Hello to All
My homework question is –
“A 3uF and a 6uF capacitor are connected in series to a 24 V a.c. 50Hz supply. Calculate the voltage across each capacitor”
I researched how to answer this one and found this –
https://answers.yahoo.com/question/index?qid=20100219094359AADPjZR
The best answer states that -
Xc = 1/(2pi fc) = 1/[(6.28) x (20) x (.0001) = 79.6 ohm
Impedance angle = arc tan Xc/R = arc tan (79.6 ohm) / (56 ohm) = 54.87 degrees
Voltage across the capacitor = (sine of impedance angle) x (Input Voltage)
Vc = (sin 54.87 degrees) x 10 VRMS = 8.18 VRMS
I have done the first line of this equation and I have –
1/2xPiex50x0.000003= 1,061.03 Ohm
Now when I come to the second line of the solution I am totally stuck, I don’t have two impedance values to get the impedance angle from, so I don't really know how to proceed.
Or if anyone knows a simplier formula for calculating this one I would appreciate it - I have a feel I have missed something with my approach.
Thanks,
Matt.
My homework question is –
“A 3uF and a 6uF capacitor are connected in series to a 24 V a.c. 50Hz supply. Calculate the voltage across each capacitor”
I researched how to answer this one and found this –
https://answers.yahoo.com/question/index?qid=20100219094359AADPjZR
The best answer states that -
Xc = 1/(2pi fc) = 1/[(6.28) x (20) x (.0001) = 79.6 ohm
Impedance angle = arc tan Xc/R = arc tan (79.6 ohm) / (56 ohm) = 54.87 degrees
Voltage across the capacitor = (sine of impedance angle) x (Input Voltage)
Vc = (sin 54.87 degrees) x 10 VRMS = 8.18 VRMS
I have done the first line of this equation and I have –
1/2xPiex50x0.000003= 1,061.03 Ohm
Now when I come to the second line of the solution I am totally stuck, I don’t have two impedance values to get the impedance angle from, so I don't really know how to proceed.
Or if anyone knows a simplier formula for calculating this one I would appreciate it - I have a feel I have missed something with my approach.
Thanks,
Matt.