For the four capacitors in the circuit shown in the figure below,
CA = 1.00 µF,
CB = 4.50 µF,
CC = 2.40 µF,
and
CD = 3.40 µF.
What is the equivalent capacitance between points a and b?...µF
www.webassign.net/katzpse1/27-p-030.png
call intersection at top also a because just a wire from left to top
call intersection at bottom also b, beacuse just a wire again
well, first deal with Cb andCc in series
C = Q/V in general
same charge q on Cb and Cc
so voltage on Cb =q/4.5 and voltage on Cc = q/2.4
so total voltage top to bottom = V = q/4.5 + q/2.4 = .6387 q
Cbc = q/V = 1/.6387 = 1.57µF
NOW we just have three capacitors in parallel between a and b
Ca = 1.0
Cbc = 1.57
Cd = 3.4
Just add them :)