# AC - RC/RLC Question Help

#### RysterHD

Joined Dec 15, 2015
8
Hello,

I've been doing work for an exam that's coming up and actually happen to have lost my notes 2 days before the exam, now i'm in a state of panic and know i won't do well, but i need to pass at least, so i've asked for some practice questions but i don't have much of an idea on how to do them. Any help is appreciated,

Find the resonant frequency of the graphs : http://prntscr.com/ax7dzc
Edit: On the left axis is the current, on the x axis, is the frequency

By looking at the graphs, i guess
I Supply series, 65Hz
I Supply parallel, 1Hz

Define the term resonant frequency - Resonant frequency is the point at which the impedence of the circuit is at it's lowest or highest point (Lowest point for a series circuit and highest point for a parallel circuit)

For each circuit define how the current of the circuit changes at the resonant frequency *And also the impedence* - For the series circuit, the current of the circuit peaks, and from this point onwards will begin to decrease as it has the lowest impedence at this point, which is visible by looking at the graph
For the parallel circuit, the impedence is increasing, and once it reaches this resonant frequency, the impedance is at it's highest point and thus cannot go higher, this means that after the resonant frequency point, the current will begin to decrease as the impedance will be getting lower, resulting in less current.

There are other questions which are longer and am struggling on, but i don't think there's enough time left so i'll have to fail them, please let me know if the above is correct, thanks

#### WBahn

Joined Mar 31, 2012
26,398
How do you get 65 Hz off of that graph? Remember that the x-axis is logarithmic, not linear.

For parallel resonant circuit, what is the current at resonance (compared to the current away from resonance)?

Does the claim that the current will decrease as the impedance decreases make any sense?

Are you claims even consistent? You say, for the series circuit, that the current will decrease away from resonance because the impedance is lowest at resonance. But then you say, for the parallel circuit, that the current will decrease away from resonance because the impedance is decreasing. Think about it.

#### RysterHD

Joined Dec 15, 2015
8
How do you get 65 Hz off of that graph? Remember that the x-axis is logarithmic, not linear.

For parallel resonant circuit, what is the current at resonance (compared to the current away from resonance)?

Does the claim that the current will decrease as the impedance decreases make any sense?

Are you claims even consistent? You say, for the series circuit, that the current will decrease away from resonance because the impedance is lowest at resonance. But then you say, for the parallel circuit, that the current will decrease away from resonance because the impedance is decreasing. Think about it.
Thanks for your reply, i was doing this kind of stuff all day so my brain sort of shut off, i've corrected the bottom statement about the parallel and series and the current increasing statement,

For the graphs I have looked at them again and this is what i seem to get

http://prntscr.com/axj2p2

Please aid me if i'm wrong, many thanks

#### WBahn

Joined Mar 31, 2012
26,398
The locations of the resonant frequencies look reasonable. Now what do you get numerically?

#### RysterHD

Joined Dec 15, 2015
8
The locations of the resonant frequencies look reasonable. Now what do you get numerically?
40 Hz for both

#### RysterHD

Joined Dec 15, 2015
8