Circuits
posted by Heather on .
I need to calculate the voltage through the resistor and the voltage through the inductor of an RL series circuit for different frequencies. The voltage of the source is 2.5 V. The inductor is at the top and has an inductance of .200 H and the resistor is at the right with a resistance of 5.1 kilo ohms.
To calculate VR and VL I was going to use VR=Z2/(Z2+Z1) and VL=Z1/(Z2+Z1).
I read that Z for the resistor is equal to R and i don't know what Z for the inductor is. i know that the total Z is sqrt(R^2+w^2L^2) so i don't know if Z for the inductor (or Z1) will be ZtotZR.
I want to know if I'm approaching this correctly. What I did for VL when the frequency was 1000 Hz was:
Z2= ZtotZ1
Z2=3.91
VL= Z2/(Z1+Z2)=1.92 mV.
Am I doing this correctly?
How can the angles for Vr be calculated?

The current flows THROUGH the resistor.
The voltage is ACROSS the resistor.
R = 5.1k Ohms = 5100 Ohms.
L = 0.20 H.
Xl = 2pi*F*L = 6.28 * 1000 * 0.20 = 1256 Ohms = = The inductive reactance.
Z = R + jXl = sqrt(R^2+Xl^2)=The circuit impedance.
Z = sqrt((5100)^2+(1256)^2)=5,252 Ohms.
I = V/Z = 2.5 / 5252 = 4.76*10^4 Amps.
VR = I*R = 4.76*10^4 * 5100 = 2.43 Volts.
VL = I*Xl = 4.76*10^4 * 1256 = 0.598
Volts.
CHECK: E = sqrt(VR^2+VL^2).
E=sqrt((2.43)^2+(0.598)^2)=2.50 Volts.
NOTE: We calculated the vector sum of VR and VL.
Phase Shift Angle:
tanA = Xi/R = 1256 / 5100 = 0.24627.
A = 13.84 Deg.