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Hi there!
Basically I'm ploughing through my assignment and trying to work out the phase angles of a capacitor, of the coil and of the supply.
I was given the values as follows:
Supply: 260V @ 50Hz
Inductor: 31.831mH
Resistor: 24 Ohm
I had to work out the value of the cap which I worked out as: 47.1 uF
My attempt
Note: Results given rounded to 2dp.
I worked out that the current through the coil and its phase angle as follows:
Current: 10A
Angle: 23º
( I did this by working out Xl, then the impedance then using ohms law and results were similar from simulation)
I also worked out the current in the cap and its phase angle as follows:
Current: 3.85A
Angle= 90º (as it leads)
( I did this by working out Xc, then using ohms law and results were similar from simulation)
Now for the supply angle and its current I did the following.
I worked out the the total horizontal component which was = 9.21A
Then did the same for the vertical components and that calculated at = -0.06
Then calculated I by doing the following I= √((9.21)²+(-0.06)²)= 9.21
To get the angle I used inverse tan as follows:
tan-1(0.06/9.21)=0.37º
To me this angle seems way to small?! Have I gone wrong somewhere?
Many thanks,
Basically I'm ploughing through my assignment and trying to work out the phase angles of a capacitor, of the coil and of the supply.
I was given the values as follows:
Supply: 260V @ 50Hz
Inductor: 31.831mH
Resistor: 24 Ohm
I had to work out the value of the cap which I worked out as: 47.1 uF
My attempt
Note: Results given rounded to 2dp.
I worked out that the current through the coil and its phase angle as follows:
Current: 10A
Angle: 23º
( I did this by working out Xl, then the impedance then using ohms law and results were similar from simulation)
I also worked out the current in the cap and its phase angle as follows:
Current: 3.85A
Angle= 90º (as it leads)
( I did this by working out Xc, then using ohms law and results were similar from simulation)
Now for the supply angle and its current I did the following.
I worked out the the total horizontal component which was = 9.21A
Then did the same for the vertical components and that calculated at = -0.06
Then calculated I by doing the following I= √((9.21)²+(-0.06)²)= 9.21
To get the angle I used inverse tan as follows:
tan-1(0.06/9.21)=0.37º
To me this angle seems way to small?! Have I gone wrong somewhere?
Many thanks,
