For question 1.b if we add -180 it will also change the sign but we will have a different phase angle. We will get V laggs by 180 and it is another answer no?
To convert a negative sine angle to a positive sine ,you must add 180 degree to it . that's why he added 180 degree to the original phase angle 90 degree and changed the -sine into +sine. Hope it cleared your doubt😊
something i dont understand. in example 2.a at 15:45 I can be I = 10cos(wt+180+50) [adding 180 to make the sign positive] and then v = 12cos(wt-10-90) [we subtract by 90 to convert positive sin to positive cos] so that makes it v = 12cos(wt-100) and I = 10cos(wt+230) that make the phase difference 330 right. whats wrong about this approach
With the last question, when you change the angle of the voltage to sine, and not the current to cosine, will you still get the current leading the Voltage by 155 degrees as the answer?
@@SkanCityAcademy_SirJohn thank you so much, I actually got a different answer...the current leading the Voltage by 65 degrees when the voltage is changed to sine.
Hi! I tried changing voltage to sine and making the current's amplitude positive, and this is my solution: I = 4 sin (377t + 25 + 180) so I = 4 sin (377t + 205) and then, V = sin (377t -40 + 90) so V = sin (377t + 50) finally, to compare the two phasors, I = 4 sin (377t + 50 + 155) V = 4 sin (377t + 50) Therefore, the phase change is still 155 degrees.
For question 1.b if we add -180 it will also change the sign but we will have a different phase angle. We will get V laggs by 180 and it is another answer no?
I don't think so, but first check if your trig conversion is correct or not, I believe that is where the problem comes from.
Excellent explanation 👏👏👏
Thanks so so much Evelyn.
Where do you watch from?
Can you please explain 1b example once again?
I need explanation
To convert a negative sine angle to a positive sine ,you must add 180 degree to it . that's why he added 180 degree to the original phase angle 90 degree and changed the -sine into +sine. Hope it cleared your doubt😊
@hrithishree145 thanks so much for your contribution
👍👍👍👍👍👍
Very easy to understand God bless you so much
Thanks do much Sharon
Nigerian 😂
Oh okay... Thanks for watching
In 1(b) what to consider if there is v=-20cos(wt+90)
Can you please reframe your question well
@@SkanCityAcademy_SirJohn Thank you got your Point
Most welcome
Where do you watch from?
@@SkanCityAcademy_SirJohn from your video only, good keep going the videos
@gate-oz2gt thanks so much
something i dont understand. in example 2.a at 15:45 I can be I = 10cos(wt+180+50) [adding 180 to make the sign positive] and then v = 12cos(wt-10-90) [we subtract by 90 to convert positive sin to positive cos] so that makes it v = 12cos(wt-100) and I = 10cos(wt+230) that make the phase difference 330 right. whats wrong about this approach
That's also correct, hence I leads V by 330
With the last question, when you change the angle of the voltage to sine, and not the current to cosine, will you still get the current leading the Voltage by 155 degrees as the answer?
It could, but always ensure that the amplitudes of the two signals are positive. That way will be safe.
@@SkanCityAcademy_SirJohn thank you so much, I actually got a different answer...the current leading the Voltage by 65 degrees when the voltage is changed to sine.
Okay... Thanks so much, Teena for coming back
Hi! I tried changing voltage to sine and making the current's amplitude positive, and this is my solution:
I = 4 sin (377t + 25 + 180)
so I = 4 sin (377t + 205)
and then,
V = sin (377t -40 + 90)
so V = sin (377t + 50)
finally, to compare the two phasors,
I = 4 sin (377t + 50 + 155)
V = 4 sin (377t + 50)
Therefore, the phase change is still 155 degrees.