Currents through ammeters A₂ and A₃ in the figure are 1∠10° and 1∠70°, respectively. The reading of the ammeter A₁ (rounded off to 3 decimal places) is _____ A.

The correct answer is 1.732.

Step 1: Understand the Given Circuit and Data
From the given circuit diagram, we observe a node where the main branch carrying current $I_1$ splits into two parallel branches containing ammeters A₂ and A₃, carrying currents $I_2$ and $I_3$ respectively.

The given branch currents in phasor form are:
$I_2=1\angle {10}^{\circ }\text{ A}$
$I_3=1\angle {70}^{\circ }\text{ A}$

Step 2: Apply Kirchhoff's Current Law (KCL)
According to Kirchhoff's Current Law at the junction node:
$I_1=I_2+I_3$

Step 3: Convert Phasors to Rectangular Form
Express each phasor current in terms of its real (cos) and imaginary (sin) components:
$I_2=\cos \left({10}^{\circ }\right)+j\sin \left({10}^{\circ }\right)\approx 0.9848+j0.1736\text{ A}$
$I_3=\cos \left({70}^{\circ }\right)+j\sin \left({70}^{\circ }\right)\approx 0.3420+j0.9397\text{ A}$

Adding the real parts and imaginary parts together:
$I_1=(0.9848+0.3420)+j(0.1736+0.9397)$
$I_1=1.3268+j1.1133\text{ A}$

Step 4: Calculate the Magnitude (Ammeter A₁ Reading)
An ammeter reads the RMS magnitude of the alternating current. The magnitude of phasor $I_1$ is:
$\left|I_1\right|=\sqrt{{\left(1.3268\right)}^2+{\left(1.1133\right)}^2}$
$\left|I_1\right|=\sqrt{1.7604+1.2394}=\sqrt{3}\approx 1.732\text{ A}$

Alternatively, using the formula for the magnitude of two equal phasors separated by angle $\theta ={70}^{\circ }-{10}^{\circ }={60}^{\circ }$:
$\left|I_1\right|=2\times 1\times \cos \left(\frac{{60}^{\circ }}{2}\right)=2\times \cos \left({30}^{\circ }\right)=2\times \frac{\sqrt{3}}{2}=\sqrt{3}\approx 1.732\text{ A}$

Thus, the reading of ammeter A₁ rounded off to three decimal places is 1.732 A.