The circuit consists of two blocks of resistors connected in series. The ammeters $A_1, A_2, A_3$ form a mesh between the nodes of these blocks but contain small internal resistances. Since the problem states the ammeter resistances are “much smaller” than the resistors, we first calculate the main circuit currents assuming the connection points are equipotential (ideal case), and then apply a correction for the ammeter loop.

Left Block: Three $3\,\text{k}\Omega$ resistors in parallel. $$ R_{\text{left}} = \frac{3\,\text{k}\Omega}{3} = 1\,\text{k}\Omega $$

Right Block: Resistors $1\,\text{k}\Omega, 2\,\text{k}\Omega, 4\,\text{k}\Omega$ in parallel. $$ \frac{1}{R_{\text{right}}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} = \frac{4+2+1}{4} = \frac{7}{4}\,\text{mS} $$ $$ R_{\text{right}} = \frac{4}{7}\,\text{k}\Omega $$

Total Resistance: $$ R_{\text{eq}} = R_{\text{left}} + R_{\text{right}} = 1 + \frac{4}{7} = \frac{11}{7}\,\text{k}\Omega $$

Total Main Current ($I$): $$ I = \frac{V}{R_{\text{eq}}} = \frac{3.3\,\text{V}}{11/7\,\text{k}\Omega} = \frac{3.3 \times 7}{11}\,\text{mA} = 2.1\,\text{mA} $$

We determine how this total current of $2.1\,\text{mA}$ splits in each branch.

• Left Side (Input): By symmetry, the current splits equally among the three identical $3\,\text{k}\Omega$ resistors. $$ i_{\text{left}} = \frac{2.1}{3} = 0.7\,\text{mA per branch} $$
• Right Side (Output): The voltage drop across the right block is $V_{\text{right}} = I \times R_{\text{right}} = 2.1 \times \frac{4}{7} = 1.2\,\text{V}$. The currents are: $$ i_{1\text{k}} = \frac{1.2\,\text{V}}{1\,\text{k}\Omega} = 1.2\,\text{mA} $$ $$ i_{2\text{k}} = \frac{1.2\,\text{V}}{2\,\text{k}\Omega} = 0.6\,\text{mA} $$ $$ i_{4\text{k}} = \frac{1.2\,\text{V}}{4\,\text{k}\Omega} = 0.3\,\text{mA} $$

We now check the current balance at the three junction nodes (Top, Middle, Bottom). Let $i$ be a circulating correction current flowing through the loop formed by the three ammeters ($A_1, A_2, A_3$). We define the upward direction as positive for the vertical ammeters.

• Top Node: Input $= 0.7\,\text{mA}$, Output $= 1.2\,\text{mA}$.
  Deficit $= 0.5\,\text{mA}$.
  This current must be supplied from the node below.
  $\Rightarrow$ Current in $A_1 = 0.5 + i$ (Upwards)
• Bottom Node: Input $= 0.7\,\text{mA}$, Output $= 0.3\,\text{mA}$.
  Excess $= 0.4\,\text{mA}$.
  This excess flows to the node above.
  $\Rightarrow$ Current in $A_2 = 0.4 + i$ (Upwards)
• Loop Closure ($A_3$): Since $A_1$ and $A_2$ push current up, the return path is through $A_3$. The sum of voltage drops across identical ammeters in a closed loop must be zero.
  Current in $A_3 = i$

Since the ammeters are identical with internal resistance $r$, Kirchhoff’s Voltage Law (KVL) around the ammeter loop ($A_1 \to A_2 \to A_3$) gives:

We substitute $i = -0.3$ back into the current expressions to find the magnitude of current in each ammeter.

Reading of $A_1$: $$ |0.5 + (-0.3)| = |0.2| = \mathbf{0.2\,\text{mA}} $$

Reading of $A_2$: $$ |0.4 + (-0.3)| = |0.1| = \mathbf{0.1\,\text{mA}} $$

Reading of $A_3$: $$ |-0.3| = \mathbf{0.3\,\text{mA}} $$