# Finding phase angle of circuit using phasors.

#### danielb33

Joined Aug 20, 2012
105
See circuit in file attached (and problem statement).

I first found the voltage across the inductor - I * Z = 5.6*<-28.
Next I simply used V1 - delta V = V2. I found that 100 - 5.6*cos(-28)+ i*sin(-28)*5.6 = ... and simplified the expression and brought back to phasor notation. I did not get the right answer...at all!

Note that my answer in the image attached was my first attempt. The way I solved above gave a different answer.

Any help as to where I am going wrong? Thanks!

#### WBahn

Joined Mar 31, 2012
26,398
What's the angle of the impedance of the inductor?

#### shteii01

Joined Feb 19, 2010
4,644
i would change all the values to the complex notation

sum all the voltages is equal to zero: (real v1+ imaginary v1) + (real v inductor+ imaginary v inductor) + (real v2+ imaginary v2)=0

solve for real v2 and imaginary v2

use trig to find magnitude and angle of v2

#### danielb33

Joined Aug 20, 2012
105
shteii01 - 100 + 0.5<90 + V2<theta = 0 --> 100 + 0.5*cos(90) + i*sin(90)0.5+V2*cos(theta)+i*sin(theta)*V2 = 0

How doe this help me?

Wbahn - impedence of inductor is 0.5<90

#### WBahn

Joined Mar 31, 2012
26,398
I first found the voltage across the inductor - I * Z = 5.6*<-28.
Wbahn - impedence of inductor is 0.5<90
So now find the voltage across the inductor a bit more carefully.

#### shteii01

Joined Feb 19, 2010
4,644
shteii01 - 100 + 0.5<90 + V2<theta = 0 --> 100 + 0.5*cos(90) + i*sin(90)0.5+V2*cos(theta)+i*sin(theta)*V2 = 0

How doe this help me?
Those are not complex numbers.

You can do your own thing... invent your own math... if you like. I aint gonna stop u.