# (R+Xl)||(R+Xc) circuit

Discussion in 'Homework Help' started by Tommer Stein, Nov 27, 2015.

1. ### Tommer Stein Thread Starter New Member

Nov 27, 2015
4
0
hello
I need help with this circuit, i need to find the Q of this circuit
Q=W0/Bw=?
thanks for all who could help

by the way how Q call in english?

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2. ### Papabravo Expert

Feb 24, 2006
10,340
1,850
1. Do you think the resonant frequency, ω0, depends on L and C alone, or do the resistors have any effect?
2. Do you know how to find the bandwidth (BW)?
3. In English, Q is just Q.

Sep 25, 2015
22
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4. ### RBR1317 Active Member

Nov 13, 2010
270
54
I would think the place to start is to excite the circuit with a voltage E, and then calculate the current flowing in the circuit. I=E/Z where Z is the parallel combination of the RL & RC branches. Z=(Rl+sL)║(Rc+1/sC)

Plot the magnitude of the current vs frequency. At some point the current will be at a minimum - that point is ω0. At two nearby points the current will be 3dB above the minimum value. BW = ω2-ω1

So Q = ω0/(ω2-ω1)

5. ### Tommer Stein Thread Starter New Member

Nov 27, 2015
4
0
i already have ω0. the problem is how i found the Bandwidth, its not the case of [im=re], i think its suppost to be with energies but i have some mistakes probably because i cant found an answer of wH and wL..

6. ### Tommer Stein Thread Starter New Member

Nov 27, 2015
4
0
how i can i find the bw?

Nov 27, 2015
4
0
8. ### hsazerty2 New Member

Sep 25, 2015
22
1
Why exactly would we need E and I ?
Just plot the magnitude of Z vs frequency, and to get a plot, we need to have values for the components, no ?

@Tommer Stein: can you give us ω0 that you found ?

9. ### RBR1317 Active Member

Nov 13, 2010
270
54
Because there seemed to be a conceptual issue. The concept of finding the bandwidth of the current response seemed like it would be easier to understand than finding the bandwidth of the impedance.

10. ### e-learner New Member

Apr 25, 2015
28
0
did u try using the series-parallel equivalent circuit? I mean converting each of the series branch into a parallel ,which will give you two resistances , a capcitor and an induction all in parallel and u can easily combine two parallel resistances into single resistance. this will give a simple parallel RLC circuit.