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

#### Tommer Stein

Joined Nov 27, 2015
4
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?

#### Papabravo

Joined Feb 24, 2006
20,383
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.

#### RBR1317

Joined Nov 13, 2010
709
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)

#### Tommer Stein

Joined Nov 27, 2015
4
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.
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..

#### Tommer Stein

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

#### hsazerty2

Joined Sep 25, 2015
30
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)
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 ?

#### RBR1317

Joined Nov 13, 2010
709
Why exactly would we need E and I ?
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.

#### e-learner

Joined Apr 25, 2015
30
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.