Kvant Physics Problem 54
The circuit in Figure 2 consists of an ideal DC source, a network of resistors, and a single ammeter inserted in one branch.
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Problem
What will the ammeter show in the circuit shown in Figure 2? The resistance of the ammeter is very small.
A. R. Zilberman
Setup and Assumptions
The circuit in Figure 2 consists of an ideal DC source, a network of resistors, and a single ammeter inserted in one branch. The ammeter is modeled as an ideal current measuring device with negligible internal resistance, so it does not introduce a voltage drop and does not disturb the currents in the surrounding branches.
The task is to determine the current $I_A$ indicated by the ammeter. The remaining circuit elements are assumed to be ohmic resistors with constant resistances independent of current and temperature. The system is analyzed in the steady state so that all currents are constant in time and Kirchhoff’s laws apply.
The circuit topology is treated as a linear resistive network, so node voltages uniquely determine all branch currents.
Physical Principles
The analysis rests on Kirchhoff’s current law, which states that the algebraic sum of currents at any node is zero, expressed as $\sum I_k = 0$.
It also uses Kirchhoff’s voltage law, which states that the algebraic sum of potential differences around any closed loop is zero, written as $\sum \Delta V_k = 0$.
Ohm’s law relates current and voltage in each resistor as $\Delta V = IR$.
Because the ammeter resistance is negligible, the potential difference across it is $\Delta V_A = 0$, which implies it enforces equality of the electric potentials of the two nodes it connects.
Derivation
The ammeter connects two nodes of the circuit that are otherwise linked by symmetric resistive branches. The central step is to compare the potentials of these two nodes by writing Kirchhoff’s voltage law along two different paths connecting the same pair of nodes.
Let the two nodes connected by the ammeter be $P$ and $Q$. The ammeter current is $I_A = (V_P - V_Q)/R_A$, where $R_A \to 0$. Since $R_A$ is negligible, any nonzero current would require a finite potential difference. The potential difference is therefore determined entirely by the surrounding resistive network.
Writing Kirchhoff’s voltage law along the upper branch from $P$ to $Q$ gives
$$V_P - V_Q = \sum_i I_i R_i.$$
Writing the same relation along the lower branch from $P$ to $Q$ gives an identical expression because the resistances in the corresponding segments are equal, so the two path integrals produce the same total voltage drop. Denoting the upper path drop by $\Delta V_{\text{top}}$ and the lower by $\Delta V_{\text{bottom}}$, the circuit symmetry gives
$$\Delta V_{\text{top}} = \Delta V_{\text{bottom}}.$$
Subtracting these two expressions yields
$$(V_P - V_Q) - (V_P - V_Q) = 0,$$
so
$$V_P - V_Q = 0.$$
The ammeter current follows from Ohm’s law for the meter branch,
$$I_A = \frac{V_P - V_Q}{R_A}.$$
Substituting $V_P - V_Q = 0$ gives
$$I_A = 0.$$
Result
The ammeter reads
$$I_A = 0.$$
Numerically,
$$\boxed{I_A = 0~\text{A}}.$$
Sanity Checks
The result is consistent with dimensional analysis since the only possible scale for current is set by source voltage divided by resistance, but symmetry eliminates any net potential difference between the ammeter terminals, forcing the current to vanish independently of parameter values.
In the limiting case where one branch resistance is perturbed slightly so that the symmetry is broken, the node potentials become unequal and a nonzero current appears proportional to the mismatch, confirming that $I_A = 0$ is the degenerate symmetric limit of a standard bridge circuit.
The key step where an error would most easily occur is the assumption of equal potential drops along the two alternative paths between the ammeter terminals, since overlooking hidden series or parallel combinations would incorrectly predict a finite $V_P - V_Q$ and therefore an incorrect nonzero ammeter reading.