NEET · Physics Electrostatics Capacitors

Adjacent corners of a square loop: the short edge plus the long detour

Four equal capacitors in a ring, with P and Q on neighbouring corners. One route is the single edge between them; the other is the rest of the loop in series. They run in parallel.

Question

Four capacitors, each $1\,\mu\text{F}$, are connected in a square as shown. Find the equivalent capacitance between P and Q.

Fig 1 · Square ring of four 1 μF capacitors; P and Q are adjacent corners.

(a) 4 μF
(b) 1/4 μF
(c) 3/4 μF
(d) 4/3 μF

Given

Each capacitor1 μF
Count4 (square ring)
CornersP · Q · M · N
TerminalsP–Q (adjacent)

Asked

The equivalent capacitance $C_{PQ}$ across the two adjacent corners P and Q.

Concept

A ring gives two parallel routes between any two terminals. From P to Q:

Short route — the single capacitor on the bottom edge, $1\,\mu\text{F}$.

Long route — the other three capacitors (left, top, right) in series: $P\!-\!N\!-\!M\!-\!Q$. Three $1\,\mu\text{F}$ in series give $\tfrac{1}{3}\,\mu\text{F}$.

Same endpoints, so the two routes add in parallel.

Method & Steps

Long route (series of three). $$\frac{1}{C_{long}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3 \;\Rightarrow\; C_{long}=\frac{1}{3}\,\mu\text{F}$$
Add the short route (parallel). $$C_{PQ}=1+\frac{1}{3}=\frac{4}{3}\,\mu\text{F}$$

1 μF ∥ (three in series = 1/3 μF) → 4/3 μF

Easy Trick

Equal-C square, adjacent corners → $C+\tfrac{C}{3}=\tfrac{4C}{3}$. With $C=1$: straight to $\tfrac{4}{3}$. (Diagonal corners would instead give $C$ — worth keeping both in your pocket.)

Sanity: the answer must beat the single direct edge (1 μF) but only slightly, since the detour adds just $\tfrac13$. That rules out the inflated 4 and the too-small 1/4, 3/4.

Solution

Direct edge 1 μF in parallel with the three-in-series detour $\tfrac13$ μF:

$$C_{PQ}=1+\frac{1}{3}=\frac{4}{3}\,\mu\text{F}$$

d 43 μF ≈ 1.33 μF