Series shrinks, parallel swells — and the gap is fifteen-fold
The same three capacitors, wired two ways. In series they collapse below their smallest member; in parallel they simply add. The ratio between the two outcomes is the whole question.
Three capacitors of \(3\,\mu\text{F}\), \(9\,\mu\text{F}\) and \(18\,\mu\text{F}\) are connected first in series and then in parallel. Find the ratio of equivalent capacitances in the two cases, \(\dfrac{C_S}{C_P}\).
- Capacitor 1C₁ = 3 μF
- Capacitor 2C₂ = 9 μF
- Capacitor 3C₃ = 18 μF
- ArrangementsSeries & Parallel
The ratio \(\dfrac{C_S}{C_P}\), where \(C_S\) is the series equivalent and \(C_P\) the parallel equivalent of the same three capacitors.
Capacitors are the opposite of resistors. Watch the formulas flip:
Series — reciprocals add, so the total is smaller than the smallest:
$$\frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$Parallel — capacitances add directly, so the total is the sum:
$$C_P=C_1+C_2+C_3$$Because series always pulls down and parallel always piles up, \(C_P\) is the largest possible value and \(C_S\) the smallest — the ratio is always tiny.
- Series — use LCM 18. $$\frac{1}{C_S}=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}=\frac{6+2+1}{18}=\frac{9}{18}=\frac{1}{2}$$ $$\Rightarrow\;C_S=2\,\mu\text{F}$$
- Parallel — just add. $$C_P=3+9+18=30\,\mu\text{F}$$
- Take the ratio. $$\frac{C_S}{C_P}=\frac{2}{30}=\frac{1}{15}$$
Parallel is free arithmetic — just add: \(3+9+18=30\). That's nearly always the fast half.
For the series half, pick the LCM of the values as your common denominator (here 18). The numerators \(6,2,1\) sum to \(9\), giving \(\tfrac{9}{18}=\tfrac12\), so \(C_S=2\). Then \(C_S:C_P = 2:30 = \mathbf{1:15}\).
Sanity check the direction: series < smallest member (2 < 3 ✓), parallel = sum (30 ✓). The ratio must be \(C_S:C_P\), a number well below 1 — that alone kills options (b) and (c).
With \(C_S = 2\,\mu\text{F}\) and \(C_P = 30\,\mu\text{F}\):
$$\frac{C_S}{C_P}=\frac{2}{30}=\frac{1}{15}$$