## kanwal32 one year ago Two capacitor having capacitance 8uF and 16uF have breaking voltage 20V & 80V. They are combined in series. The maximum charge they can store individually in the combination is

1. kanwal32

2. kanwal32

@Michele_Laino

3. kanwal32

@Michele_Laino

4. anonymous

5. kanwal32

@IrishBoy123 hlp

6. kanwal32

@UnkleRhaukus hlp

7. kanwal32

@Michele_Laino hlp

8. IrishBoy123

same charge on each, right? so for given charge Q, the voltage across each is Q / 8 microF and Q/ 16 microF so breaks at either Q / 8 microF = 20 V or Q / 16 microF = 80V so Q_break = 160 microC do you agree?

9. anonymous

this should help with that. http://bosscade.com/c?r=CKUEDE4sd

10. IrishBoy123

then C_t = 16/3 microF, meaning V_t = 160 microC / 16/3microF = 30V so V1 = 20V, V2 = 10V agree?

11. IrishBoy123

sorry i'm doing this on hoof or i would draw it and use some latex to make it more clear

12. kanwal32

@IrishBoy123

13. kanwal32

14. IrishBoy123

i think i have given you what you need bigger picture: because they are in series, charges on each will be same but the smaller 8microF capacitance will require a bigger voltage. here the smaller capacitance also has the smaller breaking voltage so it will be the one to break when its drop is 20V. because in series, the total voltage will be split between the capacitors in ratio 1/8 : 1/16 = 2:1 so there will be 10V on 16microF other capacitor and 30V overall. they'll each store 20V*8microF and 10V*16microF = 160micro C

15. kanwal32

16. kanwal32

17. IrishBoy123

What is voltage across 8microF if it holds 200microC Isn't it 25v and aren't you saying that breaks that capacitor so it all shorts out? Makes no sense to me. Wait for someone else to swing by?

As you have learned when capacitors are connected in series, it is the smallest value capacitor that will have the largest voltage drop (voltage across the capacitor). Then the limiting value of applied voltage is the 8 Micro-farad @ WVDC 20. So lets determine what voltage is applied to create the 20 volt drop. Can you do this?

With 20 volts across the 8UF, you would have half that across the 16 uF. That means the max voltage (with in-spec parameters) is 30 Volts. Can you now solve your problem?

The max charge is limited by the max voltage that can be placed on this combination. @Greg_D Is this the way you perceive this problem?

21. Michele_Laino

if we consider the subsequent circuit: |dw:1434638264835:dw| then the electrostatic charge on each capacitor is: $\Large Q = V\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = \frac{2}{3}{C_1}V$ where:$\Large {C_1} = 8\mu F$

22. Michele_Laino

so these conditions hold: $\Large \frac{Q}{{{C_1}}} < 20,\quad \frac{Q}{{{C_2}}} < 80$

23. Michele_Laino

substituting the expression for Q, we get: $\Large V < 30,\quad V < 240$ respectively, so we have to pick the first condition, namely $\Large V < 30$

24. Michele_Laino

substituting into the expression for Q, we get: $\Large \begin{gathered} Q < \frac{2}{3}{C_1}V = \frac{2}{3} \times 8 \times 20 \hfill \\ Q < \frac{{320}}{3}\mu Coulombs \hfill \\ \end{gathered}$

25. anonymous

@radar i agree with your reasoning, maybe @kanwal32 can tell us were did he get his answer from...

I was taught that capacitors connected in series had the same current flowing through them regardless of their values, consequently they would have the same charge. The following is from "Electronics Tutorial" "Although the voltage drops across each capacitor will be different for different values of capacitance, the coulomb charge across the plates will be equal because the same amount of current flow exists throughout a series circuit as all the capacitors are being supplied with the same number or quantity of electrons." Now discussing the circuit with 30 volts applied: 1. The voltage drops will add up to 30 volts. The voltage drops will be as follows the value of the drop will inversely related to capacitor values. In other words, the largest drop will be across the smallest capacitor.....in this case 20 volts across the 8 uF capacitor and 10 volts across the 16 uF capacitor. The charge of each capacitor will be equal as stated above in the quote. This charge will be Q=VC or for the 8 uF, Q=8uF*20 = 160 micro coulombs. The charge across the 16uF will be: Q= VC = 10 * 16 uF = 160 micro coulombs. This checks out. Now how does this check out with the total circuit. First determine the total capacitance Ct= (8 * 16)/(8 + 16) =16/3 uF Now for total charge Q = VC = 30 (16/3) = 160 micro coulombs as the charges on one plate of a capacitor is from the plate from the adjacent capacitor. (which is to be expected.)