Two balanced three-phase loads, as shown in the figure, are connected to a 100 √3 V , three phase, 50 Hz main supply. Given : Z₁= (18 +j24) Ω and Z₂= (6 +j8) Ω . The ammeter reading, in amperes, is ______. (round off to nearest integer)

The correct answer is 20.

1. Analysis of the Circuit:
From the given circuit diagram:

The three-phase supply is balanced with a line-to-line voltage of:
$V\_L\; =\; 100\backslash sqrt\{3\}\; \backslash text\{\; V\}$
The ammeter $A$ is connected in the line $R$. It measures the total line current $I\_R$ flowing from the source into the two parallel balanced loads:
- A delta-connected load with phase impedance $Z\_1\; =\; 18\; +\; j24\; \backslash text\{\; \}\backslash Omega$
- A star (wye)-connected load with phase impedance $Z\_2\; =\; 6\; +\; j8\; \backslash text\{\; \}\backslash Omega$

2. Calculation for Delta-Connected Load ($Z\_1$):
For a delta connection, the phase voltage is equal to the line voltage:
$V\_\{ph1\}\; =\; V\_L\; =\; 100\backslash sqrt\{3\}\; \backslash text\{\; V\}$
The magnitude of the impedance $Z\_1$ is:
$|Z\_1|\; =\; \backslash sqrt\{18^2\; +\; 24^2\}\; =\; \backslash sqrt\{324\; +\; 576\}\; =\; \backslash sqrt\{900\}\; =\; 30\; \backslash text\{\; \}\backslash Omega$
The phase current is:
$I\_\{ph1\}\; =\; \backslash frac\{V\_\{ph1\}\}\{|Z\_1|\}\; =\; \backslash frac\{100\backslash sqrt\{3\}\}\{30\}\; =\; \backslash frac\{10\backslash sqrt\{3\}\}\{3\}\; \backslash text\{\; A\}$
For a balanced delta-connected load, the line current magnitude is:
$I\_\{L1\}\; =\; \backslash sqrt\{3\}\; I\_\{ph1\}\; =\; \backslash sqrt\{3\}\; \backslash times\; \backslash frac\{10\backslash sqrt\{3\}\}\{3\}\; =\; 10\; \backslash text\{\; A\}$
The phase angle of $Z\_1$ is:
$\backslash theta\_1\; =\; \backslash tan^\{-1\}\backslash left(\backslash frac\{24\}\{18\}\backslash right)\; =\; \backslash tan^\{-1\}(1.333)\; \backslash approx\; 53.13^\backslash circ\; \backslash text\{\; (lagging)\}$

3. Calculation for Star-Connected Load ($Z\_2$):
For a star connection, the phase voltage is:
$V\_\{ph2\}\; =\; \backslash frac\{V\_L\}\{\backslash sqrt\{3\}\}\; =\; \backslash frac\{100\backslash sqrt\{3\}\}\{\backslash sqrt\{3\}\}\; =\; 100\; \backslash text\{\; V\}$
The magnitude of the impedance $Z\_2$ is:
$|Z\_2|\; =\; \backslash sqrt\{6^2\; +\; 8^2\}\; =\; \backslash sqrt\{36\; +\; 64\}\; =\; 10\; \backslash text\{\; \}\backslash Omega$
The line current magnitude (which equals the phase current in a star connection) is:
$I\_\{L2\}\; =\; \backslash frac\{V\_\{ph2\}\}\{|Z\_2|\}\; =\; \backslash frac\{100\}\{10\}\; =\; 10\; \backslash text\{\; A\}$
The phase angle of $Z\_2$ is:
$\backslash theta\_2\; =\; \backslash tan^\{-1\}\backslash left(\backslash frac\{8\}\{6\}\backslash right)\; =\; \backslash tan^\{-1\}(1.333)\; \backslash approx\; 53.13^\backslash circ\; \backslash text\{\; (lagging)\}$

4. Calculation of the Total Line Current ($I\_R$):
Since both loads are balanced and have the same power factor angle ($\backslash theta\_1\; =\; \backslash theta\_2\; \backslash approx\; 53.13^\backslash circ$), the line currents drawn by both loads are in phase with each other.
Therefore, the total current in line $R$ is the direct algebraic sum of the line current magnitudes:
$I\_R\; =\; I\_\{L1\}\; +\; I\_\{L2\}\; =\; 10\; +\; 10\; =\; 20\; \backslash text\{\; A\}$

Thus, the reading of the ammeter is 20 A.