The problem provides information about a sinusoidal waveform: a peak value of 20 A and a frequency of 50 Hz. It then asks four questions: (I) Write the mathematical expression for that sin wave. (II) Find the value of the current at time $t = 0.0075$ s after starting. (III) Find the time from the starting when the current value is $17.15$ A. (IV) Define the term RMS value for a sin wave, and write the expression.

Applied MathematicsTrigonometrySinusoidal WaveAC CircuitsRMS ValueWaveform Analysis
2025/6/29

1. Problem Description

The problem provides information about a sinusoidal waveform: a peak value of 20 A and a frequency of 50 Hz. It then asks four questions:
(I) Write the mathematical expression for that sin wave.
(II) Find the value of the current at time t=0.0075t = 0.0075 s after starting.
(III) Find the time from the starting when the current value is 17.1517.15 A.
(IV) Define the term RMS value for a sin wave, and write the expression.

2. Solution Steps

(I)
The general form of a sinusoidal current is:
i(t)=Imsin(ωt)i(t) = I_m \sin(\omega t)
where ImI_m is the peak value and ω\omega is the angular frequency.
We are given Im=20I_m = 20 A and f=50f = 50 Hz. The angular frequency ω\omega is related to the frequency ff by the formula:
ω=2πf\omega = 2\pi f
ω=2π(50)=100π\omega = 2\pi (50) = 100\pi rad/s
Therefore, the mathematical expression for the sinusoidal current is:
i(t)=20sin(100πt)i(t) = 20 \sin(100\pi t)
(II)
To find the current at t=0.0075t = 0.0075 s, we substitute t=0.0075t = 0.0075 into the equation:
i(0.0075)=20sin(100π(0.0075))i(0.0075) = 20 \sin(100\pi (0.0075))
i(0.0075)=20sin(0.75π)i(0.0075) = 20 \sin(0.75\pi)
i(0.0075)=20sin(3π4)i(0.0075) = 20 \sin(\frac{3\pi}{4})
i(0.0075)=20sin(135)i(0.0075) = 20 \sin(135^\circ)
i(0.0075)=2022=102i(0.0075) = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} A
i(0.0075)14.14i(0.0075) \approx 14.14 A
(III)
We need to find the time tt when i(t)=17.15i(t) = 17.15 A.
17.15=20sin(100πt)17.15 = 20 \sin(100\pi t)
sin(100πt)=17.1520=0.8575\sin(100\pi t) = \frac{17.15}{20} = 0.8575
100πt=arcsin(0.8575)100\pi t = \arcsin(0.8575)
100πt=1.0306100\pi t = 1.0306 radians (approximately 59.0559.05^\circ)
t=1.0306100π0.00328t = \frac{1.0306}{100\pi} \approx 0.00328 s
There will be multiple times where the current equals 17.1517.15 A, so we should consider the solution in the second quadrant as well. The reference angle is approximately 59.0559.05^\circ, thus the other angle is 18059.05=120.95180^\circ - 59.05^\circ = 120.95^\circ, or 2.112.11 radians.
100πt=2.11100\pi t = 2.11
t=2.11100π0.00672t = \frac{2.11}{100\pi} \approx 0.00672 s
(IV)
The RMS value (Root Mean Square) of a sinusoidal waveform is the effective value of the waveform. For current, it's the DC current that would produce the same heating effect as the AC current.
For a sinusoidal waveform i(t)=Imsin(ωt)i(t) = I_m \sin(\omega t), the RMS value is given by:
Irms=Im2I_{rms} = \frac{I_m}{\sqrt{2}}
Irms=202=10214.14I_{rms} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \approx 14.14 A

3. Final Answer

(I) i(t)=20sin(100πt)i(t) = 20 \sin(100\pi t)
(II) i(0.0075)14.14i(0.0075) \approx 14.14 A
(III) t0.00328t \approx 0.00328 s and t0.00672t \approx 0.00672 s.
(IV) The RMS value of a sinusoidal waveform is the effective value of the waveform. Irms=Im2I_{rms} = \frac{I_m}{\sqrt{2}}.

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