AC Homework: Total Current, 6A Phase Angle, Voltage Angle

[jayryap]

Homework Statement

in figure shown below, a lamp load taking 6A and a single-phase motor taking 4A from 120v/60Hz source. the current to the lamp load is in phase with the voltage and the current to the motor logs 50 degrees. determine:
a) value of total current IT (rms)
b)angle by which current logs 6A
c)angle by which total logs voltage

Homework Equations

i_T = i₁+ i₂

The Attempt at a Solution

i_T = i₁+ i₂

i am confused on how they get this

i_T =8.49(sin wt cos 40 - cos wt sin 40)+ 5.66(sinwt cos 90 - cos wt sin 90)

please help ...tnx!

 

[KataKoniK]

it is important to approach problems with a critical and analytical mindset. In this case, it is important to understand the concepts of Ohm's Law and power in order to solve this problem.

First, let's break down the given information. We have a lamp load with a current of 6A and a single-phase motor with a current of 4A. Both are connected to a 120V/60Hz source. We are also given the information that the current to the lamp load is in phase with the voltage, while the current to the motor lags by 50 degrees.

To solve this problem, we need to use the equation iT = i1+ i2, where iT is the total current and i1 and i2 are the individual currents.

To find the value of iT, we need to use the concept of power. Power is equal to voltage multiplied by current (P=VI). Since we know the voltage and current for both the lamp load and the motor, we can calculate their individual powers. The power for the lamp load is P1 = 120V x 6A = 720W and the power for the motor is P2 = 120V x 4A = 480W.

Next, we need to use the concept of phasors to determine the total current. A phasor is a vector representation of a complex number that is used to represent the magnitude and phase of a sinusoidal quantity. In this case, we can represent the current as a phasor with a magnitude of 6A and a phase angle of 0 degrees for the lamp load and a magnitude of 4A and a phase angle of -50 degrees for the motor.

Using the phasor diagram, we can add the two individual currents (i1 and i2) to find the total current iT. This can be done graphically by drawing the phasors to scale and adding them tip to tail, or mathematically using trigonometric functions.

a) To find the value of iT, we can use the Pythagorean theorem to calculate the magnitude of the vector addition of i1 and i2. This gives us iT = √(6^2 + 4^2) = 7.21A. This is the value of the total current in rms.

b) To find the angle by which the current lags 6A, we can use the inverse tangent

 

[Mene]

I would approach this problem by first understanding the basic principles of AC circuits and how they differ from DC circuits. In AC circuits, the current and voltage are constantly changing in magnitude and direction, which is why we use the root mean square (rms) values to calculate the total current and voltage.

To determine the total current in this circuit, we can use the formula iT = i1 + i2, which means that the total current is the sum of the current through the lamp load (i1) and the current through the motor (i2). In this case, i1 = 6A and i2 = 4A, so iT = 6A + 4A = 10A.

Next, we need to determine the phase angle of the current to the motor, which is given as 50 degrees. This means that the current to the motor lags the voltage by 50 degrees. To find the angle by which the current lags the voltage, we can use the formula tanθ = i2/iT, where θ is the phase angle and iT is the total current. Plugging in the values, we get tanθ = 4/10, which gives us θ = 22.6 degrees. This means that the current to the motor lags the voltage by 22.6 degrees.

Finally, to find the angle by which the total current lags the voltage, we can use the formula tanθ = i1/iT. Plugging in the values, we get tanθ = 6/10, which gives us θ = 33.7 degrees. This means that the total current lags the voltage by 33.7 degrees.

In conclusion, the total current in this circuit is 10A, the current to the motor lags the voltage by 22.6 degrees, and the total current lags the voltage by 33.7 degrees. These values can be used to analyze and understand the behavior of the circuit and make any necessary adjustments or improvements.