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An ac voltage of 220 Vrms is applied to the primary of a 10:1 step-down transformer. The secondary of the transformer is centre tapped and connected to a full-wave rectifier with a load resistance. This dc voltage appearing in the load is.
- a)44/π
- b)31/π
- c)62/π
- d)22/π
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
An ac voltage of 220 Vrms is applied to the primary of a 10:1 step-dow...
For step-down transformer,
E < E1
(44/π)
E < E1
(44/π)
Most Upvoted Answer
An ac voltage of 220 Vrms is applied to the primary of a 10:1 step-dow...
To find the DC voltage appearing in the load, we need to consider the voltage transformation ratio of the transformer and the rectification process.
Given:
AC voltage applied to the primary of the transformer = 220 Vrms
Step-down transformer ratio = 10:1
The secondary voltage of the transformer can be calculated using the transformation ratio:
Secondary voltage = Primary voltage / Transformation ratio
Secondary voltage = 220 Vrms / 10
Secondary voltage = 22 Vrms
Since the secondary is center-tapped, we have two equal halves of the secondary voltage.
Now, let's consider the full-wave rectification process. In a full-wave rectifier, the output voltage is the absolute value of the input voltage. So, the negative half of the voltage is flipped to become positive.
So, the peak voltage of the rectified waveform is equal to the peak voltage of the secondary voltage.
Peak voltage = Secondary voltage = 22 Vrms
To find the DC voltage, we need to convert the peak voltage to the average voltage by multiplying it by 0.637 (the peak-to-average conversion factor for a full-wave rectifier).
Average voltage = Peak voltage * 0.637
Average voltage = 22 V * 0.637
Average voltage ≈ 14.014 V
Therefore, the DC voltage appearing in the load is approximately 14.014 V.
Given:
AC voltage applied to the primary of the transformer = 220 Vrms
Step-down transformer ratio = 10:1
The secondary voltage of the transformer can be calculated using the transformation ratio:
Secondary voltage = Primary voltage / Transformation ratio
Secondary voltage = 220 Vrms / 10
Secondary voltage = 22 Vrms
Since the secondary is center-tapped, we have two equal halves of the secondary voltage.
Now, let's consider the full-wave rectification process. In a full-wave rectifier, the output voltage is the absolute value of the input voltage. So, the negative half of the voltage is flipped to become positive.
So, the peak voltage of the rectified waveform is equal to the peak voltage of the secondary voltage.
Peak voltage = Secondary voltage = 22 Vrms
To find the DC voltage, we need to convert the peak voltage to the average voltage by multiplying it by 0.637 (the peak-to-average conversion factor for a full-wave rectifier).
Average voltage = Peak voltage * 0.637
Average voltage = 22 V * 0.637
Average voltage ≈ 14.014 V
Therefore, the DC voltage appearing in the load is approximately 14.014 V.
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An ac voltage of 220 Vrms is applied to the primary of a 10:1 step-dow...
Correct answer is option (c)
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An ac voltage of 220 Vrms is applied to the primary of a 10:1 step-down transformer. The secondary of the transformer is centre tapped and connected to a full-wave rectifier with a load resistance. This dc voltage appearing in the load is.a)44/πb)31/πc)62/πd)22/πCorrect answer is option 'A'. Can you explain this answer?
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