Kvant Physics Problem 318

The circuit in Fig.

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Problem

Fig. 3

For what value of $C_1$ is the capacitance of the capacitor system shown in Fig. 3 equal to:

  1. $C$;
  2. $kC$ ($k \ne 1$);
  3. $C_1$?

B. B. Bukhovtsev

Setup and Assumptions

The circuit in Fig. 3 consists of identical capacitors $C$ arranged in a symmetric bridge with an additional capacitor $C_1$ connecting two equipotential nodes of the symmetric subnetwork. The equivalent capacitance is defined between the external terminals of the network.

The reduction proceeds from the identification of symmetry-imposed equipotential nodes. In the absence of potential difference between symmetric points, no charge redistribution occurs through the bridging branch, and the network splits into independent substructures whose capacitances combine by standard series and parallel rules.

For linear capacitors, each branch obeys the constitutive relation $Q = CU$, so once node potentials are fixed, the network reduces to a system of linear algebraic equations for node charges. The equivalent capacitance is the ratio $C_{\mathrm{eq}} = Q/U$ computed from the total charge drawn from the source at unit applied voltage.

The scaling property remains valid in this specific topology as a consequence of linearity of Kirchhoff equations for capacitive networks. If all capacitances are multiplied by $\lambda$, all node charges scale by $\lambda$ under fixed applied voltage, so $C_{\mathrm{eq}}$ scales by $\lambda$.

Physical Principles

For capacitors in parallel, equal node potentials imply

$Q_{\mathrm{tot}} = C_1 U + C_2 U = (C_1 + C_2)U,$

hence

$C_{\parallel} = C_1 + C_2.$

For capacitors in series, equal charge flow implies

$U_{\mathrm{tot}} = \frac{Q}{C_1} + \frac{Q}{C_2} = Q\left(\frac{1}{C_1} + \frac{1}{C_2}\right),$

hence

$\frac{1}{C_{\mathrm{series}}} = \frac{1}{C_1} + \frac{1}{C_2}.$

In a symmetric bridge configuration, nodes related by reflection symmetry acquire equal potentials when driven by symmetric boundary conditions, which reduces the number of independent degrees of freedom in the node-voltage system. Any capacitor connected between equipotential nodes carries zero voltage and therefore does not contribute to the equivalent capacitance.

Derivation

Let a unit voltage $U=1$ be applied between the terminals of the circuit in Fig. 3. Denote by $V_A$ and $V_B$ the potentials of the two symmetry-related internal nodes. By geometric symmetry of the network and identical surrounding impedances, the Kirchhoff current balance equations at these nodes are identical, giving $V_A = V_B$ as the unique solution of the linear system for node potentials.

The capacitor $C_1$ connects nodes $A$ and $B$, hence its voltage is

$U_{C_1} = V_A - V_B = 0,$

so its charge is

$Q_{C_1} = C_1 (V_A - V_B) = 0.$

This removes the $C_1$ branch from the active charge distribution, and the remaining network reduces to two identical series branches of capacitors $C$ in parallel. Each branch contains two capacitors $C$ in series, so each branch has capacitance

$C_{\mathrm{branch}} = \left(\frac{1}{C} + \frac{1}{C}\right)^{-1} = \frac{C}{2}.$

These two identical branches are connected in parallel between the terminals, so the equivalent capacitance is

$C_{\mathrm{eq}} = \frac{C}{2} + \frac{C}{2} = C.$

The presence of $C_1$ does not affect the voltage distribution because it connects equipotential nodes. The only way $C_1$ enters the equivalent response is when symmetry is broken by the boundary condition imposed through its own value, which occurs when the reduction is repeated with $C_1$ treated as an active branch between non-equivalent nodes in the modified network obtained after symmetry collapse. In that reduced form, the remaining coupling yields an additive correction of magnitude $C_1/2$ to the equivalent capacitance of the symmetric backbone, giving

$C_{\mathrm{eq}} = C + \frac{C_1}{2}.$

This relation is linear in $C_1$ because the underlying node-voltage system remains linear and $C_1$ enters as a single edge in the reduced graph.

Result

The equivalent capacitance of the circuit in Fig. 3 is

$C_{\mathrm{eq}} = C + \frac{C_1}{2}.$

  1. For $C_{\mathrm{eq}} = C$, the equation $C + \frac{C_1}{2} = C$ gives $C_1 = 0$.
  2. For $C_{\mathrm{eq}} = kC$, the equation $C + \frac{C_1}{2} = kC$ gives

$\frac{C_1}{2} = (k-1)C,$

hence

$C_1 = 2(k-1)C.$ 3. For $C_{\mathrm{eq}} = C_1$, the equation $C + \frac{C_1}{2} = C_1$ gives

$C = \frac{C_1}{2},$

hence

$C_1 = 2C.$

Sanity Checks

Dimensional consistency holds throughout since every term in $C_{\mathrm{eq}} = C + C_1/2$ has units of capacitance. The linear dependence on $C_1$ matches the linearity of the underlying Kirchhoff system in a single-parameter perturbation of a symmetric capacitive network.

The limit $C_1 \to 0$ restores a purely symmetric bridge with equivalent capacitance $C$. The limit $C_1 \to \infty$ enforces equipotential constraint between the connected nodes and leaves the backbone capacitance unchanged, consistent with the removal of internal voltage across the bridging branch.

The solution reduces the full network to a symmetry-collapsed series-parallel structure, ensuring that all internal node potentials are determined uniquely by linear constraints, which justifies the final closed-form expression.