Write the equation of 25 cycle current sine wave having rms value of 30 A.
Answer
653.4k+ views
Hint
A curve that represents periodic oscillation of a wave during the movement of energy from one point to the other is referred to as a sinusoidal wave or sine wave. Rms or root mean square current is used in expressing alternating current that is equivalent to the same form as direct current.
Complete step by step answer
Given,
Frequency of the current sine wave, $\nu = 25$ Hz
Rms value of current, ${{\rm{I}}_{{\rm{rms}}}} = 30$ A
The required equation of the current sine wave is determined using,
${\rm{I}} = {{\rm{I}}_0}\sin {\rm{\omega t}}$ … (1)
Since the peak value of current i.e. ${{\rm{I}}_0}$ is square root times the rms value of current,
${{\rm{I}}_0} = \sqrt 2 {{\rm{I}}_{{\rm{rms}}}}$
On putting the given value of rms current, we get
${{\rm{I}}_0} = \sqrt 2 \times 30$
∴ ${{\rm{I}}_0} = 42.24$ A … (2)
On using the relation between angular frequency and regular frequency, we get
${\rm{\omega }} = 2{\rm{\pi \nu }}$
On putting the value of regular frequency, we get
${\rm{\omega }} = 2{\rm{\pi }} \times 25$
∴ ${\rm{\omega }} = 50{\rm{\pi }}\;{\rm{rad\;}}{{\rm{s}}^{ - 1}}$ … (3)
On substituting values of equations (2) and (3) in equation (1), we get
${\rm{I}} = 42.24\sin 50{\rm{\pi }}\;{\rm{t}}$
Therefore, ${\rm{I}} = 42.42\sin 50{\rm{\pi t}}$ is the required equation.
Note
The term 25 cycle current in the question represents the rate at which the direction of current changes. The 25 cycle thus represents the frequency of the current sine wave.
A curve that represents periodic oscillation of a wave during the movement of energy from one point to the other is referred to as a sinusoidal wave or sine wave. Rms or root mean square current is used in expressing alternating current that is equivalent to the same form as direct current.
Complete step by step answer
Given,
Frequency of the current sine wave, $\nu = 25$ Hz
Rms value of current, ${{\rm{I}}_{{\rm{rms}}}} = 30$ A
The required equation of the current sine wave is determined using,
${\rm{I}} = {{\rm{I}}_0}\sin {\rm{\omega t}}$ … (1)
Since the peak value of current i.e. ${{\rm{I}}_0}$ is square root times the rms value of current,
${{\rm{I}}_0} = \sqrt 2 {{\rm{I}}_{{\rm{rms}}}}$
On putting the given value of rms current, we get
${{\rm{I}}_0} = \sqrt 2 \times 30$
∴ ${{\rm{I}}_0} = 42.24$ A … (2)
On using the relation between angular frequency and regular frequency, we get
${\rm{\omega }} = 2{\rm{\pi \nu }}$
On putting the value of regular frequency, we get
${\rm{\omega }} = 2{\rm{\pi }} \times 25$
∴ ${\rm{\omega }} = 50{\rm{\pi }}\;{\rm{rad\;}}{{\rm{s}}^{ - 1}}$ … (3)
On substituting values of equations (2) and (3) in equation (1), we get
${\rm{I}} = 42.24\sin 50{\rm{\pi }}\;{\rm{t}}$
Therefore, ${\rm{I}} = 42.42\sin 50{\rm{\pi t}}$ is the required equation.
Note
The term 25 cycle current in the question represents the rate at which the direction of current changes. The 25 cycle thus represents the frequency of the current sine wave.
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