# Steady-State AC Response of RL Circuits

Discussion in 'Homework Help' started by TheBloke, May 23, 2014.

1. ### TheBloke Thread Starter New Member

May 23, 2014
2
0
Hi,
Wondering if anyone can help me out with determining expression for the resistor voltage and inductor voltage, with an AC input 5V (RMS) ??

Any help is much appreciated

Feb 17, 2009
3,936
1,088
3. ### TheBloke Thread Starter New Member

May 23, 2014
2
0
Yes in Series
Thanks for the reply, I have used the equations you have provided, and I am trying to plot the magnitude of the voltage VR and VL against the frequency, however, even though the source is 5v (RMS) I am seeing answers as high as 50V......
Any idea?
Thanks

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,936
1,088
Or show us your work so we can check it.

5. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,294
482
If they are in series, it is just a voltage divider.

Replace inductor with its impedance, now you have a voltage divider formed by two "resistors":
Vresistor=5[R/(R+Zinductor)]
Vinductor=5[Zinductor/(R+Zinductor)]

As you can see the denominator is always larger than nominator. So you will always have a number smaller than 1 in the square brackets. Then you multiply this number by 5 and you will have result that is smaller than 5.
Example.
Let us say that you have:
R=1 kOhm
Zinductor= 100j
V= 5 V
Then:
Vresistor=5[1k/(1k+100j)]=4.95-0.495j
|Vresistor|=4.975 V
Vinductor=5[100j/(1k+100j)]=0.0495+0.495j
|Vinductor|=0.497 V

Also note that increasing the frequency will increase denominator even more. So there is no way for you to exceed 5 volts.

6. ### WBahn Moderator

Mar 31, 2012
17,446
4,699
It would help if you showed your work. Otherwise you are expecting use to be mind readers (i.e., guess), when if you simply showed your work we could determine where you are going wrong. Doesn't that sound like a better approach?