13 Sine wave quantities (A)

13 Sine wave quantities (A)

Vocational training

8 Qs

quiz-placeholder

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13 Sine wave quantities (A)

13 Sine wave quantities (A)

Assessment

Quiz

Construction

Vocational training

Practice Problem

Medium

Created by

Paul Ward

Used 2+ times

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8 questions

Show all answers

1.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

1. A sine wave has a peak voltage of 100V; calculate the RMS value. (1dp)

2.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

2. A sine wave has a peak voltage of 100V; calculate the average value. (1dp)

Answer explanation

Average Voltage = Peak Voltage × 0.636

Average Voltage= 100 × 0.636= 63.6 Volts

3.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

3.A sine wave has a peak voltage of 565.7V; calculate the RMS value.

(Nearest whole number)

4.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

4.A sine wave has a peak voltage of 565.7V; calculate the average value.

(Nearest whole number)

Answer explanation

Average Voltage = Peak Voltage × 0.636

Average Voltage= 565.7 × 0.636= 359.79 Volts

∴ 360 volts to the nearest whole number

5.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

5. A sine wave has a peak voltage of 90V; calculate the RMS value. (2dp)

6.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

6.A sine wave has a peak voltage of 90V; calculate the average value.

(2dp)

Answer explanation

Average Voltage = Peak Voltage × 0.636

Average Voltage= 90 × 0.636= 57.24 Volts

∴ The answer is 57.24 volts

7.

FILL IN THE BLANK QUESTION

30 sec • 1 pt

7.A sine wave has an RMS voltage of 40V; calculate the peak value. (1dp)

Answer explanation

Peak Voltage =RMS Voltage/0.707

or

Peak Voltage=RMS Voltage × √2

40∕0.707 = 56.6 Volts (1dp)

40 × √2 = 56.6 Volts (1dp)

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