The total capacitance of the circuit above is?

NPXT

5 years ago

3 equal capacitances of 2mF are connected in parallel, therefore equivalent capacitance for parallel:
C(parallel) = 2mF+2mF+2mF=6mF
Observe in the diagram that the parallel capacitors are then connected to another of 2mF in series, therefore total capacitance in the system:
1/C (series) = 1/C +1/C
1/C = 1/6mF + 1/2mF
1/C = 4mF/6mF
1/C = 2/3 (mF cancels out)
Therefore C=3/2
C=1.5

hernyhorba

2 years ago

Diagram

zion400

2 years ago

No diagram

DiamondBlack

8 months ago

because I thought series is c1 + c2 + c3... and parallel is 1/c1+ 1/ c2 + 1/ c3...
but you guys are saying the opposite of what I was thought and what is in my physics text book

Spongebobandoscar

2 years ago

mF was not included

Tesshunters

4 years ago

The answer to the question is not in the option because the solution is meant to be
Considering 2mF in series
1/CT=1/c1+1/c2
=1/2+1/2=1+1/2=2/2=1
CT=1mF
In parallel
Ct=c1+c2+c3
2+1+2=5mF or 5×10^-6F

Therefore the final answer is meant to be 5mF