kanwal32 one year ago equivalent capacitance of

1. kanwal32

|dw:1434350864779:dw|

2. kanwal32

@Michele_Laino @IrishBoy123

3. kanwal32

@Ehsan18

4. anonymous

|dw:1434365430719:dw|

5. anonymous

the capacitors are labelled on the right side with alphabets

6. kanwal32

ok

7. kanwal32

how to find equivalent capacitance

8. anonymous

first A+C=C1 E+G=C2 then 1/D+1/C2=C3 C3+B=C4 1/C1+1/C4=C5 C5+F=Ce(Equivalent capacitance) just put the values yourself

9. anonymous

The answer is 10.1144

10. kanwal32

11. kanwal32

@IrishBoy123

12. kanwal32

@UnkleRhaukus hlp

13. kanwal32

@IrishBoy123

14. UnkleRhaukus

|dw:1434371149308:dw|

15. UnkleRhaukus

First. lets look at the section A–C

16. UnkleRhaukus

First. lets look at the section A–C |dw:1434371313237:dw|

17. UnkleRhaukus

the capacitance in this section can be found with the formula for parallel capacitance $\frac1{C_\text{qe}}=\frac1C_1+\frac1{C_2}$ ie $C_\text{eq}= \left(\frac1{12}+\frac1{23}\right )^{-1} = 276/35\approx7.89$

18. UnkleRhaukus

so we can rewrite the diagram... |dw:1434371666596:dw|

19. UnkleRhaukus

similarly, with the section D–B... |dw:1434371905223:dw|

20. UnkleRhaukus

|dw:1434372308311:dw|

21. UnkleRhaukus

No two capacitors in this circuit are either directly in series or directly in parallel, you might need to use simultaneous equations from here

22. anonymous

@UnkleRhaukus you didn't add the two capacitors in parallel correctly it is Ce=C1+C2

@Ehsan18 You're right, The total capacitance value of two capacitors connected in parallel is simply the sum of each capacitor value. It is the series configuration that is similar to resistors in parallel.

|dw:1434387900902:dw| Does this help?

Remove the 13 uF (mentally) You now have a pair of series capacitors one with a value of 35/6 uF and the other with a value of 10/6 uF connected in parallel. Now add those two values.

I probably should have said an effective value of 35/6 and an effective value of 10/6 micro-farads. Any way when you add them you will get the approved solution of 7.5 uF.

27. UnkleRhaukus

you are quite right. i added those capacitors wrong, —sorry

28. UnkleRhaukus

@radar i have followed your method and got the same result, and it now makes sense to me why the 13µF cap can be effectively removed, on both branches either side of the 13, the ratio of capacitors is equal 2/10 = 7/35 = 1/5; the charge flows will be equal, and so there will be no potential difference between C and D. - thank you

29. kanwal32

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