Irodov 3.112: Wires That Fold a Chain Flat – Amit & Deepali Sinha Foundation

Problem Irodov 3.112

Three capacitors sit in a row along a straight line from the point $A$ to the point $B$ , with two junctions between them. One link joins $A$ to the junction nearer $B$ , and a second link joins the junction nearer $A$ to $B$ . Find the capacitance between $A$ and $B$ in two cases:

the three capacitors in the row have capacitances $C_1$ , $C_2$ , $C_3$ (from $A$ ), and both links are plain wires;
every capacitor has the same capacitance $C$ , and each of the two links also carries a capacitor $C$ , five capacitors in all.

(a) Three capacitors in a row from A to B; plain wires join A to the second junction and the first junction to B. — (a) Three capacitors in a row from $A$ to $B$ ; plain wires join $A$ to the second junction and the first junction to $B$ .

(b) The same layout with every capacitor equal to C and a capacitor C in each link. — (b) The same layout with every capacitor equal to $C$ and a capacitor $C$ in each link.

Given

(a) $C_1, C_2, C_3$ , plain links
(b) five capacitors $C$

Find

C_{AB}

in (a) and in (b)

Solution

Step 1Network (a): the wires merge the junctions with the terminals

A plain wire holds its two ends at one potential, so each wire turns two points into one node. Call the junction nearer $A$ node $2$ and the one nearer $B$ node $3$ :

\varphi_3 = \varphi_A, \qquad \varphi_2 = \varphi_B

Read each capacitor’s two ends in terms of the merged nodes: $C_1$ runs from $A$ to $2 = B$ , $C_2$ from $2 = B$ to $3 = A$ , and $C_3$ from $3 = A$ to $B$ . All three lie between the same pair of nodes.

Network (a) redrawn: with the junctions merged into the terminals, each capacitor sits directly between A and B. — Network (a) redrawn: with the junctions merged into the terminals, each capacitor sits directly between $A$ and $B$ .

Step 2Capacitances in parallel add

With a voltage $V$ between $A$ and $B$ , every capacitor holds the full $V$ , and the charge delivered by the terminals is the sum of the three plate charges:

q = C_1V + C_2V + C_3V = (C_1 + C_2 + C_3)\,V

Dividing by $V$ :

C_a = C_1 + C_2 + C_3

(1)

Step 3Network (b): the same five capacitors form a bridge

Now the links carry capacitors, so no node merges. Listing the five capacitors by their ends shows the topology: $A$ to $2$ , $A$ to $3$ (the top link), $2$ to $3$ , $2$ to $B$ (the bottom link), $3$ to $B$ . That is a bridge, with the capacitor between $2$ and $3$ in the rung.

Network (b) redrawn as a bridge: junctions 2 and 3 at the corners, the capacitor joining them in the rung, the four equal arms between them and the terminals. — Network (b) redrawn as a bridge: junctions $2$ and $3$ at the corners, the capacitor joining them in the rung, the four equal arms between them and the terminals.

Step 4The balanced bridge leaves the rung uncharged

Give $A$ the potential $V$ and $B$ zero. The two paths $A \to 2 \to B$ and $A \to 3 \to B$ are identical capacitors in the same order, so by symmetry

\varphi_2 = \varphi_3 = \frac{V}{2}

The rung therefore holds no voltage and no charge, and it can be removed. What remains is two chains in parallel, each of two capacitors $C$ in series:

C_b = \frac{C \cdot C}{C + C} + \frac{C \cdot C}{C + C} = \frac{C}{2} + \frac{C}{2}

C_b = C

(2)

Check

Count the charge leaving $A$ in (b). With $\varphi_2 = \varphi_3 = V/2$ , each of the two capacitors touching $A$ holds the voltage $V/2$ and the charge $CV/2$ , so $A$ delivers $CV$ , which is $C_b = C$ as in (2). Each of the junctions $2$ and $3$ is an isolated conductor and must stay neutral: junction $2$ carries $-CV/2$ on its face toward $A$ , $0$ in the rung, and $+CV/2$ on its face toward $B$ , total zero. In (a) the result exceeds each single capacitance and is symmetric in $C_1$ , $C_2$ , $C_3$ , as it must be once all three join the same two nodes.

Remark

The two networks share one layout and differ only in what the links carry, yet (a) gives $3C$ for equal capacitors and (b) gives $C$ . A wire is a capacitor of infinite capacitance, so (a) is the limit of a bridge whose two links are made very large. The general rule behind (b): a bridge whose arms satisfy $C_{A2}/C_{2B} = C_{A3}/C_{3B}$ leaves its rung uncharged, whatever the rung’s own capacitance, which is why its value never entered the answer.