Irodov 3.126: Crossing the Arms Wakes the Rung – Amit & Deepali Sinha Foundation

Problem Irodov 3.126

Two paths join the terminal $A$ to the terminal $B$ . The upper path has a capacitor $C_1$ followed by a capacitor $C_2$ ; the lower path has a capacitor $C_2$ followed by a capacitor $C_1$ . A capacitor $C_3$ joins the junction in the middle of the upper path to the junction in the middle of the lower path. Find the capacitance between $A$ and $B$ .

The network. The two paths are crossed: C_1 comes first from A on the upper path and C_2 first on the lower one, with C_3 between their middle junctions. — The network. The two paths are crossed: $C_1$ comes first from $A$ on the upper path and $C_2$ first on the lower one, with $C_3$ between their middle junctions.

Given

$C_1$ , $C_2$ , $C_3$
capacitors uncharged before a voltage is applied (assumed)

Find

the capacitance

C

between

A

and

B

Solution

Step 1Name the junctions and use the symmetry

Apply a voltage $V$ with $\varphi_A = V$ and $\varphi_B = 0$ , and call the middle junctions $X$ (upper path) and $Y$ (lower path).

The middle junctions X and Y. Turning the network end for end swaps A with B and X with Y. — The middle junctions $X$ and $Y$ . Turning the network end for end swaps $A$ with $B$ and $X$ with $Y$ .

The symmetry fixes $Y$ from $X$ :

\varphi_Y = V - \varphi_X \tag{1}

Step 2The island at

X

Junction $X$ and its three plates form an isolated conductor, neutral since the capacitors start uncharged. Its plates face $A$ across $C_1$ , $B$ across $C_2$ and $Y$ across $C_3$ :

C_1(\varphi_X - V) + C_2(\varphi_X - 0) + C_3(\varphi_X - \varphi_Y) = 0

Substituting (1) and solving for $\varphi_X$ :

\begin{aligned} C_1(\varphi_X - V) + C_2\varphi_X + C_3(2\varphi_X - V) &= 0\\[2pt] (C_1 + C_2 + 2C_3)\,\varphi_X &= (C_1 + C_3)\,V \end{aligned}

\varphi_X = \frac{C_1 + C_3}{C_1 + C_2 + 2C_3}\,V

(2)

Step 3The charge drawn from

A

Terminal $A$ holds the plates of the upper $C_1$ (facing $X$ ) and of the lower $C_2$ (facing $Y$ ); (1) turns the voltage of the lower one into $\varphi_X$ :

\begin{aligned} q &= C_1(V - \varphi_X) + C_2(V - \varphi_Y)\\[2pt] &= C_1V + (C_2 - C_1)\,\varphi_X \end{aligned}

Substituting (2) and putting everything over the common denominator:

\begin{aligned} \frac{q}{V} &= C_1 + \frac{(C_2 - C_1)(C_1 + C_3)}{C_1 + C_2 + 2C_3}\\[2pt] &= \frac{C_1^2 + C_1C_2 + 2C_1C_3 + C_1C_2 + C_2C_3 - C_1^2 - C_1C_3}{C_1 + C_2 + 2C_3} \end{aligned}

The $C_1^2$ terms cancel, and the capacitance is $q/V$ :

C = \frac{2C_1C_2 + C_3(C_1 + C_2)}{C_1 + C_2 + 2C_3}

(3)

Check

Three special cases. With $C_3 = 0$ the rung is gone and each path is $C_1$ and $C_2$ in series, so the two in parallel give $2C_1C_2/(C_1 + C_2)$ , which is (3) at $C_3 = 0$ . With $C_3 \to \infty$ the junctions $X$ and $Y$ merge, leaving $C_1 + C_2$ (from $A$ ) in series with $C_2 + C_1$ (to $B$ ), that is $(C_1 + C_2)/2$ , which is the limit of (3), $C_3(C_1 + C_2)/(2C_3)$ . With $C_1 = C_2$ the bridge is balanced, both middle junctions sit at $V/2$ , and the formula gives $(2C_1^2 + 2C_1C_3)/(2C_1 + 2C_3) = C_1$ for every $C_3$ .

Remark

Had both paths been drawn $C_1$ then $C_2$ , the two middle junctions would sit at the same potential, $C_3$ would carry no charge, and it would drop out. Crossing the arms is what makes $C_3$ matter: from (2), $\varphi_X - \varphi_Y = (C_1 - C_2)V/(C_1 + C_2 + 2C_3)$ , which vanishes only for $C_1 = C_2$ .