Question 1.20: Determine the voltage developed across inductor terminals wh...
Determine the voltage developed across inductor terminals when a sinusoidal current that is a function of time (t) flows through it. The sinusoidal current is depicted in Figure 1.33, and three key parameters determine the sinusoidal current:
• Frequency (f): the inverse of the sinusoidal waveform period T.
• Phase angle (φ): the angle with respect to the origin of the time axis that
the waveform is shifted.
• Amplitude: also called the peak value of the waveform.
Figure 1.33 depicts a sinusoidal current waveform that conforms to Equation
(108):
i(t)=I_{Peak}sin(2\pi ft+ϕ) (1.108)
For this example we assume that
I_{Peak}=1 A, f = 1000 Hz, and ϕ is zero degrees
Where the unit of peak current is the ampere, the unit of frequency is second^{-1}(s^{-1}),also called hertz and abbreviated Hz. The unit of phase angle is either degrees or radians. When the phase is expressed in degrees, the“°” has to be explicitly shown next to the number of degrees. If radians are used, then no units are indicated next to the phase angle in radians. The radian is considered to be dimensionless. From high school geometry, let us remember that 2π radians or simply 2π equals 360° . So, for example, a 45° angle equals to π/4 (i.e., π/4 radians).
Rewriting Equation (1.108) with the parameters given by Equation (1.109) leads to
I_{Peak}=1 A, f = 1000 Hz, and ϕ is zero degrees (1.109)
i (t) = 1 sin( 2000πt + 0°) . (1.110)
This can be simply stated as
i (t) = sin(2000πt) A . (1.111)
The argument of the sinusoidal waveform in Equation (1.111), 2πft is alsoreferred to as ωt; where ω (the Greek letter omega) is called the sinusoidal waveform angular frequency or pulsation. Its units are hertz or s^{-1}. In some electrical engineering literature, the units of ω are also referred to as radians per second (rad/s). Since the radian is a dimensionless unit, hertz and rad/s are basically the same thing.
Figure 1.33 depicts the waveform described by Equation (1.111) with a zero phase angle.
Solution to Example 1.20
V_{L}(t)=Ldi/dt (1.112)
The voltage across the inductor produced by the sinusoidal current given by Equation (1.112) is easily calculated plugging Equation (1.111) into Equation (1.112).
Thus,
V_{L}(t)=L\frac{d}{dt} [sin(2000\pi t] (1.113)
=V_{L}(t)=2000\pi Lcos(2000\pi t)
Figure 1.34 shows both the sinusoidal current through the inductor and the sinusoidal voltage developed across the inductor terminals. It is important to note that the voltage waveform leads the current waveform by 90° . Note: Cosine and sine waveforms are both referred to as sinusoidal waveforms in a general sense.
