Respuesta :
Answer:
the phasor form of voltage [tex]\( E \) is \( 60 \cdot (\cos(70^\circ) - j \cdot \sin(70^\circ)) \) and the phasor form of current \( I \) is \( 15 \cdot (\sin(50^\circ) + j \cdot \cos(50^\circ)) \).[/tex]
Explanation:
To express the voltage and current in phasor form, we need to convert the given sinusoidal expressions into complex exponential forms. The complex exponential form is useful for analyzing sinusoidal signals in the frequency domain.
The complex exponential form of a sinusoidal signal is given by:
[tex]\[ x(t) = A \cdot e^{j(\omega t + \phi)} \][/tex]
where:
[tex]- \( x(t) \) is the sinusoidal signal,[/tex]
[tex]- \( A \) is the amplitude of the signal,[/tex]
[tex]- \( \omega \) is the angular frequency (in radians per second),[/tex]
[tex]- \( t \) is time (in seconds),[/tex]
[tex]- \( \phi \) is the phase angle (in radians), and[/tex]
[tex]- \( j \) is the imaginary unit (\( j = \sqrt{-1} \)).[/tex]
Given that:
- Instantaneous voltage [tex]\( E = 60 \cos(\omega t - 70^\circ) \)[/tex]
- Instantaneous current [tex]\( I = 15 \sin(\omega t + 50^\circ) \)[/tex]
We can rewrite these expressions in phasor form:
For voltage E:
[tex]\[ E = 60 \cos(\omega t - 70^\circ) \][/tex]
[tex]\[ E = 60 \cdot \cos(-70^\circ) \cdot \cos(\omega t) + 60 \cdot \sin(-70^\circ) \cdot \sin(\omega t) \][/tex]
[tex]\[ E = 60 \cdot \cos(70^\circ) \cdot \cos(\omega t) - 60 \cdot \sin(70^\circ) \cdot \sin(\omega t) \][/tex]
[tex]\[ E = 60 \cdot (\cos(70^\circ) - j \cdot \sin(70^\circ)) \cdot e^{j\omega t} \][/tex]
For current \( I \):
[tex]\[ I = 15 \sin(\omega t + 50^\circ) \][/tex]
[tex]\[ I = 15 \cdot \cos(50^\circ) \cdot \sin(\omega t) + 15 \cdot \sin(50^\circ) \cdot \cos(\omega t) \][/tex]
[tex]\[ I = 15 \cdot \sin(50^\circ) \cdot \sin(\omega t) + 15 \cdot \cos(50^\circ) \cdot \cos(\omega t) \][/tex]
[tex]\[ I = 15 \cdot (\sin(50^\circ) + j \cdot \cos(50^\circ)) \cdot e^{j\omega t} \][/tex]
Therefore, the phasor form of voltage [tex]\( E \) is \( 60 \cdot (\cos(70^\circ) - j \cdot \sin(70^\circ)) \) and the phasor form of current \( I \) is \( 15 \cdot (\sin(50^\circ) + j \cdot \cos(50^\circ)) \).[/tex]
Final answer:
The voltage and current in phasor form are E = 60∠-70° V and I = 15∠-40° A, respectively, after converting the current from sine to cosine form.
Explanation:
The instantaneous voltage and current in a circuit are given as E = 60 cos (ωt-70°) and I = 15 sin (ωt + 50°) respectively. To express these in phasor form, we first convert them into a standard cosine function for phasors. The voltage phasor can be directly written as E = 60∠-70° V since it's already in cosine form. The current, however, needs to be converted from sine to cosine, noting that sin(θ) = cos(θ - 90°), which gives us I = 15 cos(ωt + 50° - 90°), simplifying to I = 15∠-40° A.
