C1 and C2 are connected in series and so their equivalent capacitance will be:
1/C = 1/10u + 1/3u
=> C = 3.75 x 10-6 F
there are two of these branches and they are parallel to each other. Alongside them, is another capacitor C3 will is parallel to both.
So Equivalent Capacitance will now be: C' = 3.75 x 10-6 + 3.75 x 10-6 + 3 x 10-6 = 10.5 x 10-6 F
this is in series with the lower branch of two C2 capacitors connected in parallel. Therefore, the equivalent capacitance will finally be:
1/Ceq = 1/10.5u + 1/(10u + 10u)
=> Ceq = 6.885 x 10-6 F
this is the equivalent capacitance between a and b.
Consider the following figure. C2 Find the equivalent capacitance between points a and b for the...
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