capacitance will be 1.8μF
total charge =1.8 X 200 =360μC

Voltage across A = Q/C(A)
= 360μC / 3μF
= 120V

Voltage across B = Q/C(B)
= 360μC / 5μF
= 72V

how do i apply this logic to the above if both A and B are unknown? where i only have the p.d across A 120V, and when 3rd capacitor is applied, it increases to 140V.

Last edited: Nov 7, 2012
6. Nov 7, 2012

### bsbs

attached is the diagram i have prepared.

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7. Nov 8, 2012

### bsbs

really hope someone can help to explain. Am i posting in the correct section? I am new here.

8. Nov 8, 2012

### aralbrec

From the first circuit you have voltages across capacitors A and B. Given that information you can find the ratio of the two capacitors A and B.

The second circuit, again, you can find the ratio of the capacitors A and (B in parallel with 3uF).

Two equations, two unknowns, solve :)

9. Nov 8, 2012

### bsbs

care to show how to solve these 2 unknown? :)

10. Nov 8, 2012

### aralbrec

Ah, check back to post #5. You solved this once already assuming two concrete capacitance values for Ca and Cb. Do it again but this time use unknown capacitance values and call them Ca and Cb.

If the voltage across A is 120V, the charge on cap A is: Q=(Ca)120
Capacitor B is in series so the same charge appears across it. Q=(Cb)80

The second case could be more complicated (a little). You'll have to decide if the 3uF capacitor is added to a live circuit or if the caps are discharged before the new cap is inserted. I think it is the former but it may not matter (you'll have to think on that if you want to solve the simpler problem). Otherwise you will have to begin with the initial charges on Ca and Cb and see what happens when some charge change occurs as a result of the added cap. Follow where this dQ flows, keeping in mind no charges will be destroyed. The final voltage on cap A probably tells you something about dQ.

Last edited: Nov 8, 2012