What is the relationship between voltage and impedance in electrical circuits?

[nhrock3]

http://i49.tinypic.com/ejyka9.jpg

i know that there is a phase difference between the current and its voltage for each
component
but here its between the voltage of the source and the impidance of each components

how did they get the angle in each equation?

in each equation we have Vm divided by some number
and the resolt is Vm and angle

how they get the coefficient of Vm
?

 

[cepheid]

For the inductor, you know that the impedance is (jωL), and in this case, ωL = (5 ⨉ 10³ rad/s )(0.2 ⨉ 10^-3 H) = 1 Ω.

We're assuming a magnitude V_m for the (phasor) voltage, and taking its phase to be zero. Therefore, since the (phasor) current is given by the (phasor) voltage divided by the impedance:

I_L = (V_m∠0) /(jωL) = (V_m) / (j)​

Now, dividing by j is the same as multiplying by -j. Multplying by -j is the same as introducing a phase shift of -90°. To see this explicitly, you can write j in complex exponential form (if you are familiar with it):

-j = e^-j(π/2)​

The phase angles of the other currents are determined in a similar way, by noting that current = voltage / impedance, and taking careful account of the phases of these two quantities.

 

[nhrock3]

thanks :)