NEET · Physics Electrostatics Capacitors

Three networks, one habit: reduce the simple node first

A bridge-ladder, a parallel-then-series trio, and a symmetric chain. Each yields to the same routine — spot what's parallel, spot what's series, collapse one step at a time.

7 Bridge-ladder between A and BStep-by-step reduction
Question

Find the total capacitance of the system between points A and B.

Fig 7 · 2 μF top, right & bottom; 1 μF on the left edge and the diagonal
Given
Asked

The equivalent capacitance \(C_{AB}\) between A and B.

Concept

Peel from the far corner inward. The corner TR sits between just two capacitors (A→TR and TR→BR), so that pair is a plain series link from A to BR. Fold it in, then the corner BR becomes a simple series link from A to B. Each fold leaves a parallel pair to add.

Method & Steps
  1. Fold TR. A→TR→BR is \(2\) series \(2 = 1\,\mu\text{F}\).
  2. Add the diagonal (parallel). A→BR \(= 1 + 1 = 2\,\mu\text{F}\).
  3. Fold BR. A→BR→B is \(2\) series \(2 = 1\,\mu\text{F}\).
  4. Add the left edge (parallel). $$C_{AB}=1+1=2\,\mu\text{F}$$
Easy Trick

Two 2's in series make a 1; a 1 beside a 1 makes a 2. This circuit just alternates that pair twice. No simultaneous equations needed — peel one corner at a time and the value bounces 1 → 2 → 1 → 2.

Solution

Folding both corners and adding the parallel 1 μF caps:

$$C_{AB}=2\,\mu\text{F}$$
b2 μF total A–B
9 C₁, C₂ in parallel, then C₃ in seriesParallel-then-series
Question

In the network, \(C_1=10\,\mu\text{F}\), \(C_2=5\,\mu\text{F}\) and \(C_3=4\,\mu\text{F}\). Find the resultant capacitance between A and B (approximately).

Fig 9 · C₁ and C₂ share both ends, then C₃ leads to B
Given
Asked

The resultant capacitance \(C_{AB}\) (nearest option).

Concept

C₁ and C₂ share the same two nodes — the top rail (A) and the junction M — so they're in parallel. That bundle then meets C₃ at a single node M, so the bundle and C₃ are in series down to B.

Method & Steps
  1. Parallel top pair. \(C_1+C_2 = 10+5 = 15\,\mu\text{F}\).
  2. Series with C₃. $$C_{AB}=\frac{15\times 4}{15+4}=\frac{60}{19}\approx 3.2\,\mu\text{F}$$
Easy Trick

Add the shared pair, then product-over-sum with the lone one. \(15\) and \(4\) give \(\tfrac{60}{19}\approx 3.16\). A series result must fall below the smaller member (below 4), so 4.7 is impossible — that's the trap for series-vs-parallel slips.

Solution

\(C_1\parallel C_2 = 15\,\mu\text{F}\), in series with \(C_3 = 4\,\mu\text{F}\):

$$C_{AB}=\frac{60}{19}\approx 3.2\,\mu\text{F}$$
b≈ 3.2 μF 60/19
10 Symmetric 2–1–2 chainCollapse the ends first
Question

All capacitors are \(1\,\mu\text{F}\). Find the equivalent capacitance between A and B.

Fig 10 · A parallel pair, a single cap, then another parallel pair
Given
Asked

The equivalent capacitance \(C_{AB}\).

Concept

Collapse the two parallel bundles first. Each pair shares both ends, so each becomes \(1+1 = 2\,\mu\text{F}\). The circuit is then three capacitors in a single line — \(2,\,1,\,2\) — all in series.

Method & Steps
  1. End bundles (parallel). Left \(= 2\,\mu\text{F}\), right \(= 2\,\mu\text{F}\).
  2. Whole line (series). $$\frac{1}{C_{AB}}=\frac{1}{2}+\frac{1}{1}+\frac{1}{2}=2$$
  3. Invert. \(C_{AB}=0.5\,\mu\text{F}\).
Easy Trick

2, 1, 2 in series → reciprocals \(\tfrac12+1+\tfrac12 = 2\) → answer \(\tfrac12\). Series always undershoots the smallest member (1 μF here), so only the sub-1 option survives — 0.5 μF.

Solution

Ends collapse to 2 μF each; series of 2, 1, 2:

$$C_{AB}=\frac{1}{\tfrac12+1+\tfrac12}=0.5\,\mu\text{F}$$
d0.5 μF equivalent A–B