83μF
- Akash Anand Excellent ..try to give a crisp solution of this.Upvote·0· Reply ·2013-06-06 02:54:48
83μF
We can join the two points connected by the plain wire as there is no potential drop there. The resulting two 1 μF capacitors are in parallel.
c(parallel)=1+1=2
This is in series with the other 1μF capacitor
1/c = 1 + 1/2 = 3/2
c(series)=2/3
This whole thing is in parallel with the 2μF capacitor
c(total)= 2/3 + 2 = 8/3