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A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectively
- a)10 A, 5 A
- b)5 A, 7.07 A
- c)7.07 A, 10 A
- d)7.07 A, 5 A
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A single-phase full-bridge diode rectifier delivers a load current of ...
Given, I0 = 10A
The rms value of diode current,
The average value of diode current,
The rms value of diode current,
The average value of diode current,
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A single-phase full-bridge diode rectifier delivers a load current of ...
Calculation of RMS and Average Values of Diode Currents in a Single-Phase Full-Bridge Diode Rectifier
Given:
Load current = 10 A (ripple free)
To find:
RMS and average values of diode currents
Solution:
The circuit diagram of a single-phase full-bridge diode rectifier is as follows:
For the positive half-cycle of the input voltage, the diodes D1 and D2 conduct, and for the negative half-cycle, the diodes D3 and D4 conduct.
During the positive half-cycle, the current flows from the source through the transformer, diode D1, load, and diode D2 back to the source. Therefore, the average value of diode current is:
Id_avg = (I_load + I_D1 + I_D2) / 2
Since the load current is ripple-free, I_load = 10 A. The voltage drop across the conducting diode is typically 0.7 V. Therefore, the voltage drop across D1 and D2 is 1.4 V.
The transformer secondary voltage is given by:
V_sec = V_p / N
where V_p is the peak value of the input voltage, and N is the turns ratio of the transformer.
Since the load current is ripple-free, the transformer secondary current is also ripple-free. Therefore, the rms value of the transformer secondary current is:
I_sec_rms = I_load / √2
The current through diode D1 is given by:
I_D1 = I_sec_rms * (V_sec / V_D1)
where V_D1 is the voltage drop across D1, which is 0.7 V.
Similarly, the current through diode D2 is given by:
I_D2 = I_sec_rms * (V_sec / V_D2)
where V_D2 is the voltage drop across D2, which is also 0.7 V.
Substituting the values, we get:
I_D1 = I_D2 = 7.07 A
Therefore, the average value of diode current is:
Id_avg = (10 A + 7.07 A + 7.07 A) / 2 = 12.07 A
The rms value of diode current is the same as the rms value of transformer secondary current, which is:
I_D_rms = I_sec_rms = 10 / √2 = 7.07 A
Therefore, the correct answer is option D: 7.07 A (rms value) and 5 A (average value).
Given:
Load current = 10 A (ripple free)
To find:
RMS and average values of diode currents
Solution:
The circuit diagram of a single-phase full-bridge diode rectifier is as follows:
For the positive half-cycle of the input voltage, the diodes D1 and D2 conduct, and for the negative half-cycle, the diodes D3 and D4 conduct.
During the positive half-cycle, the current flows from the source through the transformer, diode D1, load, and diode D2 back to the source. Therefore, the average value of diode current is:
Id_avg = (I_load + I_D1 + I_D2) / 2
Since the load current is ripple-free, I_load = 10 A. The voltage drop across the conducting diode is typically 0.7 V. Therefore, the voltage drop across D1 and D2 is 1.4 V.
The transformer secondary voltage is given by:
V_sec = V_p / N
where V_p is the peak value of the input voltage, and N is the turns ratio of the transformer.
Since the load current is ripple-free, the transformer secondary current is also ripple-free. Therefore, the rms value of the transformer secondary current is:
I_sec_rms = I_load / √2
The current through diode D1 is given by:
I_D1 = I_sec_rms * (V_sec / V_D1)
where V_D1 is the voltage drop across D1, which is 0.7 V.
Similarly, the current through diode D2 is given by:
I_D2 = I_sec_rms * (V_sec / V_D2)
where V_D2 is the voltage drop across D2, which is also 0.7 V.
Substituting the values, we get:
I_D1 = I_D2 = 7.07 A
Therefore, the average value of diode current is:
Id_avg = (10 A + 7.07 A + 7.07 A) / 2 = 12.07 A
The rms value of diode current is the same as the rms value of transformer secondary current, which is:
I_D_rms = I_sec_rms = 10 / √2 = 7.07 A
Therefore, the correct answer is option D: 7.07 A (rms value) and 5 A (average value).
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A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer?.
A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer? for Electrical Engineering (EE) 2025 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 diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer?.
Solutions for A single-phase full-bridge diode rectifier delivers a load current of 10 A, which is ripple free. The rms and average values of diode currents are respectivelya)10 A, 5 Ab)5 A, 7.07 Ac)7.07 A, 10 Ad)7.07 A, 5 ACorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
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