I did another online course for this a few years ago. What a challenge it was to try this again cold, after not thinking about it for years. To my happy surprise it mostly came back, ha ha.
This is something I struggled with last year. Not so much struggled, I guess. It's pretty straightforward. But when you're trying to cram all this into your brain plus code, plus labs, plus HVAC and Mags.. At some point the spaghetti stops sticking to the wall :p really straightforward info, thanks!
Dear My Friend, Save some nerve with using complex method! :D The result comes with 3 steps! Step1: R=15ohms, X(RL)=(14.4+j10.8)ohms and X(C)=(-j40)ohms Step2: I(R)=16A, I(RL)=(10.66-j8)A and I(C)=(j6)A so I(total)=(26.66-j2)A. From this I(pol)= 26.73A with -4.29 degree. Step3: Draw all of this in a complex coordinate system and the phasor is there within 10 seconds! Happy days!
The 240V of the AC voltage source is RMS in this video? Sometimes it is given like Acos(wt+teta), so should i convert A peak voltage to RMS in this situation? Thank for answer.
In a four branch parallel circuit, 10mA of current flows in each branch . If one of the branch is opens, the current in each of the other branches.. ? A) decrease B) increase C) remaining unaffected D) doubles Please say the answer sir
It's often something you're provided in one way or another and always depends on the load. It is the relationship between current and voltage across a load
It's great to have these online in lecture format. Thank you! And nice job on writing backwards!
its a camera trick
I did another online course for this a few years ago. What a challenge it was to try this again cold, after not thinking about it for years. To my happy surprise it mostly came back, ha ha.
This is something I struggled with last year. Not so much struggled, I guess. It's pretty straightforward. But when you're trying to cram all this into your brain plus code, plus labs, plus HVAC and Mags.. At some point the spaghetti stops sticking to the wall :p really straightforward info, thanks!
Glad it helped! IT's a very busy time when in school with a ton of information. I understand the struggle!
Shouldn't the inductance angle be positive because inductuance is postitive j
Thanks for solving this learning problem for me!
Dear My Friend,
Save some nerve with using complex method! :D The result comes with 3 steps!
Step1:
R=15ohms, X(RL)=(14.4+j10.8)ohms and X(C)=(-j40)ohms
Step2:
I(R)=16A, I(RL)=(10.66-j8)A and I(C)=(j6)A so I(total)=(26.66-j2)A. From this I(pol)= 26.73A with -4.29 degree.
Step3:
Draw all of this in a complex coordinate system and the phasor is there within 10 seconds!
Happy days!
Awesome Video Zack!
Thanks Sir, nicely explained,
Keep making videos for us.
The 240V of the AC voltage source is RMS in this video? Sometimes it is given like Acos(wt+teta), so should i convert A peak voltage to RMS in this situation? Thank for answer.
Thanks 👍
Great!
What is zth is it divide
Saves my ass in school
In a four branch parallel circuit, 10mA of current flows in each branch . If one of the branch is opens, the current in each of the other branches.. ?
A) decrease
B) increase
C) remaining unaffected
D) doubles
Please say the answer sir
C - Remains uneffected
@@ZackHartle thank you sir.. it's very useful for my exam preparation..thanks a lot
It's really hard to write reverse,
After making this video, perhaps you see the world in reverse too 😂
im not sure how to get the phase angle thats the one thing i find difficult to wrap my brain around
It's often something you're provided in one way or another and always depends on the load.
It is the relationship between current and voltage across a load
im lost