IMAT 2012 Q74 [Capacitators]

Three identical capacitors are connected as follows:


Which of the following shows the order of increasing capacitance (smallest first)?

A. 3, 2, 1
B. 1, 3, 2
C. 2, 1, 3
D. 1, 2, 3
E. 2, 3, 1

This is a simple, straightforward question. The only information you would need to know are 2 facts:

  • When connected in parallel, the capacitance of the capacitors is added:
    ^Ct= ^C1+^C2 +^C3

  • When connected in series, the capacitance of the capacitors is calculated by finding sum of the reciprocals inverted:
    \frac{1}{^Cn} = \frac{1}{^C1}+ \frac{1}{^C2} + \frac{1}{^C3}

Therefore it is easy to conclude that the largest capacitance (for an identical number of identical capacitors) would be connected in parallel.

1 is connected in series, making it smallest. Between 2 and 3, 3 has more capacitors connected in parallel. So our ranking becomes 1-2-3, option D.