A two-phase generator is connected to two 90 Ω load resistors. Each coil generates 120 V ac. A common neutral line exists. How much current flows through the common neutral line?

Solution(By Examveda Team)

Okay, let's break down this problem step-by-step:

First, let's understand the basics of a two-phase system.
In a two-phase system, you have two AC voltages that are out of phase with each other.

Here's how to solve for the current flowing through the neutral line:

Step 1: Calculate the current in each phase.
We use Ohm's Law, which is Voltage (V) = Current (I) * Resistance (R).
Rearranging for current, we get I = V / R.
Since each coil generates 120V and is connected to a 90Ω resistor, the current in each phase is:
I = 120V / 90Ω = 1.33 A
So, each phase carries 1.33 Amperes.

Step 2: Understand the phase relationship and neutral current.
In a two-phase system, the two voltages are 90 degrees out of phase.
The neutral current is the vector sum of the currents in the two phases.
Because the phases are 90 degrees apart, we can use the Pythagorean theorem to find the magnitude of the neutral current.

Step 3: Apply the Pythagorean theorem.
The neutral current (In) is the square root of the sum of the squares of the individual phase currents:
In = √(I1² + I2²)
In = √((1.33 A)² + (1.33 A)²)
In = √(1.77 + 1.77)
In = √3.54
In ≈ 1.88 A

Therefore, the current flowing through the common neutral line is approximately 1.88 A.
The correct answer is Option B: 1.88 A.