i = 4 sin(10t + 10◦)
Which phasor correct for the sinusoidal above?
1. magnitude = 4 phase = -80◦
2. maginitude = 4 phase = 100◦ (this is my answer, is it wrong ? why?)

2nd problem

dv/dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦)

= Re(ωVm * e^jωt * e^jφ * e^j90◦ ) = Re(j * ω * V * e^jωt )

how did j appear in Re(j ω V e^jωt )?

 

i = 4 sin(10t + 10◦)
Which phasor correct for the sinusoidal above?
1. magnitude = 4 phase = -80◦
2. maginitude = 4 phase = 100◦ (this is my answer, is it wrong ? why?)

2nd problem

dv/dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦)

= Re(ωVm * e^jωt * e^jφ * e^j90◦ ) = Re(j * ω * V * e^jωt )

how did j appear in Re(j ω V e^jωt )?

For the first portion of your question, phasors are usually written in the following form,

\(V_{m}sin( \omega t + \phi)\)

Where \(V_{m}\) is the magnitude, omega is the frequency (in rad/s) and \(\phi\) is the phase shift.

For the second part of your question, are you trying to figure out what the derivative of this phasor is, like so,

\(\frac{d}{dt}\left( -\omega V_{m}sin( \omega t + \phi) \right)\)

?

 

For the first portion of your question, phasors are usually written in the following form,

\(V_{m}sin( \omega t + \phi)\)

Where \(V_{m}\) is the magnitude, omega is the frequency (in rad/s) and \(\phi\) is the phase shift.

For the second part of your question, are you trying to figure out what the derivative of this phasor is, like so,

\(\frac{d}{dt}\left( -\omega V_{m}sin( \omega t + \phi) \right)\)

?

1st question - Very sorry , its polar form, i cudnt find that angle symbol here, so i wrote magnitude = .. phase =..

2nd question - Sorry Again,
v(t) = Re(V * e^jωt ) = Vm cos (ωt+φ),

dv/dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦)

= Re(ωVm * e^jωt * e^jφ * e^j90◦ ) = Re(j * ω * V * e^jωt )

how did j appear in Re(j ω V e^jωt )?

 

1st question - Very sorry , its polar form, i cudnt find that angle symbol here, so i wrote magnitude = .. phase =..

2nd question - Sorry Again,
v(t) = Re(V * e^jωt ) = Vm cos (ωt+φ),

dv/dt = −ωVm sin(ωt + φ) = ωVm cos(ωt + φ + 90◦)

= Re(ωVm * e^jωt * e^jφ * e^j90◦ ) = Re(j * ω * V * e^jωt )

how did j appear in Re(j ω V e^jωt )?

Okay so for the first question again, it will be of the form,

\(V_{m} \quad \angle \phi \)

For the second question, I don't see how you get that j down there.

\(\frac{dv}{dt} = \omega V_{m}cos(\omega t + \phi + \beta), \quad \text{ Where, } \beta = 90^{o}\)

So,

\(Re \left{ V_{m}e^{j(\phi + \beta)}e^{jwt} \right}\)

Note, \(e^{jwt}\) in simply a complex factor, the information we seek is representated by,

\(\overline{V} = V_{m}e^{j(\phi - \beta)}} = V_{m} \angle (\phi - \beta)\)

 

Okay so for the first question again, it will be of the form,

\(V_{m} \quad \angle \phi \)

For the second question, I don't see how you get that j down there.

\(\frac{dv}{dt} = \omega V_{m}cos(\omega t + \phi + \beta), \quad \text{ Where, } \beta = 90^{o}\)

So,

\(Re \left{ V_{m}e^{j(\phi + \beta)}e^{jwt} \right}\)

Note, \(e^{jwt}\) in simply a complex factor, the information we seek is representated by,

\(\overline{V} = V_{m}e^{j(\phi - \beta)}} = V_{m} \angle (\phi - \beta)\)

Its solved 1st one i have took it like 90-θ when i should take θ-90 ((ωt+10)-90)

2nd one,
\(e^{j90}\) = cos90 + jsin90 = 0+j = j

Thanks