I wan't to write the instantaneous value expression from the curves below.

The voltage is the bigger one and the other is the current and the scale for the voltage is 1 V and 10 mA for the current.

f = 1/(20E-3) = 50 Hz

For the voltage I get |U| = 2 V and α = 0° => u(t) = 2 Sin (2π50t) V

For the current I get |I| 10 mA and from the graphs it looks like the current is ~1,25 ms behind so I get α =(-1,25/20)*360 = -21,6° => i(t) = 10 Sin(2π50t - 21,6°) mA
The awnser for the current is wrong and it should be 10 Sin(2π50t - 30°) mA. What am I doing wrong when calculating the phase shift?

http://s32.postimg.org/yhmu67ryd/moment.png
It appears that you've incorrectly estimated the voltage-current lag. -30° implies α=-30=(lag/20)*360 = -5/3 ~= -1.6667 milliseconds. Surely you were given the equations for those two so you could find the derivatives of each, compute the exact peaks, and determine the lag from that?
Thanks!
KermMartian wrote:
It appears that you've incorrectly estimated the voltage-current lag. -30° implies α=-30=(lag/20)*360 = -5/3 ~= -1.6667 milliseconds. Surely you were given the equations for those two so you could find the derivatives of each, compute the exact peaks, and determine the lag from that?

I don't seem te get it now when I look at it again. I was not given the equations for thoes two and from the graphs it looks like the lag is about a half of a half which would be 5/4 = -1.25 ms.
Can you show me how you get -1,67 just from looking at the graphs?
From this grap I get: 50/4 and (-12/50)*360 =- 90 deg but if I look at the solution it uses -60 deg.

http://s33.postimg.org/ij47d6vgv/grafer222.png
Solved it... pi/3 = 60....

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