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A half-wave rectifier employs a transformer with turn ratio n1 : n2 = 12 : 1. If primary coil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?
Correct answer is '25 .9'. Can you explain this answer?
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A half-wave rectifier employs a transformer with turn ratio n1 : n2= 1...
The turn ratio of transform ation is 12:1. So if primary coil is connected to power mains o f 220V , then potential of secondary coil is 
Volts. In a half-wave rectifier,
is the maximum voltage appearing across the transformer secondary coil. Now. peak inverse voltage, 
for half wave rectifier in given as V0
= 25.9V
The correct answer is: 25 .9
Volts. In a half-wave rectifier,is the maximum voltage appearing across the transformer secondary coil. Now. peak inverse voltage, for half wave rectifier in given as V0= 25.9V
The correct answer is: 25 .9
Most Upvoted Answer
A half-wave rectifier employs a transformer with turn ratio n1 : n2= 1...
Given information:
- Transformer turn ratio: n1 : n2 = 12 : 1
- Input voltage: 220V, 50Hz
- Diode resistance in forward bias: negligible
Calculating peak voltage:
The peak voltage of the input signal can be calculated using the formula:
Vp = Vm × √2
Where Vp is the peak voltage and Vm is the maximum voltage.
The maximum voltage can be calculated using the formula:
Vm = Vrms × √2
Where Vrms is the root mean square (RMS) voltage.
The RMS voltage can be calculated using the formula:
Vrms = Vdc / √2
Where Vdc is the DC voltage.
Calculating DC voltage:
The DC voltage can be calculated by multiplying the peak voltage of the input signal by the duty cycle.
In a half-wave rectifier, the duty cycle is 0.5 since only half of the input waveform is rectified.
Therefore, the DC voltage is given by:
Vdc = Vp × 0.5
Calculating peak inverse voltage (PIV):
The PIV is the maximum voltage that appears across the diode when it is in reverse bias.
In a half-wave rectifier, the PIV is equal to the peak voltage of the input signal.
Calculating the turn ratio:
The turn ratio of the transformer is given as n1 : n2 = 12 : 1.
This means that for every 12 turns on the primary coil, there is 1 turn on the secondary coil.
Calculating the secondary voltage:
The secondary voltage can be calculated using the turn ratio formula:
V2 / V1 = n2 / n1
Where V2 is the secondary voltage and V1 is the primary voltage.
In this case, V1 is the input voltage of 220V.
Plugging in the values, we get:
V2 / 220 = 1 / 12
Cross-multiplying, we get:
V2 = 220 / 12 = 18.33V
Calculating the peak voltage:
The peak voltage of the secondary voltage can be calculated using the formula:
Vp2 = V2 × √2
Plugging in the value of V2, we get:
Vp2 = 18.33 × √2 = 25.9V
Therefore, the peak inverse voltage (PIV) of the diode is approximately 25.9V.
- Transformer turn ratio: n1 : n2 = 12 : 1
- Input voltage: 220V, 50Hz
- Diode resistance in forward bias: negligible
Calculating peak voltage:
The peak voltage of the input signal can be calculated using the formula:
Vp = Vm × √2
Where Vp is the peak voltage and Vm is the maximum voltage.
The maximum voltage can be calculated using the formula:
Vm = Vrms × √2
Where Vrms is the root mean square (RMS) voltage.
The RMS voltage can be calculated using the formula:
Vrms = Vdc / √2
Where Vdc is the DC voltage.
Calculating DC voltage:
The DC voltage can be calculated by multiplying the peak voltage of the input signal by the duty cycle.
In a half-wave rectifier, the duty cycle is 0.5 since only half of the input waveform is rectified.
Therefore, the DC voltage is given by:
Vdc = Vp × 0.5
Calculating peak inverse voltage (PIV):
The PIV is the maximum voltage that appears across the diode when it is in reverse bias.
In a half-wave rectifier, the PIV is equal to the peak voltage of the input signal.
Calculating the turn ratio:
The turn ratio of the transformer is given as n1 : n2 = 12 : 1.
This means that for every 12 turns on the primary coil, there is 1 turn on the secondary coil.
Calculating the secondary voltage:
The secondary voltage can be calculated using the turn ratio formula:
V2 / V1 = n2 / n1
Where V2 is the secondary voltage and V1 is the primary voltage.
In this case, V1 is the input voltage of 220V.
Plugging in the values, we get:
V2 / 220 = 1 / 12
Cross-multiplying, we get:
V2 = 220 / 12 = 18.33V
Calculating the peak voltage:
The peak voltage of the secondary voltage can be calculated using the formula:
Vp2 = V2 × √2
Plugging in the value of V2, we get:
Vp2 = 18.33 × √2 = 25.9V
Therefore, the peak inverse voltage (PIV) of the diode is approximately 25.9V.
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A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer?
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A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer?.
A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer?.
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A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer?, a detailed solution for A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer? has been provided alongside types of A half-wave rectifier employs a transformer with turn ratio n1 : n2= 12 : 1.If primarycoil is connected to the power mains 220V, 50Hz. If the diode resistance in forward bias is negligible, what is the peak inverse voltage (in Volts) of the diode?Correct answer is '25 .9'. Can you explain this answer? theory, EduRev gives you an
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