NEET · Physics Electrostatics Capacitors

The bridge where one capacitor goes to sleep

Five identical capacitors wired as a Wheatstone bridge. The trick is spotting that the bridge is balanced — once you do, the whole problem collapses into a single line of arithmetic.

Question

Five capacitors of $10\,\mu\text{F}$ each are connected to a $100\,\text{V}$ DC supply as shown. Find the equivalent capacitance between points A and B.

Fig 1 · The five-capacitor bridge between A and B

(a) 40 μF
(b) 20 μF
(c) 30 μF
(d) 10 μF

Given

Each capacitorC = 10 μF
Count5 capacitors
SupplyV = 100 V (DC)
TopologyWheatstone bridge

Asked

The single equivalent capacitance $C_{eq}$ seen between terminals A and B. (The 100 V value is a distractor — it sets up charge/energy, but equivalent capacitance never depends on voltage.)

Concept

Re-draw the network as a Wheatstone bridge: two arms on top, two on the bottom, and a fifth capacitor bridging the middle two nodes (P and Q).

A capacitor bridge is balanced when the ratio of the two left/right arm pairs match:

$$\frac{C_1}{C_2}=\frac{C_3}{C_4}$$

When balanced, nodes P and Q sit at the same potential, so the middle capacitor $C_5$ has zero voltage across it — it holds no charge and can be removed entirely.

Here every arm is $10\,\mu\text{F}$:

$$\frac{C_1}{C_2}=\frac{10}{10}=1=\frac{10}{10}=\frac{C_3}{C_4}\quad\Rightarrow\quad\textbf{balanced}$$

Method & Steps

Check balance. $C_1/C_2 = C_3/C_4$ holds, so the bridge capacitor $C_5$ carries no charge — delete it.
Top branch (series). $C_1$ and $C_2$ in series: $$C_{top}=\frac{C_1 C_2}{C_1+C_2}=\frac{10\times 10}{20}=5\,\mu\text{F}$$
Bottom branch (series). Identically, $C_3$ and $C_4$: $$C_{bot}=\frac{10\times 10}{20}=5\,\mu\text{F}$$
Combine branches (parallel). The two 5 μF branches both run A → B: $$C_{eq}=C_{top}+C_{bot}=5+5=10\,\mu\text{F}$$

Fig 2 · After dropping C₅ — two 5 μF branches in parallel

Easy Trick

Equal arms → middle one sleeps. The instant all four outer capacitors are equal, the bridge is balanced and the centre one drops out — don't compute it.

What remains is symmetric: two equal capacitors in series make half (10 → 5), and two equal branches in parallel make double (5 → 10). For a fully symmetric bridge of equal $C$, the answer is simply $C_{eq}=C$. Same value you started with: $10\,\mu\text{F}$.

Solution

The bridge is balanced, the middle capacitor is uncharged, and the two series branches (5 μF each) combine in parallel:

$$C_{eq}=\frac{C}{2}+\frac{C}{2}=5+5=10\,\mu\text{F}$$

d 10 μF between A and B