kanwal32
  • kanwal32
equivalent capacitance of
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
kanwal32
  • kanwal32
|dw:1434350864779:dw|
kanwal32
  • kanwal32
@Michele_Laino @IrishBoy123
kanwal32
  • kanwal32
@Ehsan18

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More answers

Ehsan18
  • Ehsan18
|dw:1434365430719:dw|
Ehsan18
  • Ehsan18
the capacitors are labelled on the right side with alphabets
kanwal32
  • kanwal32
ok
kanwal32
  • kanwal32
how to find equivalent capacitance
Ehsan18
  • Ehsan18
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
Ehsan18
  • Ehsan18
The answer is 10.1144
kanwal32
  • kanwal32
answer is 7.5 @Ehsan18 your answer is coming wrong
kanwal32
  • kanwal32
@IrishBoy123
kanwal32
  • kanwal32
@UnkleRhaukus hlp
kanwal32
  • kanwal32
@IrishBoy123
UnkleRhaukus
  • UnkleRhaukus
|dw:1434371149308:dw|
UnkleRhaukus
  • UnkleRhaukus
First. lets look at the section A–C
UnkleRhaukus
  • UnkleRhaukus
First. lets look at the section A–C |dw:1434371313237:dw|
UnkleRhaukus
  • 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 \[ C_\text{eq}= \left(\frac1{12}+\frac1{23}\right )^{-1} = 276/35\approx7.89 \]
UnkleRhaukus
  • UnkleRhaukus
so we can rewrite the diagram... |dw:1434371666596:dw|
UnkleRhaukus
  • UnkleRhaukus
similarly, with the section D–B... |dw:1434371905223:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1434372308311:dw|
UnkleRhaukus
  • 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
Ehsan18
  • Ehsan18
@UnkleRhaukus you didn't add the two capacitors in parallel correctly it is Ce=C1+C2
radar
  • radar
@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.
radar
  • radar
|dw:1434387900902:dw| Does this help?
radar
  • radar
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.
radar
  • radar
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.
UnkleRhaukus
  • UnkleRhaukus
you are quite right. i added those capacitors wrong, —sorry
UnkleRhaukus
  • 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
kanwal32
  • kanwal32
|dw:1434454183746:dw|
radar
  • radar
@UnkleRhaukus I actually used your diagram, you obviously understood the configuration, just a momentary slip on effective capacitance when capacitors are connected in parallel. I have followed many of your responses in the past, I know this was just a "momentary slip" and you do an excellent job of explaining your methods. Keep up the good work.

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