A square loop has two ways round — always count both
Four equal capacitors form a ring. Between any two terminals there's a short hop and a long detour, sitting in parallel. Adjacent corners and diagonal corners give different totals — and their ratio is fixed.
Four capacitors, each \(3\,\mu\text{F}\), are connected in a square loop as shown. Find the ratio of the equivalent capacitance between A and B to that between A and C.
- Each capacitorC = 3 μF
- Count4 (square ring)
- CornersA · B · C · D
- Terminals usedA–B and A–C
The ratio \(\dfrac{C_{AB}}{C_{AC}}\): equivalent capacitance across adjacent corners A–B versus across diagonal corners A–C.
A ring always offers two routes in parallel. Pick the two terminals, then trace the short way and the long way around the loop — those two routes share the same endpoints, so they add.
A–B are adjacent. Short route is the single capacitor between them; long route is the other three in series.
A–C are diagonal. Both routes are equal: two capacitors in series each way.
- CAB — adjacent corners. Direct \(A\!-\!B = 3\). Long way \(A\!-\!D\!-\!C\!-\!B\) is three in series: \(\tfrac{3}{3}=1\). In parallel: $$C_{AB}=3+1=4\,\mu\text{F}$$
- CAC — diagonal corners. Path \(A\!-\!B\!-\!C\) is \(3\) series \(3 = 1.5\); path \(A\!-\!D\!-\!C\) is also \(1.5\). In parallel: $$C_{AC}=1.5+1.5=3\,\mu\text{F}$$
- Ratio. $$\frac{C_{AB}}{C_{AC}}=\frac{4}{3}\quad\Rightarrow\quad 4:3$$
Memorise the two ring results for equal C. Across adjacent corners: \(C + \tfrac{C}{3} = \tfrac{4C}{3}\). Across diagonal corners: \(\tfrac{C}{2}+\tfrac{C}{2}=C\).
So the ratio is \(\tfrac{4C}{3} : C = \mathbf{4:3}\) — the same answer whatever the capacitor value is. The 3 μF here was never needed for the ratio; it cancels out.
\(C_{AB}=4\,\mu\text{F}\) (adjacent) and \(C_{AC}=3\,\mu\text{F}\) (diagonal):
$$\frac{C_{AB}}{C_{AC}}=\frac{4}{3}=4:3$$