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I am trying to solve a LCR related exercise, but I ran into a problem.

Here is the exercise task and information:

An alternating current circuit is composed of a 50 Ω resistor, an ideal inductor with an inductance of 10 mH and 1 μF capacitor. Find out the frequencies of the AC for which a phase difference of \$\frac{π}{4}\$ is present between the supply current and the supply voltage.

So:
\$R=50\ Ω\$
\$L=10\ mH=10^{-2}\ H\$
\$C=1\ μF=10^{-6}\ F\$

I also know know the value of the resonant frequency if it helps.

schematic

simulate this circuit – Schematic created using CircuitLab

Since it is not specified which of the current and voltage leads, I've figured that there will be only two values, one below the resonant frequency when the entire circuit acts as a capacitor and the other bigger than the resonant frequency when the circuit works as an inductor.

schematic

simulate this circuit

schematic

simulate this circuit

For the first diagram: \$tan\ φ=\frac{X_L-X_C}{R}\$
\$tan\ \frac{π}{4} =\frac{X_L-X_C}{R}\$

\$\frac{X_L-X_C}{R}=1\$
φ is the angle between the voltage phasor and the current phasor. The phasor values are all peaks.

Something similar can be written for the second phasor diagram.

The problem is that if I continue solving the above equation I will be obliged to deal with a quadratic equation which makes me think I'm overcomplicating everything, given the numbers I have to use in order to calculate \$\Delta\$ and then find \$\omega_1\$ and \$\omega_2\$. I'm thinking that there is an easier way to solve the task.

How should I go about solving this exercise?

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  • \$\begingroup\$ Solving a quadratic isn't a major issue. \$\endgroup\$ – Chu Mar 19 '17 at 9:31
  • \$\begingroup\$ It can become so when you have to work with big numbers. \$\endgroup\$ – Daniel Tork Mar 19 '17 at 9:34
  • \$\begingroup\$ Are you afraid of running out of ink? \$\endgroup\$ – Chu Mar 19 '17 at 20:06
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I see no problem solving the two quadratic formulas.

You are seeking the half-power points.

$$f_1, f_2 = f_0 \left ( \sqrt {1 + \frac {1}{4Q^2}}\ \mp\ \frac {1}{2Q}\right ) \ \ \ \ [1]$$ Q = 2 and \$f_0\$ = 1591.549Hz.

So \$f_1 = 1591.549Hz \times \left ( \sqrt {1 + \frac {1}{4 \times 2^2}}\ -\ \frac {1}{2 \times 2}\right )\$ = 1242.644Hz and \$f_2\$ = 2038.419Hz.

Or: $$f_1, f_2 = f_0 \left ( \sqrt {\left ({\frac {R}{2L}} \right )^2 + \frac {1}{LC}}\ \mp\ \frac {R}{2L}\right ) \ \ \ \ [2]$$

Resonance Chapter

The rest is not part of the answer, but lots of references do not show how the formulas are derived.

Deriving half-power points from quality factor.

Quality factor: $$Q = \frac {f_0}{BW} = \frac {f_0}{f_2\ -\ f_1} = \frac {1}{R} \sqrt {\frac {L}{C}} = \frac {\omega_0 \ L}{R}$$

Resonance: $$\omega_0 = \frac {1}{\sqrt {LC}}$$

At \$\theta = 45^{\circ}, Z = \sqrt {2} R\$.

$$\sqrt {R^2 +\left ({\omega L - \frac {1}{\omega C}} \right )^2} = \sqrt {2} R$$ $$R^2 +\left ({\omega L - \frac {1}{\omega C}} \right )^2 = 2 R^2$$ $$\left ({\omega L - \frac {1}{\omega C}}\right )^2 = R^2$$ $$\omega L - \frac {1}{\omega C} = R$$ $$\omega^2 LC - \omega RC - 1 = 0$$ $$\omega^2 - \omega \frac {R}{L} - \frac {1}{LC} = 0$$ Which gives \$[2]\$ when applied to quadratic formula. $$\omega^2 - \omega \frac {\omega_0}{Q} - \omega_0^2 = 0$$ Which gives \$[1]\$ when applied to quadratic formula.

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  • \$\begingroup\$ This is what I was expecting. Thank you, I will analyse what you posted, so be patient, it will take a small while before I may accept your answer. \$\endgroup\$ – Daniel Tork Mar 20 '17 at 17:19
  • \$\begingroup\$ Fixed intermediate math. \$\endgroup\$ – StainlessSteelRat Mar 20 '17 at 17:56
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Since the difference between the voltage and the current phase angles is pi/4, the angle of the equivalent impedance, Z, is equal to pi/4. So you must write the equation:

Z = jwL + 1/(jwC) + 50 = 50 + j (wL + 1/(wC))

Then solve the equation for the angle (tangent(pi/4) = "opp"/"adj"): enter image description here

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  • \$\begingroup\$ I still have to use the quadratic by following this method. \$\endgroup\$ – Daniel Tork Mar 19 '17 at 11:46
  • \$\begingroup\$ Using the quadratic formula is common for RLC circuit problems. I don't think you should think that you are not onto the right solution when you have to utilize it. \$\endgroup\$ – Campbell Mar 19 '17 at 19:29
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Impedance is R + jwL + 1/jwC. Phase of Z should be pi/4 (or -pi/4 the question is not clear about it).
So, phase{(jwRC - w^2LC + 1 ) / jwC} should be +- pi/4. Therfore:

arctan(wRC / (1 - w^2LC)) - pi / 2 = +- pi/4.  
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