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For a sine wave with peak value Emax the average value is –when the phase difference between the current and voltage is –
- a)0.636 Emax
- b)0.707 Emax
- c)0.434 Emax
- d)1.414 Emax
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
For a sine wave with peak value Emax the average value is –when ...
Average Value of a Sine Wave with Phase Difference
Explanation:
When a sine wave is applied to a circuit, the current and voltage may not be in phase with each other. The phase difference between the current and voltage is represented by the angle Φ.
The average value of a sine wave with peak value Emax can be calculated using the formula:
Vavg = (2/π) * Emax * sin(Φ/2)
Where Vavg is the average value of the sine wave.
To find the phase difference at which the average value is maximum, we need to differentiate this equation with respect to Φ and equate it to zero.
dVavg/dΦ = (Emax/π) * cos(Φ/2) = 0
cos(Φ/2) = 0
Φ/2 = π/2, 3π/2, 5π/2, ...
Φ = π, 3π, 5π, ...
The phase difference at which the average value is maximum is therefore Φ = π.
Substituting this value into the equation for Vavg gives:
Vavg = (2/π) * Emax * sin(π/2)
Vavg = 0.636 * Emax
Therefore, the correct option is A) 0.636 * Emax.
Conclusion:
The average value of a sine wave with peak value Emax is maximum when the phase difference between the current and voltage is π. At this phase difference, the average value is 0.636 * Emax.
Explanation:
When a sine wave is applied to a circuit, the current and voltage may not be in phase with each other. The phase difference between the current and voltage is represented by the angle Φ.
The average value of a sine wave with peak value Emax can be calculated using the formula:
Vavg = (2/π) * Emax * sin(Φ/2)
Where Vavg is the average value of the sine wave.
To find the phase difference at which the average value is maximum, we need to differentiate this equation with respect to Φ and equate it to zero.
dVavg/dΦ = (Emax/π) * cos(Φ/2) = 0
cos(Φ/2) = 0
Φ/2 = π/2, 3π/2, 5π/2, ...
Φ = π, 3π, 5π, ...
The phase difference at which the average value is maximum is therefore Φ = π.
Substituting this value into the equation for Vavg gives:
Vavg = (2/π) * Emax * sin(π/2)
Vavg = 0.636 * Emax
Therefore, the correct option is A) 0.636 * Emax.
Conclusion:
The average value of a sine wave with peak value Emax is maximum when the phase difference between the current and voltage is π. At this phase difference, the average value is 0.636 * Emax.
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For a sine wave with peak value Emax the average value is –when ...
Vaverage=(2/pi)*Emax
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