NEET · Physics Electrostatics Capacitors

A tidy series chain, and a bypass that adds on top

First a symmetric three-link chain that collapses to a round number; then four equal capacitors where the bottom one quietly shortcuts the whole top path.

11 Three 3 μF links with a parallel middleEqual-series collapse

Question

Find the capacitance between A and B in the given circuit.

Fig 11 · 3 μF, then 1.5 μF ∥ 1.5 μF, then 3 μF

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) 4 μF

Given

Left3 μF
Middle (×2)1.5 μF each
Right3 μF

Asked

The equivalent capacitance $C_{AB}$.

Concept

The two middle 1.5 μF capacitors share both nodes, so they're in parallel → $1.5+1.5 = 3\,\mu\text{F}$. The circuit is then three $3\,\mu\text{F}$ capacitors in a single line: pure series.

Method & Steps

Middle (parallel). $1.5+1.5 = 3\,\mu\text{F}$.
Three in series. $$\frac{1}{C_{AB}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1 \;\Rightarrow\; C_{AB}=1\,\mu\text{F}$$

Easy Trick

n equal capacitors in series = value ÷ n. The middle pair adds to 3, matching the outer caps, so it's three identical 3 μF in series → $3/3 = 1\,\mu\text{F}$. No reciprocal arithmetic needed once they're all equal.

Solution

Middle pair → 3 μF; three 3 μF in series:

$$C_{AB}=\frac{3}{3}=1\,\mu\text{F}$$

a1 μF between A and B

12 Four equal capacitors with a bottom bypassTwo paths in parallel

Question

Four equal capacitors, each of capacity $C$, are arranged as shown. Find the effective capacitance between A and B.

Fig 12 · Top: C then C ∥ C. Bottom: a single C straight from A to B.

(a) 5C/8
(b) 3C/5
(c) 5C/3
(d) C

Given

Each capacitorC
Input cap1 (A → M)
Middle pair2 in parallel
Bottom1 (A → B bypass)

Asked

The effective capacitance $C_{AB}$ in terms of $C$.

Concept

There are two separate routes from A to B. The top route runs through the input C and then the parallel pair. The bottom route is a single C that bypasses everything. Two routes between the same terminals → add them in parallel.

Method & Steps

Middle pair (parallel). $C+C = 2C$.
Top route (series). Input $C$ in series with $2C$: $$\frac{C\times 2C}{C+2C}=\frac{2C}{3}$$
Add the bottom bypass (parallel). $$C_{AB}=\frac{2C}{3}+C=\frac{5C}{3}$$

Easy Trick

Reduce the top route fully, then just add the bypass. Top gives $\tfrac{2C}{3}$; the lone bottom C sits straight across A–B and adds on top, lifting the total above C to $\tfrac{5C}{3}$. Any answer below C (the 5C/8, 3C/5 traps) ignores that the bypass alone already supplies a full C.

Solution

Top route $=\tfrac{2C}{3}$, in parallel with the bottom $C$:

$$C_{AB}=\frac{2C}{3}+C=\frac{5C}{3}$$

c 5C3≈ 1.67 C