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kanwal32
 one year ago
equivalent capacitance of
kanwal32
 one year ago
equivalent capacitance of

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kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434350864779:dw

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino @IrishBoy123

Ehsan18
 one year ago
Best ResponseYou've already chosen the best response.0the capacitors are labelled on the right side with alphabets

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0how to find equivalent capacitance

Ehsan18
 one year ago
Best ResponseYou've already chosen the best response.0first 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

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0answer is 7.5 @Ehsan18 your answer is coming wrong

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434371149308:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1First. lets look at the section A–C

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1First. lets look at the section A–C dw:1434371313237:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1so we can rewrite the diagram... dw:1434371666596:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1similarly, with the section D–B... dw:1434371905223:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1dw:1434372308311:dw

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1No two capacitors in this circuit are either directly in series or directly in parallel, you might need to use simultaneous equations from here

Ehsan18
 one year ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus you didn't add the two capacitors in parallel correctly it is Ce=C1+C2

radar
 one year ago
Best ResponseYou've already chosen the best response.4@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
 one year ago
Best ResponseYou've already chosen the best response.4dw:1434387900902:dw Does this help?

radar
 one year ago
Best ResponseYou've already chosen the best response.4Remove 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
 one year ago
Best ResponseYou've already chosen the best response.4I probably should have said an effective value of 35/6 and an effective value of 10/6 microfarads. Any way when you add them you will get the approved solution of 7.5 uF.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1you are quite right. i added those capacitors wrong, —sorry

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434454183746:dw

radar
 one year ago
Best ResponseYou've already chosen the best response.4@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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