2. A two-winding transformer with a laminated core is shown in the Figure. The winding with N1=200 turns is connected to a voltage to produce a flux density in the core B=1.2Âsin377t. The second winding, with N2=400 turns, is left open-circuited. The stacking factor of the core is 0.95, i.e. the core occupies 95% of the gross core volume. The gross cross-sectional area of the core is 25cm2 , and ȝr for the core is 10,000. The core length is lc=90cm. a) Determine the rms value of the applied voltage E1. b) Determine the current in the winding. c) Determine the rms voltage Er induced in the second winding.
a) The rms value of the applied voltage E1 is given by:
E1 = (B*N1*lc*S)/(4.44*sqrt(2)*ȝr)
E1 = (1.2*200*90*0.95)/(4.44*sqrt(2)*10000)
E1 = 4.8 V
b) The current in the winding is given by:
I1 = (E1*sqrt(2))/(N1*lc*S)
I1 = (4.8*sqrt(2))/(200*90*0.95)
I1 = 0.0267 A
c) The rms voltage Er induced in the second winding is given by:
Er = (B*N2*lc*S)/(4.44*sqrt(2)*ȝr)
Er = (1.2*400*90*0.95)/(4.44*sqrt(2)*10000)
Er = 9.6 V
To solve this problem, we can follow these steps:
a) Determine the rms value of the applied voltage E1:
Step 1: Calculate the average value of the applied voltage.
The average value of a sinusoidal waveform is zero, so the average value in this case will be 0.
Step 2: Determine the peak value of the applied voltage.
The peak value of a sinusoidal waveform can be found by multiplying the maximum amplitude by the square root of 2. In this case, the peak value will be:
Vpeak = B * max(Ac * lc) * sqrt(2)
B: Flux density = 1.2 A/m
Ac: Core cross-sectional area = 25 cm² = 0.0025 m²
lc: Core length = 90 cm = 0.9 m
Vpeak = 1.2 * 0.0025 * 0.9 * sqrt(2) = 0.00243 V
Step 3: Calculate the rms value of the applied voltage.
The rms value of a sinusoidal waveform can be found by dividing the peak value by the square root of 2. In this case, the rms value will be:
Erms = Vpeak / sqrt(2)
Erms = 0.00243 / sqrt(2) = 0.00172 V
Therefore, the rms value of the applied voltage E1 is 0.00172 V.
b) Determine the current in the winding:
The current in the winding can be calculated using the formula:
I = E1 / (N1 * ƒ)
E1: Applied voltage = 0.00172 V (rms value)
N1: Number of turns in the winding = 200
ƒ: Frequency of the waveform = 377 Hz
I = 0.00172 / (200 * 377) = 2.28 × 10^(-8) A
Therefore, the current in the winding is 2.28 × 10^(-8) A.
c) Determine the rms voltage Er induced in the second winding:
The rms voltage induced in the second winding can be calculated using the formula:
Er = (N2 / N1) * E1
N2: Number of turns in the second winding = 400
N1: Number of turns in the first winding = 200
E1: Applied voltage = 0.00172 V (rms value)
Er = (400 / 200) * 0.00172 = 0.00344 V
Therefore, the rms voltage Er induced in the second winding is 0.00344 V.
To calculate the answers to the given questions, we need to follow a set of steps. Let's go through each step one by one:
Step 1: Determine the maximum value of the flux density Bmax:
Given B = 1.2 × sin(377t) and the flux density waveform is sinusoidal, we know that the maximum value of the flux density Bmax is equal to the peak value of the waveform. The peak value can be found by taking the absolute value of the coefficient of sin(377t), which is 1.2. Therefore, Bmax = 1.2 Tesla.
Step 2: Calculate the effective core area Ac:
The effective core area Ac is obtained by multiplying the gross cross-sectional area of the core by the stacking factor. In this case, the gross cross-sectional area is given as 25 cm² and the stacking factor is 0.95.
Ac = 0.95 × 25 cm² = 23.75 cm²
Step 3: Convert Ac to square meters
To perform further calculations, we need to convert the area from cm² to m². Since 1 m² = 10,000 cm², we can calculate Ac in square meters:
Ac = 23.75 cm² × (1 m² / 10,000 cm²) = 0.002375 m²
Step 4: Calculate the maximum flux Φmax:
The maximum flux Φmax can be determined by multiplying the maximum value of the flux density Bmax by the effective core area Ac:
Φmax = Bmax × Ac = 1.2 T × 0.002375 m² = 0.00285 Wb
Step 5: Calculate the peak value of the induced emf:
The peak value of the induced emf is obtained by multiplying the maximum flux Φmax by the number of turns N1 of the winding connected to the voltage source:
E1peak = 4.44 × f × N1 × Φmax (where f is the frequency of the flux density)
Given the flux density waveform B = 1.2 × sin(377t), we can see that the frequency is 377 Hz. Substituting the values, we get:
E1peak = 4.44 × 377 Hz × 200 turns × 0.00285 Wb = 303.15 V
Step 6: Calculate the RMS value of the applied voltage E1:
The RMS value of a sinusoidal waveform is equal to the peak value divided by the square root of 2.
E1rms = E1peak / √2 = 303.15 V / √2 ≈ 214.53 V
Therefore, the RMS value of the applied voltage E1 is approximately 214.53 V.
Step 7: Calculate the current in the winding:
The current flowing in the winding connected to the voltage source can be determined using Ohm's law. Since the second winding is left open-circuited, we can assume that the winding impedance is purely inductive.
Z = jωL = j2πfL (where Z is the impedance and ω is the angular frequency)
The impedance of the winding is given by:
Z = (V / I)
Rearranging the above equation, we get:
I = V / Z
Since the winding impedance is inductive, we can assume that the power factor is 1 (i.e., purely reactive).
Therefore, we can calculate the current as:
I = E1rms / (j2πfL)
Step 8: Determine the RMS voltage Er induced in the second winding:
The RMS voltage induced in the second winding can be determined using the formula:
Er = (N2 / N1) × E1rms
Given N2 = 400 turns and N1 = 200 turns, we can substitute the values:
Er = (400 / 200) × 214.53 V = 429.06 V
Therefore, the RMS voltage Er induced in the second winding is approximately 429.06 V.