# Voltage drops in resonant tank circuit

#### phenohm

Joined Nov 12, 2006
10
i am having trouble figuring out the voltage drops across individual components in a resonant circuit... the question is posted below. i have correctly computed the total current through the circuit, and i can correctly compute the resonant frequency.

here is what i have, i'll walk through my method for one of the AC frequencies in the table:
first i simplify based on the constants for all f -
Xl = (2 pi f L) = 2.073f
Xc = 1/(2 pi f C) = 15915/f

then work it out for the individual frequencies -
@50 Hz:
Xl = 103.65@90deg
Xc = 318.3@-90deg
Z = 214.65@-90deg. + 5@0deg. [5ohm resistor] = 214.708
I = E/Z = 250/214.708 = 1.16A

resonant freq:
f = sqrt(1/(2 pi L C)) = sqrt(7675.8) = 87.61Hz

but, i'm not sure where that puts me since according to the answers the voltages dropped do not seem to add up to 250. i guess that makes sense, since it's a resonant circuit, but given that i don't know how to calculate the drops across individual components. the answers show VERY small voltage drops across the inductor and capacitor. anybody know how to calculate this?

thanks a ton!

#### kc8ljh

Joined Nov 25, 2006
15
for starters, KVL states the voltage around a closed loop is zero. Stated another way Vsource = Vc + VL + Vr and its a series circuit so the current is the same throughout the circuit ( Ir = Ic = IL)

and VL and Vc oppose each other, so VL + (-Vc)

then add vectorally the voltage arcross the resistor

#### kc8ljh

Joined Nov 25, 2006
15
1st, find Zt, total impedance for each of the frequencies.

ill start with 50 Hz ZL = j104 Zc = -j318 Zr = 5

combining ZL and Zc = ZL + ZC = 104 + -j318 = -j214

Zt = sqrt(5sq + 214sq) = 214.05 -90deg

find I I = Vt/Zt 250mV 0deg / 214ohms -90deg = 1.168 mA +90deg

#### kc8ljh

Joined Nov 25, 2006
15
its been a while since i had to calculate that stuff, i hope i am correct, feel free to correct me if necessary

#### JoeJester

Joined Apr 26, 2005
4,259
Attached is a plot of Vx and Vr and the resultant KVL summation done in Excel.

#### Attachments

• 17 KB Views: 32

#### phenohm

Joined Nov 12, 2006
10
oh, holy crap. i just wasted your time guys, so sorry. i had just written this long post about how i totally didn't get it because it was a huge 250 volts and the total resistances would never add up since they weren't even a whole volt, but i'm a dumbass. it's 250*milli*V.

k, your work was exactly right, although i had put all the calculations for the total current in my original post. i was pretty pleased with myself that i figured out the pythagorean part too, and then my voltage drop calculations were way off because my decimal point on the total voltage was 3 places off so my totals were coming in looking way out of whack. i suck. i am still developing the "common sense" arm of my circuit analysis battle plan...

and thanks for the plot joe! that's great, what program is it?

#### JoeJester

Joined Apr 26, 2005
4,259
The plot was made with Microsoft Excel, a spreadsheet, using their graphing routines.

My simulation program is TINA. I could have used the AC transfer function in the simulation program but I wanted to do it with the spreadsheet.