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A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______
Correct answer is between '4.51,5.10'. Can you explain this answer?
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A single-phase, full-bridge, fully controlled thyristor rectifier feed...
KW resistive heater. The supply voltage is 230 V, 50 Hz. Determine the firing angle of the thyristors required to achieve the rated power output.
To determine the firing angle of the thyristors, we need to calculate the average output voltage of the rectifier.
The average output voltage of a single-phase full-bridge thyristor rectifier can be calculated using the following formula:
Vavg = (2 * √2 * Vm / π) * (1 - cos(α))
Where:
Vavg = Average output voltage
Vm = Peak value of supply voltage
α = Firing angle of the thyristors
The peak value of the supply voltage (Vm) can be calculated as:
Vm = √2 * Vrms
Where:
Vrms = RMS value of the supply voltage
The RMS value of the supply voltage (Vrms) can be calculated as:
Vrms = Vm / √2
Given:
Supply voltage (V) = 230 V
Frequency (f) = 50 Hz
Power (P) = 10 kW
First, let's calculate the RMS value of the supply voltage:
Vrms = V / √2
= 230 / √2
= 162.63 V
Next, let's calculate the peak value of the supply voltage:
Vm = √2 * Vrms
= √2 * 162.63
= 230 V
Now, let's calculate the average output voltage at the rated power:
Vavg = (2 * √2 * Vm / π) * (1 - cos(α))
= (2 * √2 * 230 / π) * (1 - cos(α))
We know that power (P) is equal to the product of average voltage (Vavg) and current (I):
P = Vavg * I
Since we have a resistive load, the current (I) can be calculated as:
I = P / Vavg
Substituting the value of power (P) and average voltage (Vavg) in the above equation, we get:
10,000 = [(2 * √2 * 230 / π) * (1 - cos(α))] * (10,000 / [(2 * √2 * 230 / π) * (1 - cos(α))])
The above equation simplifies to:
1 = 1 - cos(α)
Simplifying further, we get:
cos(α) = 0
To achieve the rated power output of 10 kW, the firing angle (α) of the thyristors required is 90 degrees.
To determine the firing angle of the thyristors, we need to calculate the average output voltage of the rectifier.
The average output voltage of a single-phase full-bridge thyristor rectifier can be calculated using the following formula:
Vavg = (2 * √2 * Vm / π) * (1 - cos(α))
Where:
Vavg = Average output voltage
Vm = Peak value of supply voltage
α = Firing angle of the thyristors
The peak value of the supply voltage (Vm) can be calculated as:
Vm = √2 * Vrms
Where:
Vrms = RMS value of the supply voltage
The RMS value of the supply voltage (Vrms) can be calculated as:
Vrms = Vm / √2
Given:
Supply voltage (V) = 230 V
Frequency (f) = 50 Hz
Power (P) = 10 kW
First, let's calculate the RMS value of the supply voltage:
Vrms = V / √2
= 230 / √2
= 162.63 V
Next, let's calculate the peak value of the supply voltage:
Vm = √2 * Vrms
= √2 * 162.63
= 230 V
Now, let's calculate the average output voltage at the rated power:
Vavg = (2 * √2 * Vm / π) * (1 - cos(α))
= (2 * √2 * 230 / π) * (1 - cos(α))
We know that power (P) is equal to the product of average voltage (Vavg) and current (I):
P = Vavg * I
Since we have a resistive load, the current (I) can be calculated as:
I = P / Vavg
Substituting the value of power (P) and average voltage (Vavg) in the above equation, we get:
10,000 = [(2 * √2 * 230 / π) * (1 - cos(α))] * (10,000 / [(2 * √2 * 230 / π) * (1 - cos(α))])
The above equation simplifies to:
1 = 1 - cos(α)
Simplifying further, we get:
cos(α) = 0
To achieve the rated power output of 10 kW, the firing angle (α) of the thyristors required is 90 degrees.
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Community Answer
A single-phase, full-bridge, fully controlled thyristor rectifier feed...
Concept:
In a single-phase full wave rectifier with source inductance is given by,
Average output current
Where, Vm is the maximum value of supply voltage
α is the firing angle
μ is the overlapping angle
Ls is the source inductance
R is the load resistance
In a single-phase full wave rectifier with source inductance is given by,
Average output current
Where, Vm is the maximum value of supply voltage
α is the firing angle
μ is the overlapping angle
Ls is the source inductance
R is the load resistance
Calculation:
Given that, supply voltage (Vrms) = 230 V
Firing angle (α) = 45°
Source inductance (Ls) = 2.28 mH
Load resistance (R) = 10 Ω
Frequency (f) = 50 Hz
The average output voltage is
⇒ I0 = 14.003 A
⇒ μ = 4.8°
Given that, supply voltage (Vrms) = 230 V
Firing angle (α) = 45°
Source inductance (Ls) = 2.28 mH
Load resistance (R) = 10 Ω
Frequency (f) = 50 Hz
The average output voltage is
⇒ I0 = 14.003 A
⇒ μ = 4.8°
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A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer?
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A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer?.
A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single-phase, full-bridge, fully controlled thyristor rectifier feeds a load comprising a 10 Ω resistance in series with a very large inductance. The rectifier is fed from an ideal 230 V, 50 Hz sinusoidal source through cables which have negligible internal resistance and a total inductance of 2.28 mH. If the thyristors are triggered at an angle α = 45°, the commutation overlap angle in degree (rounded off to 2 decimal places) is _______Correct answer is between '4.51,5.10'. Can you explain this answer?.
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